problem solving of algebraic expression

The expand command is used mainly to rewrite polynomials with all brackets and whole number powers multiplied out and all like terms collected together. In the advanced section, you also have the option of expanding trigonometric functions, expanding modulo any integer and leaving certain parts of the expression untouched whilst expanding the rest.

Go to the Expand page

The factor command will try to rewrite an expression as a product of smaller expressions. It takes care of such things as taking out common factors, factoring by pairs, quadratic trinomials, differences of two squares, sums and differences of two cubes, and a whole lot more. The advanced section includes options for factoring trigonometric functions, factoring modulo any integer, factoring over the field of Gaussian integers (just the thing for those tricky sums of squares), and even extending the field over which factoring occurs with your own custom extensions.

Go to the Factor page

Simplifying is perhaps the most difficult of all the commands to describe. The way simplification is performed in QuickMath involves looking at many different combinations of transformations of an expression and choosing the one which has the smallest number of parts. Amongst other things, the Simplify command will take care of canceling common factors from the top and bottom of a fraction and collecting like terms. The advanced options allow you to simplify trigonometric functions or to instruct QuickMath to try harder to find a simplified expression.

Go to the Simplify page

The cancel command allows you to cancel out common factors in the denominator and numerator of any fraction appearing in an expression. This command works by canceling the greatest common divisor of the denominator and numerator.

Go to the Cancel page

Partial Fractions

The partial fractions command allows you to split a rational function into a sum or difference of fractions. A rational function is simply a quotient of two polynomials. Any rational function can be written as a sum of fractions, where the denominators of the fractions are powers of the factors of the denominator of the original expression. This command is especially useful if you need to integrate a rational function. By splitting it into partial fractions first, the integration can often be made much simpler.

Go to the Partial Fractions page

Join Fractions

The join fractions command essentially does the reverse of the partial fractions command. It will rewrite a number of fractions which are added or subtracted as a single fraction. The denominator of this single fraction will usually be the lowest common multiple of the denominators of all the fractions being added or subtracted. Any common factors in the numerator and denominator of the answer will automatically be cancelled out.

Go to the Join Fractions page

Introduction to Algebraic Functions

The notion of correspondence is encountered frequently in everyday life. For example, to each book in a library there corresponds the number of pages in the book. As another example, to each human being there corresponds a birth date. To cite a third example, if the temperature of the air is recorded throughout a day, then at each instant of time there is a corresponding temperature.

The examples of correspondences we have given involve two sets X and Y. In our first example, X denotes the set of books in a library and Y the set of positive integers. For each book x in X there corresponds a positive integer y, namely the number of pages in the book. In the second example, if we let X denote the set of all human beings and Y the set of all possible dates, then to each person x in X there corresponds a birth date y.

We sometimes represent correspondences by diagrams of the type shown in Figure 1.17, where the sets X and Y are represented by points within regions in a plane. The curved arrow indicates that the element y of Y corresponds to the element x of X. We have pictured X and Y as different sets. However, X and Y may have elements in common. As a matter of fact, we often have X = Y.

A function f from a set X to a set Y is a correspondence that assigns to each element x of X a unique element y of Y. The element y is called the image of x under f and is denoted by f(x). The set X is called the domain of the function. The range of the function consists of all images of elements of X.

Earlier, we introduced the notation f(x) for the element of Y which corresponds to x. This is usually read "f of x." We also call f(x) the value of f at x. In terms of the pictorial representation given earlier, we may now sketch a diagram as in Figure 1.18. The curved arrows indicate that the elements f(x), f(w), f(z), and f(a) of Y correspond to the elements x, y, z and a of X. Let us repeat the important fact that to each x in X there is assigned precisely one image f(x) in Y; however, different elements of X such as w and z in Figure 1.18 may have the same image in Y.

Solution As in Example 1, finding images under f is simply a matter of substituting the appropriate number for x in the expression for f(x). Thus:

Many formulas which occur in mathematics and the sciences determine functions. As an illustration, the formula A = pi*r 2 for the area A of a circle of radius r assigns to each positive real number r a unique value of A. This determines a function f, where f(r) = pi*r 2 , and we may write A= f(r). The letter r, which represents an arbitrary number from the domain off, is often called an independent variable. The letter A, which represents a number from the range off, is called a dependent variable, since its value depends on the number assigned tor. When two variables r and A are related in this manner, it is customary to use the phrase A is a function of r. To cite another example, if an automobile travels at a uniform rate of 50 miles per hour, then the distance d (miles) traveled in time t (hours) is given by d = 50t and hence the distance d is a function of time t.

We have seen that different elements in the domain of a function may have the same image. If images are always different, then, as in the next definition, the function is called one-to-one.  

Math Topics

More solvers.

• Add Fractions
• Simplify Fractions

Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

• For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
• For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

• Add or Subtract the same value from both sides
• Clear out any fractions by Multiplying every term by the bottom parts
• Divide every term by the same nonzero value
• Combine Like Terms
• Expanding (the opposite of factoring) may also help
• Recognizing a pattern, such as the difference of squares
• Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

• solve Quadratic Equations
• solve Radical Equations
• solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

• Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
• Show all the steps , so it can be checked later (by you or someone else)

Math Practice

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7.3 Simple Algebraic Equations and Word Problems

An algebraic equation is a mathematical sentence expressing equality between two algebraic expressions (or an algebraic expression and a number).

When two expressions are joined by an equal (=) sign, it indicates that the expression to the left of the equal sign is identical in value to the expression to the right of the equal sign.

For example, when two algebraic expressions, such as [latex]5x + 7[/latex] and [latex]x + 19[/latex], are equal, the two expressions are joined by an equal (=) sign and the equation is written as:

[latex]5x + 7 = x + 19[/latex]

‘Left side’ (LS) = ‘Right side’ (RS)

The solution to the equation is the value of the variable that makes the left side (LS) evaluate to the same number as the right side (RS).

Note: You need an equation to solve for an unknown variable – you cannot solve for a variable in an algebraic expression that is not part of an equation.

• If you have an expression , it needs to be simplified .
• If you have an equation , it needs to be solved .

In algebra, there are a variety of equations. In this section, we will learn one equation category, linear equations with one variable .

Examples of linear equations with one variable are:

[latex]2x = 8[/latex],   [latex]3x + 5 = 14[/latex],   [latex]5x + 7 = x + 19[/latex]

An equation is either true or false depending on the value of the variable.

For example, consider the equation [latex]2x = 8[/latex]:

• If [latex]x = 4[/latex],   LS = 2(4) = 8, RS = 8; therefore, the equation is true.
• If [latex]x = 3[/latex],   LS = 2(3) = 6, RS = 8; therefore, the equation is false.

Equations may be classified into the following three types:

• Conditional equation: these equations are only true when the variable has a specific value. For example, [latex]2x = 8[/latex] is a conditional equation, true if and only if [latex]x = 4[/latex].
• Identity: these equations are true for any value for the variable. For example, [latex]2x + 10 = 2(x + 5)[/latex] is an identity, true for any value of [latex]x[/latex].
• Contradiction: these equations are not true for any value of the variable. For example, [latex]x + 5 = x + 4[/latex] is a contradiction, not true for any value of [latex]x[/latex].

Equivalent Equations

Equations with the same solutions are called equivalent equations .

For example, [latex]2x + 5 = 9[/latex] and [latex]2x = 4[/latex] are equivalent equations because the solution [latex]x = 2[/latex] satisfies each equation.

Similarly, [latex]3x - 4 = 5[/latex], [latex]2x = x + 3[/latex], and [latex]x + 1 = 4[/latex] are equivalent equations because the solution [latex]x = 3[/latex] satisfies each equation.

Properties of Equality

If [latex]a = b[/latex], then,

These properties are used to solve equations.

Equations with Fractional Coefficients

If an equation contains fractional coefficients, then the fractional coefficients can be changed to whole numbers by multiplying both sides of the equation by the least common denominator (LCD) of all the fractions, using the Multiplication Property.

For example,

[latex]\displaystyle{\frac{2}{3}x = \frac{5}{2} + 4}[/latex] Since the LCD of the denominators 3 and 2 is 6, multiply both sides of the equation by 6.

[latex]\displaystyle{6\left(\frac{2}{3}x\right) = 6\left(\frac{5}{2} + 4\right)}[/latex] This is the same as multiplying each term by the LCD of 6.

[latex]\displaystyle{6\left(\frac{2}{3}x\right) = 6\left(\frac{5}{2}\right) + 6(4)}[/latex] Simplifying, [latex]4x = 15 + 24[/latex] Now, the equation has only whole number coefficients [latex]4x = 39[/latex].

Equations with Decimal Coefficients

If an equation contains decimal coefficients, then the decimal coefficients can be changed to whole numbers by multiplying both sides of the equation by an appropriate power of 10, using the Multiplication Property.

[latex]\underline{1.25}x = \underline{0.2} + 4[/latex] Since there is at most 2 decimal places in any of the coefficients or constants, multiply both sides of the equation by [latex]10^2 = 100[/latex].

[latex]100(1.25x) = 100(0.2 + 4)[/latex] This is the same as multiplying each term by 100. [latex]100(1.25x) = 100(0.2) + 100(4)[/latex] Simplifying, [latex]125x = 20 + 400[/latex] Now, the equation has only whole number coefficients [latex]125x = 420[/latex].

Steps to Solve Algebraic Equations with One Variable

If the equation contains fraction and/or decimal coefficients, it is possible to work with them as they are – in that case, proceed to Step 2. Alternatively, as explained earlier, the equation may be rewritten in whole numbers to make calculations and rearrangements easier.

If present, expand and clear brackets in the equation by following the order of arithmetic operations (BEDMAS).

Use the addition and subtraction properties to collect and group all variable terms on the left side of the equation and all constants on the right side . Then, simplify both sides.

Note: If it is more convenient to gather all the variable terms on the right side and the constants on the left side, you may do so, and then use the symmetric property and switch the sides of the equation to bring the variables over to the left side and the constants to the right side.

Use the division and multiplication properties to ensure that the coefficient of the variable is +1.

After completing Step 4, there should be a single variable with a coefficient of +1 on the left side and a single constant term on the right side – that constant term is the solution to the equation.

Verify the answer by substituting the solution from Step 5 back into the original problem.

State the answer.

Example 7.3-a: Solving Equations Using the Addition and Subtraction Properties

Solve the following equations and verify the solutions:

• [latex]x - 11 = 4[/latex]
• [latex]8 + x = 20[/latex]
• [latex]x - 11 = 4[/latex] Adding 11 to both sides, [latex]x - 11 + 11 = 4 + 11[/latex] [latex]x = 15[/latex] Verify by substituting [latex]x = 15[/latex]: LS [latex]= x - 11 = 15 - 11 = 4[/latex]     RS [latex]= 4[/latex] LS = RS Therefore, the solution is [latex]x = 15[/latex].
• [latex]8 + x = 20[/latex] Subtracting 8 from both sides, [latex]8 - 8 + x = 20 - 8[/latex] [latex]x = 12[/latex] Verify by substituting [latex]x = 12[/latex]: LS [latex]= 8 + x = 8 + 12 = 20[/latex]   RS [latex]= 20[/latex] LS = RS Therefore, the solution is [latex]x = 12[/latex].

Example 7.3-b: Solving Equations Using the Multiplication and Division Properties

• [latex]5x = 20[/latex]
• [latex]\displaystyle{\frac{3}{8}x = 12}[/latex]
• [latex]5x = 20[/latex] Dividing both sides by 5 , [latex]\displaystyle{\frac{5x}{5} = \frac{20}{5}}[/latex] [latex]x = 4[/latex] Verify by substituting [latex]x = 4[/latex]: LS [latex]= 5x = 5(4) = 20[/latex]   RS [latex]= 20[/latex] LS = RS Therefore, the solution is [latex]x = 4[/latex].
• [latex]\displaystyle{\frac{3}{8}x = 12}[/latex] Multiplying both sides by [latex]\displaystyle{\frac{8}{3}}[/latex] (the reciprocal of [latex]\displaystyle{\frac{3}{8}}[/latex]), [latex]\displaystyle{\left(\frac{8}{3}\right) \cdot \frac{3}{8}x = \left(\frac{8}{3}\right) \cdot 12}[/latex] [latex]x = 8 \times 4[/latex] [latex]x = 32[/latex]   or  [latex]\displaystyle{\frac{3}{8}x = 12}[/latex] Multiplying both sides by 5 , [latex]\displaystyle{(8) \cdot \frac{3}{8}x = (8) \cdot 12}[/latex] [latex]3x = 96[/latex]Dividing both sides by 3 , [latex]\displaystyle{\frac{3x}{3} = \cdot \frac{96}{3}}[/latex][latex]x = 32[/latex]Verify by substituting [latex]x = 32[/latex]:LS [latex]\displaystyle{= \frac{3}{8}x = \frac{3}{8} \times 32 = 12}[/latex]RS [latex]= 12[/latex]LS = RSTherefore, the solution is [latex]x = 32[/latex].

Example 7.3-c: Solving Equations with Variables on Both Sides

• [latex]3x - 8 = 12 - 2x[/latex]
• [latex]15 + 6x - 4 = 3x + 31 - x[/latex]
• [latex]3x - 8 = 12 - 2x[/latex] Adding 2x to both sides,[latex]3x + 2x - 8 = 12 - 2x + 2x[/latex] [latex]5x - 8 = 12[/latex] Adding 8 to both sides,[latex]5x - 8 + 8 = 12 + 8[/latex] [latex]5x = 20[/latex] Dividing both sides by 5 , [latex]\displaystyle{\frac{5x}{5} = \frac{20}{5}}[/latex] [latex]x = 4[/latex] Verify by substituting [latex]x = 4[/latex]: LS [latex]= 3x - 8 = 3(4) - 8 = 12 - 8 = 4[/latex]RS [latex]= 12 - 2x = 12 - 2(4) = 12 - 8 = 4[/latex]LS = RSTherefore, the solution is [latex]x = 4[/latex].
• [latex]15 + 6x - 4 = 3x + 31 - x[/latex] Combining like terms (LS: [latex]15 - 4 = 11[/latex], and RS: [latex]3x - x = 2x[/latex]), [latex]11 + 6x = 2x + 31[/latex]   Subtracting [latex]2x[/latex] from both sides, [latex]11 + 6x - 2x = 2x - 2x + 31[/latex] [latex]11 + 4x = 31[/latex] Subtracting 11 from both sides, [latex]11 - 11 + 4x = 31 - 11[/latex] [latex]4x = 20[/latex] Dividing both sides by 4 , [latex]\displaystyle{\frac{4x}{4} = \frac{20}{4}}[/latex] [latex]x = 5[/latex] Verify by substituting [latex]x = 5[/latex] back into the original equation:LS [latex]= 15 + 6x - 4 = 15 + 6(5) - 4 = 15 + 30 - 4 = 41[/latex]RS [latex]= 3x + 31 - x = 3(5) + 31 - 5 = 15 + 31 - 5 = 41[/latex]LS = RSTherefore, the solution is [latex]x = 5[/latex].

Example 7.3-d: Solving Equations with Fractions

Solve the following equation and verify the solution:

[latex]\displaystyle{\frac{x}{3} - \frac{1}{12} = \frac{1}{6} + \frac{x}{4}}[/latex]

[latex]\displaystyle{\frac{x}{3} - \frac{1}{12} = \frac{1}{6} + \frac{x}{4}}[/latex] LCD of 3, 4, 6, and 12 is 12.

Multiplying each term by 12 ,

[latex]\displaystyle{12\left(\frac{x}{3}\right) - 12\left(\frac{1}{12}\right) = 12\left(\frac{1}{6}\right) + 12\left(\frac{x}{4}\right)}[/latex]

[latex]4x - 1 = 2 + 3x[/latex]

Subtracting [latex]3x[/latex] from both sides,

[latex]4x - 3x - 1 = 2 + 3x - 3x[/latex]

[latex]x - 1 = 2[/latex]

Adding 1 to both sides,

[latex]x - 1 + 1 = 2 + 1[/latex]

[latex]x = 3[/latex]

Verify by substituting [latex]x = 3[/latex] back into the original equation:

LS [latex]\displaystyle{= \frac{x}{3} - \frac{1}{12} = \frac{3}{3} - \frac{1}{12} = \frac{12}{12} - \frac{1}{12} = \frac{11}{12}}[/latex]

RS [latex]\displaystyle{= \frac{1}{6} + \frac{x}{4} = \frac{1}{6} + \frac{3}{4} = \frac{2}{12} + \frac{9}{12} = \frac{11}{12}}[/latex]

Therefore, the solution is [latex]x = 3[/latex].

Example 7.3-e: Solving Equations with Decimals

[latex]0.15x + 1.2 = 0.4x - 0.8[/latex]

[latex]0.15x + 1.2 = 0.4x - 0.8[/latex] Greatest number of decimal places is 2 (i.e., hundredths).

Multiplying all the terms by [latex]10^2 = 100[/latex],

[latex]100(0.15x) + 100(1.2) = 100(0.4x) - 100(0.8)[/latex]

[latex]15x + 120 = 40x - 80[/latex]

Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS,

[latex]40x - 80 = 15x + 120[/latex]

Subtracting [latex]15x[/latex] from both sides,

[latex]40x - 15x - 80 = 15x - 15x + 120[/latex]

[latex]25x - 80 = 120[/latex]

Adding 80 to both sides,

[latex]25x - 80 + 80 = 120 + 80[/latex]

[latex]25x = 200[/latex]

Dividing both sides by 25 ,

[latex]\displaystyle{\frac{25x}{25} = \frac{200}{25}}[/latex]

[latex]x = 8[/latex]

Verify by substituting [latex]x = 8[/latex] back into the original equation:

LS [latex]= 0.15x + 1.2 = 0.15(8) + 1.2 = 1.2 + 1.2 = 2.4[/latex]

RS [latex]= 0.4x - 0.8 = 0.4(8) - 0.8 = 3.2 - 0.8 = 2.4[/latex]

Therefore, the solution is [latex]x = 8[/latex].

Note: For the rest of the examples in this section, we will not show the verification by substitution step.

Example 7.3-f: Solving Equations Using All the Properties

Solve the following equations by using the properties of equality, and express the answer as a fraction in its lowest terms, or as a mixed number, wherever applicable:

• [latex]8x + 7 - 3x = -6x - 15 + x[/latex]
• [latex]2(3x - 7) = 28 - 3(x + 1)[/latex]
• [latex]\displaystyle{\frac{1}{4}(x + \frac{2}{3}) = \frac{1}{2}(x - 3) + x}[/latex]
• [latex]0.45(2x + 3) - 2.55 = 0.6(3x - 5)[/latex]
• [latex]\displaystyle{\frac{x + 2}{3} = \frac{5 - 2x}{7}}[/latex]
• [latex]8x + 7 - 3x = -6x - 15 + x[/latex] Grouping like terms on both sides, [latex]8x - 3x + 7 = -6x + x - 15[/latex] [latex]5x + 7 = -5x - 15[/latex] Adding 5 to both sides, [latex]5x + 5x + 7 = -5x + 5x - 15[/latex] [latex]10x + 7 = -15[/latex] Subtracting 7 from both sides, [latex]10x + 7 - 7 = -15 - 7[/latex] [latex]10x = -22[/latex] Dividing both sides by 10 , [latex]\displaystyle{\frac{10x}{10} = -\frac{22}{10}}[/latex] [latex]\displaystyle{x = -\frac{11}{5} = -2\frac{1}{5}}[/latex]
• [latex]2(3x - 7) = 28 - 3(x + 1)[/latex] Expanding both sides,[latex]6x - 14 = 28 - 3x - 3[/latex]   Grouping like terms,[latex]6x - 14 = 28 - 3 - 3x[/latex] [latex]6x - 14 = 25 - 3x[/latex] Adding [latex]3x[/latex] to both sides,[latex]6x + 3x - 14 = 25 - 3x + 3x[/latex] [latex]9x - 14 = 25[/latex] Adding 14 to both sides, [latex]9x - 14 + 14 = 25 + 14[/latex] [latex]9x = 39[/latex] Dividing both sides by 9 ,[latex]\displaystyle{\frac{9x}{9} = \frac{39}{9}}[/latex] [latex]\displaystyle{x = \frac{13}{3} = 4\frac{1}{3}}[/latex]
• [latex]\displaystyle{\frac{1}{4}(x + \frac{2}{3}) = \frac{1}{2}(x - 3) + x}[/latex] Expanding both sides, [latex]\displaystyle{\frac{1}{4}x + \frac{1}{6} = \frac{1}{2}x - \frac{3}{2} + x}[/latex] Multiplying each term by the LCD 12 , [latex]\displaystyle{12\left(\frac{1}{4}x\right) + 12\left(\frac{1}{6}\right) = 12\left(\frac{1}{2}x\right) - 12\left(\frac{3}{2}\right) + 12(x)}[/latex] [latex]3x + 2 = 6x - 18 + 12x[/latex] Grouping like terms,[latex]3x + 2 = 6x + 12x - 18[/latex] [latex]3x + 2 = 18x - 18[/latex] Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS, [latex]18x 18 = 3x + 2[/latex]   Subtracting [latex]3x[/latex] from both sides, [latex]18x - 3x - 18 = 3x - 3x + 2[/latex] [latex]15x - 18 = 2[/latex] Adding 18 to both sides, [latex]15x - 18 + 18 = 2 + 18[/latex] [latex]15x = 20[/latex] Dividing both sides by 15 , [latex]\displaystyle{\frac{15x}{15} = \frac{20}{15}}[/latex][latex]\displaystyle{x = \frac{4}{3} = 1\frac{1}{3}}[/latex]
• [latex]0.45(2x + 3) - 2.55 = 0.6(3x - 5)[/latex] Expanding both sides, [latex]0.90x + 1.35 - 2.55 = 1.8x - 3.0[/latex] Greatest number of decimal places is 2 (i.e., hundredths). Multiplying all the terms by [latex]10^2 = 100[/latex],   [latex]100(0.90x) + 100(1.35) - 100(2.55) = 100(1.8x) - 100(3.0)[/latex] [latex]90x + 135 - 255 = 180x - 300[/latex] Grouping like terms,[latex]90x 120 = 180x 300[/latex] Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS,   [latex]180x - 300 = 90x - 120[/latex] Subtracting [latex]90x[/latex] from both sides,  [latex]180x - 90x - 300 = 90x - 90x - 120[/latex] [latex]90x - 300 = -120[/latex] Adding 300 to both sides,  [latex]90x - 300 + 300 = -120 + 300[/latex] [latex]90x = 180[/latex] Dividing both sides by 90 ,  [latex]\displaystyle{\frac{90x}{90} = \frac{180}{90}}[/latex] [latex]x = 2[/latex]
• [latex]\displaystyle{\frac{x + 2}{3} = \frac{5 - 2x}{7}}[/latex] Cross-multiplying,  [latex]7(x + 2) = 3(5 - 2x)[/latex] Expanding both sides,[latex]7x + 14 = 15 - 6x[/latex] Adding [latex]6x[/latex] to both sides,  [latex]7x + 6x + 14 = 15 - 6x + 6x[/latex] [latex]13x + 14 = 15[/latex] Subtracting 14 from both sides, [latex]13x + 14 - 14 = 15 - 14[/latex] [latex]13x = 1[/latex] Dividing both sides by 13 ,  [latex]\displaystyle{\frac{13x}{13} = \frac{1}{13}}[/latex] [latex]\displaystyle{x = \frac{1}{13}}[/latex]

Steps to Solve Word Problems

Read the entire problem and ensure you understand the situation.

Identify the given information and the question to be answered.

Look for keywords. Some words indicate certain mathematical operations (see Table 7.1).

Choose a variable to represent the unknown(s) and state what that variable represents, including the unit of measure.

Note: For now, if there is more than one unknown, try to identify all the unknowns in terms of one variable, as all the questions in this chapter can be solved with only one variable.

Where necessary, draw a simple sketch to identify the information. This helps with envisioning the question more clearly.

Create an equation (or set of equations) to describe the relationship between the variables and the constants in the question.

Group like terms, isolate the variable and solve for the unknown(s).

State the solution to the given problem.

Example 7.3-g: Solving a Word Problem Using Algebraic Equations

If Harry will be 65 years old in 5 years, how old is he today?

Let Harry’s age today be x years.

Therefore, in 5 years, Harry’s age will be:

[latex]x + 5 = 65[/latex]   Solving for [latex]x[/latex],

[latex]x = 65 - 5 = 60[/latex]

Therefore, Harry is 60 years old today.

Example 7.3-h: Solving a Geometry Problem Using Algebraic Equations

The perimeter of a rectangular garden is 50 metres. The length is 5 metres more than the width. Find the dimensions of the garden.

Hint: Perimeter = 2(length) + 2(width)

Let the width be [latex]w[/latex] metres.

Therefore, the length is (w + 5) metres.

Perimeter = 2(length) + 2(width)

[latex]50 = 2(w + 5) + 2w[/latex]

[latex]50 = 2w + 10 + 2w[/latex]

[latex]2w + 10 + 2w = 50[/latex]

[latex]4w + 10 = 50[/latex]

[latex]4w = 50 - 10[/latex]

[latex]4w = 40[/latex]

[latex]\displaystyle{w = \frac{40}{4}}[/latex]

[latex]w = 10[/latex]

Therefore, the width of the garden is 10 metres and the length is (10 + 5) = 15 metres.

Example 7.3-i: Solving a Finance Problem Using Algebraic Equations

A TV costs $190 more than a Blu-ray player. The total cost of the TV and the Blu-ray player is $688. Calculate the cost of the TV and the cost of the Blu-ray player.

Let the cost of the Blu-ray player be [latex]\$x[/latex].

Therefore, the cost of the TV is [latex]\$(x + 190.00)[/latex].

The total cost is $688.00.

[latex]x + (x + 190.00) = 688.00[/latex]

[latex]x + x + 190.00 = 688.00[/latex]

[latex]2x + 190.00 = 688.00[/latex]

[latex]2x = 688.00 - 190.00[/latex]

[latex]2x = 498.00[/latex]

[latex]\displaystyle{x = \frac{498.00}{2}}[/latex]

[latex]x = \$249.00[/latex]

Therefore, the cost of the Blu-ray player is $249.00 and the cost of the TV is (249.00 + 190.00) = $439.00.

Example 7.3-j: Solving a Mixture Problem Using Algebraic Equations

How many litres of water need to be added to 30 litres of a 15% saline solution to make a saline solution that is 10% saline?

From the last column, we get the equation for the saline mix. The number of litres of saline in the 15% solution must be the same as the number of litres in the final 10% solution, as only water is being added, which does not contribute any additional saline to the solution. Therefore,

[latex]4.5 = 0.10 \times (30 + x)[/latex]

[latex]4.5 = 3 + 0.10x[/latex]

[latex]1.5 = 0.10x[/latex]

[latex]x = 15[/latex]

Therefore, 15 litres of water need to be added to the 15% saline solution to make the solution 10% saline.

7.3 Exercises

Answers to the odd-numbered problems are available at the end of the textbook .

For problems 1 to 8, simplify and evaluate the expressions.

• The sum of a number and six is ten.
• A number decreased by fifteen is five.
• Six times a number is seventy-two.
• The product of a number and four is twenty-eight.
• A number divided by five is four.
• A number divided by three is three.
• Two-thirds of a number is twelve.
• Two-fifths of a number is six.

For problems 9 to 30, solve the algebraic equations using the properties of equality, and express the answer as a fraction in its lowest terms or as a mixed number, wherever applicable.

• [latex]x - 20 = 10[/latex]
• [latex]x - 25 = 17[/latex]
• [latex]22 = 40 - x[/latex]
• [latex]54 = 23 - x[/latex]
• [latex]21 + x = 4[/latex]
• [latex]50 + x = 45[/latex]
• [latex]16 + x = 22[/latex]
• [latex]12 + x = 38[/latex]
• [latex]11x + 4 = 17[/latex]
• [latex]7x - 16 = 22[/latex]
• [latex]\displaystyle{x - \frac{4}{5} = \frac{3}{5}}[/latex]
• [latex]\displaystyle{x - \frac{1}{6} = 1}[/latex]
• [latex]\displaystyle{\frac{10}{15} = x - \frac{4}{3}}[/latex]
• [latex]\displaystyle{\frac{x}{7} + 15 = 24}[/latex]
• [latex]\displaystyle{x + \frac{2}{5} = \frac{1}{4}}[/latex]
• [latex]\displaystyle{2x - \frac{2}{3} = \frac{5}{6}}[/latex]
• [latex]4x = 24[/latex]
• [latex]\displaystyle{\frac{2x}{3} + 1 = \frac{5x}{8} + 2}[/latex]
• [latex]\displaystyle{\frac{x}{2} - \frac{1}{6} = \frac{1}{3} + \frac{3x}{5}}[/latex]
• [latex]\displaystyle{\frac{7x}{8} - 4 = \frac{x}{4} + 6}[/latex]
• [latex]\displaystyle{\frac{8x}{3} - 5 = \frac{x}{3} + 2}[/latex]

For problems 31 to 54, solve the algebraic equations using the properties of equality, and round the answer to 2 decimal places, wherever applicable.

• [latex]10y - 0.09y = 17[/latex]
• [latex]x + 0.13x = 70[/latex]
• [latex]0.3x - 3.2 = 0.4 - 0.6x[/latex]
• [latex]4 + 0.2x = 0.7x - 0.5[/latex]
• [latex]0.4x - 1.38 = 0.3x - 1.2[/latex]
• [latex]1.2 - 0.7x = 2.7 - 0.5x[/latex]
• [latex]0.43x + 0.25 = 0.29x - 0.03[/latex]
• [latex]0.6x - 1.2 = 0.9 - 1.5x[/latex]
• [latex]x - 2 - 4x = -3x - 8 + 5x[/latex]
• [latex]4(2x - 5) = 32 - 4(x - 2)[/latex]
• [latex](4 + 6)(2 + 4x) = 45 - 2.5(x + 3)[/latex]
• [latex](5 + 0.5x)(1 + 3) = -1.2(2x + 4) + 25[/latex]
• [latex]15 + 5(x - 10) = 3(x - 1)[/latex]
• [latex]2(x - 3) + 3(x - 5) = 4[/latex]
• [latex]4(y + 7) - 2(y - 4) = 3(y - 2)[/latex]
• [latex]8(2y + 4) - 6(3y + 7) = 3y[/latex]
• [latex]\displaystyle{\frac{x - 7}{2} + \frac{x + 2}{3} = 41}[/latex]
• [latex]\displaystyle{\frac{7}{12}(2x + 1) + \frac{3}{4}(x + 1) = 3}[/latex]
• [latex]\displaystyle{\frac{5}{y + 4} = \frac{3}{y - 2}}[/latex]
• [latex]\displaystyle{\frac{3}{x + 1} = \frac{2}{x - 3}}[/latex]
• [latex]\displaystyle{\frac{7}{5x - 3} = \frac{5}{4x}}[/latex]
• [latex]\displaystyle{\frac{5}{y + 2} = \frac{3}{y}}[/latex]

For problems 55 to 76, solve the word problems using algebraic equations.

• If three times a number plus twenty is seven times that number, what is the number?
• Fifteen less than three times a number is twice that number. What is the number?
• A 25-metre-long wire is cut into two pieces. One piece is 7 metres longer than the other. Find the length of each piece.
• A 9-metre-long pipe is cut into two pieces. One piece is twice the length of the other piece. Find the length of each piece.
• $500 is shared between Andy and Becky. Andy’s share is $150 less than Becky’s share. Calculate the amount of each of their shares.
• $200 is shared between Bill and Ann. Ann’s share is $50 more than Bill’s share. Calculate the size of each of their shares.
• Movie tickets that were sold to each child were $3 cheaper than those sold to each adult. If a family of two adults and two children paid $34 to watch a movie at the cinema, what was the price of each adult ticket and each child ticket?
• Giri had twice the number of quarters (25 cents) in his bag than dimes (10 cents). If he had a total of 54 coins, how many of them were quarters? What was the total dollar value of these coins?
• A square garden, with sides of length x metres, is widened by 4 metres and lengthened by 3 metres. Write the equation for the area (A) of the expanded garden. If each side was originally 10 metres in length, find the new area. (Hint: Area of a Rectangle = Length × Width)
• A square garden, with sides of length x metres, has had its width reduced by 4 metres and its length reduced by 2 metres. Write the equation for the Area (A) of the smaller garden. If each side was originally 20 metres in length, find the new area.
• Aran bought a shirt and a pair of pants for $34.75. The pair of pants cost $9.75 more than the shirt. Calculate the cost of the shirt.
• Mythili bought a schoolbag and a toy for $30.45. The school bag cost $5.45 more than the toy. Calculate the cost of the school bag.
• Sam is paid $720 a week. He worked 9 hours of overtime last week and he received $954. Calculate his overtime rate per hour.
• Lisa is paid $840 a week. Her overtime rate is $28 per hour. Last week she received $1,036. How many hours of overtime did she work last week?
• The sum of the three angles of any triangle is 180°. If [latex]3x[/latex], [latex]7x[/latex], and [latex]8x[/latex] are the measures of the three angles of a triangle, calculate the measure of each angle of the triangle.
• The sum of the three angles of any triangle is 180°. If [latex]3x[/latex], [latex]4x[/latex], and [latex]5x[/latex] are the measures of the three angles of a triangle, calculate the measure of each angle of the triangle.
• The perimeter of a triangle is the sum of the lengths of the three sides of the triangle. The perimeter of a triangle with sides [latex]x[/latex] cm, [latex](x + 10)[/latex] cm, and [latex]2x[/latex] cm is 70 cm. Calculate the length of each side of the triangle.
• The perimeter of a triangle is the sum of the lengths of the three sides of the triangle. The perimeter of a triangle with sides [latex]x + 10[/latex], [latex]2x + 10[/latex], and [latex]3x[/latex] is 110 cm. Calculate the length of each side of the triangle.
• After completing a weight-loss program, a patient weighs 160 lb. His dietician observes that the patient has lost 15% of his original weight. What was the patient’s starting weight?
• A beaker in a chemistry lab contains 3 litres of water. While conducting an experiment, the chemistry professor removes three-fifths of the water from the beaker. He then adds three-fifths of the remaining volume to the beaker. How much water is left in the beaker at the end of the experiment?
• A researcher wants to make 4 L of a 7% acid solution. She has a beaker of 15% acid solution in stock. How much of the 15% solution does she need to use and how much water must she add in order to prepare her desired solution?
• A chemist wants to make a 10% acid solution. She has 5 L of 25% acid solution. How many litres of water should she add to the 25% solution in order to prepare her desired solution?

Unless otherwise indicated, this chapter is an adaptation of the eTextbook  Foundations of Mathe matics  (3 rd  ed.) by Thambyrajah Kugathasan, published by  Vretta-Lyryx Inc ., with permission. Adaptations include supplementing existing material and reordering chapters.

Fundamentals of Business Math Copyright © 2023 by Lisa Koster and Tracey Chase is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Algebraic Equations

Algebraic equations are two algebraic expressions that are joined together using an equal to ( = ) sign. An algebraic equation is also known as a polynomial equation because both sides of the equal sign contain polynomials. An algebraic equation is built up of variables, coefficients, constants as well as algebraic operations such as addition, subtraction, multiplication, division, exponentiation, etc.

If there is a number or a set of numbers that satisfy the algebraic equation then they are known as the roots or the solutions of that equation. In this article, we will learn more about algebraic equations, their types, examples, and how to solve algebraic equations.

What is Algebraic Equations?

An algebraic equation is a mathematical statement that contains two equated algebraic expressions. The general form of an algebraic equation is P = 0 or P = Q, where P and Q are polynomials . Algebraic equations that contain only one variable are known as univariate equations and those which contain more than one variable are known as multivariate equations. An algebraic equation will always be balanced. This means that the right-hand side of the equation will be equal to the left-hand side.

Algebraic Expressions

A polynomial expression that contains variables, coefficients, and constants joined together using operations such as addition , subtraction, multiplication, division, and non-negative exponentiation is known as an algebraic expression . An algebraic expression should not be confused with an algebraic equation. When two algebraic expressions are merged together using an "equal to" sign then they form an algebraic equation. Thus, 5x + 1 is an expression while 5x + 1 = 0 will be an equation.

Algebraic Equations Examples

x 2 - 5x = 3 is a univariate algebraic equation while y 2 x - 5z = 3x is an example of a multivariate algebraic equation.

Types of Algebraic Equations

Algebraic equations can be classified into different types based on the degree of the equation. The degree can be defined as the highest exponent of a variable in an algebraic equation. Suppose there is an equation given by x 4 + y 3 = 3 5 then the degree will be 4. In determining the degree, the exponent of the constant or coefficient is not considered. The number of roots of an algebraic equation depends on its degree. An algebraic equation where the degree equals 5 will have a maximum of 5 roots. The various types of algebraic equations are as follows:

Linear Algebraic Equations

A linear algebraic equation is one in which the degree of the polynomial is 1. The general form of a linear equation is given as a 1 x 1 +a 2 x 2 +...+a n x n = 0 where at least one coefficient is a non-zero number. These linear equations are used to represent and solve linear programming problems.

Example: 3x + 5 = 5 is a linear equation in one variable . y = 2x - 6 is a linear equation in two variables .

Quadratic Algebraic Equations

An equation where the degree of the polynomial is 2 is known as a quadratic algebraic equation . The general form of such an equation is ax 2 + bx + c = 0, where a is not equal to 0.

Example: 3x 2 + 2x - 6 = 0 is a quadratic algebraic equation. This type of equation will have a maximum of two solutions.

Cubic Algebraic Equations

An algebraic equation where the degree equals 3 will be classified as a cubic algebraic equation . ax 3 + bx 2 + cx + d = 0 is the general form of a cubic algebraic equation (a ≠ 0).

Example: x 3 + x 2 - x - 1 = 0. A cubic algebraic equation will have a maximum of three roots as the degree is 3.

Higher-Order Polynomial Algebraic Equations

Algebraic equations that have a degree greater than 3 are known as higher-order polynomial algebraic equations. Quartic (degree = 4), quintic (5), sextic (6), septic (7) equations all fall under the category of higher algebraic equations. Such equations might not be solvable using a finite number of operations.

Algebraic Equations Formulas

Algebraic equations can be simplified using several formulas and identities. These help to expedite the process of solving a given equation. Given below are some important algebraic formulas :

• (a + b) 2 = a 2 + 2ab + b 2
• (a - b) 2 = a 2 - 2ab + b 2
• (a + b)(a - b) = a 2 - b 2
• (x + a)(x + b) = x 2 + x(a + b) + ab
• (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
• (a - b) 3 = a 3 - 3a 2 b + 3ab 2 - b 3
• a 3 + b 3 = (a + b)(a 2 - ab + b 2 )
• a 3 - b 3 = (a - b)(a 2  + ab + b 2 )
• (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca
• Quadratic Formula : [-b ± √(b² - 4ac)]/2a
• Discriminant : b 2 - 4ac

How to Solve Algebraic Equations

There are many different methods that are available for solving algebraic equations depending upon the degree. If an algebraic equation has two variables then two equations will be required to find the solution. Thus, it can be said that the number of equations required to solve an algebraic equation will be equal to the number of variables present in the equation. Given below are the ways to solve algebraic equations.

A linear algebraic equation in one variable can be solved by simply applying basic arithmetic operations  on both sides of the equation.

E.g: 4x + 1 = 5.

4x = 5 - 1 (Subtracted 1 from both sides).

4x = 4 (Solve the R.H.S using algebraic operations)

x = 1 (Divided both sides by 4)

Linear algebraic equations in more than one variable will be solved using the concept of simultaneous equations .

A quadratic algebraic equation can be solved by using identities , factorizing , long division, splitting the middle term, completing the square , applying the quadratic formula, and using graphs . A quadratic equation will always have a maximum of two roots.

E.g: x 2 + 2x + 1 = 0

Using the identity (a + b) 2 = a 2 + 2ab + b 2 , we get

a = x and b = 1

(x + 1) 2 = 0

(x + 1)(x + 1) = 0

x = -1, -1.

The most effective way of solving higher-order algebraic polynomials in one variable is by using the long division method. This decomposes the higher-order polynomial into polynomials of a lower degree thus, making it easier to find the solutions.

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Important Notes on Algebraic Equations:

• An algebraic equation is an equation where two algebraic expressions are joined together using an equal sign .
• Polynomial equations are algebra equations.
• Algebraic equations can be one-step, two-step , or multi-step equations .
• Algebra equations are classified as linear, quadratic, cubic, and higher-order equations based on the degree.
• Example 1: Solve the algebraic equation x + 3 = 2x Solution: Taking the variable terms on one side of the equation and keeping the constant terms on the other side we get, 3 = 2x - x 3 = x Answer: x = 3
• Example 2: A total of 15 items can fit in a box. If the box contains 2 scales, 7 pencils, and 1 eraser then how many pens can fit in the box? Solution: Converting this problem statement in the form of an algebraic equation we get, 2 scales + 7 pencils + 1 eraser + x pens = 15 2 + 7 + 1 + x = 15 Solving the L.H.S 10 + x = 15 x = 15 - 10 x = 5 Answer: 5 pens can fit in the box
• Example 3: Find the roots of the quadratic equation x 2 + x - 6 = 0 Solution: Using the quadratic formula x = [-b ± √(b² - 4ac)]/2a. a = 1, b = 1, c = - 6 x = [-1 ± √(1² - 4 · 1 · -6)] / (2 · 1) x = [-1 ± √(25)] / 2 x = [-1 + 5] / 2,  [-1 - 5] / 2 x = 2, -3 Answer: The roots of the given algebraic equation are 2 and -3.

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Practice Questions on Algebraic Equations

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FAQs on Algebraic Equations

What are algebraic equations.

Algebraic equations are polynomial equations where two algebraic expressions are equated. Both sides of the equation must be balanced. The general form of an algebraic equation is P = 0.

What is an Example of Algebraic Equation?

An algebraic equation can be linear, quadratic, etc. Hence, an example of an algebraic equation can be 3x 2  - 6 = 0.

How Do You Solve Algebraic Equations?

There are many methods available to solve algebraic equations depending on the degree. Some techniques include applying simple algebraic operations , solving simultaneous equations , splitting the middle term, quadratic formula, long division, and so on.

What are Algebraic Expressions and Algebraic Equations?

Mathematical statements that consist of variables , coefficients , constants , and algebraic operations are known as algebraic expressions. When two algebraic expressions are equated together, they are known as algebraic equations.

How Do You Write Algebraic Equation?

We can convert real-life statement involving numbers and conditions into algebraic equation. For example, if the problem says, "the length of a rectangular field is 5 more than twice the width", then it can be written as the algebraic equations l = 2w + 5, where 'l' and 'w' are the length and width of the rectangular field.

What are Linear Algebraic Equations?

An algebraic equation where the highest exponent of the variable term is 1 is a linear algebraic equation. In other words, algebraic equations with degree 1 will be linear. For example, 3y - 9 = 1

Are Quadratic Equations Algebraic Equations?

Yes, quadratic equations are algebraic equations. It consists of an algebraic expression of the second degree.

What are the Basic Formulas of Algebraic Equations?

Some of the basic formulas of algebraic equations are listed below:

• Quadratic Formula: [-b ± √(b² - 4ac)]/2a
• Discriminant: b 2 - 4ac

What are the Rules for Algebraic Equations?

There are 5 basic rules for algebraic equations. These are as follows:

• Commutative Rule of Addition
• Commutative Rule of Multiplication
• Associative Rule of Addition
• Associative Rule of Multiplication
• Distributive Rule of Multiplication

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About solving equations

A value c c is said to be a root of a polynomial p(x) p x if p(c)=0 p c = 0 ..

. This polynomial is considered to have two roots, both equal to 3.

One learns about the "factor theorem," typically in a second course on algebra, as a way to find all roots that are rational numbers. One also learns how to find roots of all quadratic polynomials, using square roots (arising from the discriminant) when necessary. There are more advanced formulas for expressing roots of cubic and quartic polynomials, and also a number of numeric methods for approximating roots of arbitrary polynomials. These use methods from complex analysis as well as sophisticated numerical algorithms, and indeed, this is an area of ongoing research and development.

Systems of linear equations are often solved using Gaussian elimination or related methods. This too is typically encountered in secondary or college math curricula. More advanced methods are needed to find roots of simultaneous systems of nonlinear equations. Similar remarks hold for working with systems of inequalities: the linear case can be handled using methods covered in linear algebra courses, whereas higher-degree polynomial systems typically require more sophisticated computational tools.

How Wolfram|Alpha solves equations

For equation solving, Wolfram|Alpha calls the Wolfram Language's Solve and Reduce functions, which contain a broad range of methods for all kinds of algebra, from basic linear and quadratic equations to multivariate nonlinear systems. In some cases, linear algebra methods such as Gaussian elimination are used, with optimizations to increase speed and reliability. Other operations rely on theorems and algorithms from number theory, abstract algebra and other advanced fields to compute results. These methods are carefully designed and chosen to enable Wolfram|Alpha to solve the greatest variety of problems while also minimizing computation time.

Although such methods are useful for direct solutions, it is also important for the system to understand how a human would solve the same problem. As a result, Wolfram|Alpha also has separate algorithms to show algebraic operations step by step using classic techniques that are easy for humans to recognize and follow. This includes elimination, substitution, the quadratic formula, Cramer's rule and many more.

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Simple Algebra Problems – Easy Exercises with Solutions for Beginners

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Understanding Algebraic Expressions

Breaking down algebra problems, solving algebraic equations, tackling algebra word problems, types of algebraic equations, algebra for different grades.

For instance, solving the equation (3x = 7) for (x) helps us understand how to isolate the variable to find its value.

I always find it fascinating how algebra serves as the foundation for more advanced topics in mathematics and science. Starting with basic problems such as ( $(x-1)^2 = [4\sqrt{(x-4)}]^2$ ) allows us to grasp key concepts and build the skills necessary for tackling more complex challenges.

So whether you’re refreshing your algebra skills or just beginning to explore this mathematical language, let’s dive into some examples and solutions to demystify the subject. Trust me, with a bit of practice, you’ll see algebra not just as a series of problems, but as a powerful tool that helps us solve everyday puzzles.

Simple Algebra Problems and Strategies

When I approach simple algebra problems, one of the first things I do is identify the variable.

The variable is like a placeholder for a number that I’m trying to find—a mystery I’m keen to solve. Typically represented by letters like ( x ) or ( y ), variables allow me to translate real-world situations into algebraic expressions and equations.

An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like ( x ) or ( y )), and operators (like add, subtract, multiply, and divide). For example, ( 4x + 7 ) is an algebraic expression where ( x ) is the variable and the numbers ( 4 ) and ( 7 ) are terms. It’s important to manipulate these properly to maintain the equation’s balance.

Solving algebra problems often starts with simplifying expressions. Here’s a simple method to follow:

• Combine like terms : Terms that have the same variable can be combined. For instance, ( 3x + 4x = 7x ).
• Isolate the variable : Move the variable to one side of the equation. If the equation is ( 2x + 5 = 13 ), my job is to get ( x ) by itself by subtracting ( 5 ) from both sides, giving me ( 2x = 8 ).

With algebraic equations, the goal is to solve for the variable by performing the same operation on both sides. Here’s a table with an example:

Algebra word problems require translating sentences into equations. If a word problem says “I have six less than twice the number of apples than Bob,” and Bob has ( b ) apples, then I’d write the expression as ( 2b – 6 ).

Understanding these strategies helps me tackle basic algebra problems efficiently. Remember, practice makes perfect, and each problem is an opportunity to improve.

In algebra, we encounter a variety of equation types and each serves a unique role in problem-solving. Here, I’ll brief you about some typical forms.

Linear Equations : These are the simplest form, where the highest power of the variable is one. They take the general form ( ax + b = 0 ), where ( a ) and ( b ) are constants, and ( x ) is the variable. For example, ( 2x + 3 = 0 ) is a linear equation.

Polynomial Equations : Unlike for linear equations, polynomial equations can have variables raised to higher powers. The general form of a polynomial equation is ( $a_nx^n + a_{n-1}x^{n-1} + … + a_2x^2 + a_1x + a_0 = 0$ ). In this equation, ( n ) is the highest power, and ( $a_n$ ), ( $a_{n-1} $), …, ( $a_0$ ) represent the coefficients which can be any real number.

• Binomial Equations : They are a specific type of polynomial where there are exactly two terms. Like ($ x^2 – 4 $), which is also the difference of squares, a common format encountered in factoring.

To understand how equations can be solved by factoring, consider the quadratic equation ( $x^2$ – 5x + 6 = 0 ). I can factor this into ( (x-2)(x-3) = 0 ), which allows me to find the roots of the equation.

Here’s how some equations look when classified by degree:

Remember, identification and proper handling of these equations are essential in algebra as they form the basis for complex problem-solving.

In my experience with algebra, I’ve found that the journey begins as early as the 6th grade, where students get their first taste of this fascinating subject with the introduction of variables representing an unknown quantity.

I’ve created worksheets and activities aimed specifically at making this early transition engaging and educational.

6th Grade :

Moving forward, the complexity of algebraic problems increases:

7th and 8th Grades :

• Mastery of negative numbers: students practice operations like ( -3 – 4 ) or ( -5 $\times$ 2 ).
• Exploring the rules of basic arithmetic operations with negative numbers.
• Worksheets often contain numeric and literal expressions that help solidify their concepts.

Advanced topics like linear algebra are typically reserved for higher education. However, the solid foundation set in these early grades is crucial. I’ve developed materials to encourage students to understand and enjoy algebra’s logic and structure.

Remember, algebra is a tool that helps us quantify and solve problems, both numerical and abstract. My goal is to make learning these concepts, from numbers to numeric operations, as accessible as possible, while always maintaining a friendly approach to education.

I’ve walked through various simple algebra problems to help establish a foundational understanding of algebraic concepts. Through practice, you’ll find that these problems become more intuitive, allowing you to tackle more complex equations with confidence.

Remember, the key steps in solving any algebra problem include:

• Identifying variables and what they represent.
• Setting up the equation that reflects the problem statement.
• Applying algebraic rules such as the distributive property ($a(b + c) = ab + ac$), combining like terms, and inverse operations.
• Checking your solutions by substituting them back into the original equations to ensure they work.

As you continue to engage with algebra, consistently revisiting these steps will deepen your understanding and increase your proficiency. Don’t get discouraged by mistakes; they’re an important part of the learning process.

I hope that the straightforward problems I’ve presented have made algebra feel more manageable and a little less daunting. Happy solving!

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Algebraic Expressions and Word Problems

Related Topics: More Lessons for Grade 7 Math Worksheets

Examples, solutions, videos, worksheets, games and activities to help Algebra 1 or grade 7 students learn how to write algebraic expressions from word problems.

Beginning Algebra & Word Problem Steps

• Name what x is.
• Define everything in the problem in terms of x.
• Write the equation.
• Solve the equation.
• Kevin’s age is 3 years more than twice Jane’s age. The sum of their ages is 39. How old is Kevin and Jane?
• The difference between two numbers is 7. Find the two numbers if the larger number is three times the smaller.
• Mary and Jim collect baseball cards, Mary has 5 more than 3 times as many cards as Jim. The total number of cards they both have is 253. How many cards does Mary have?

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• Math Article

Algebraic Expressions Questions

Algebraic Expressions Questions and solutions are provided here to help students of Class 7 to Class 10. As we know, algebraic expressions is one of the most important concepts of mathematics as it deals with the representation of real-life situations mathematically. Also, we can perform various arithmetic operations on algebraic expressions. Let’s understand how to solve various problems related to algebraic expressions and get the practice questions to improve your problem-solving skills.

What is an algebraic expression?

An algebraic expression is formed from variables and constants using different operations. Also, algebraic expressions are made up of terms, and each term is the product of factors. These factors may be numerical or algebraic. Let us consider an example of an algebraic expression.

4x 2 + 3x – 5

Terms: 4x 2 , 3x, 5

Arithmetic operations: +, –

Coefficients: 4, 3

Constant: 5

Variable: x

Learn more about algebraic expressions .

1. Simplify the algebraic expression and write the coefficients:

2x 2 (x + 2) – 3x (x 2 – 3) – 5x(x + 5)

= 2x 3 + 4x 2 – 3x 3 + 9x – 5x 2 – 25x

= 2x 3 – 3x 3 – 5x 2 + 4x 2 + 9x – 25x

= -x 3 – x 2 – 16x

2. Evaluate algebraic expression ax 2 + by 2 – cz for x = 1, y = -1, z = 2, a = -2, b = 1, c = -2:

Given algebraic expression is:

ax 2 + by 2 – cz

Substituting x = 1, y = -1, z = 2, a = -2, b = 1 and c = -2 in the given expression, we get;

ax 2 + by 2 – cz = (-2)(1) 2 + (1)(-1) 2 – (-2)(2)

= -2 + 1 + 4

3. Add 3xy + 5yz – 7xz + 1 and -4xy + 2yz – 2xz + 5xyz + 1.

(3xy + 5yz – 7xz + 1) + (-4xy + 2yz – 2xz + 5xyz + 1)

Let us group the like terms and then add them.

= (3xy – 4xy) + (5yz + 2yz) + (-7xz – 2xz) + 5xyz + (1 + 1)

= -xy + 7yz – 9xz + 5xyz + 2

4. Write the following statements in terms of algebraic expressions.

(i) Add 4 to the product of a number and 7.

(ii) Subtract 4 from the product of a number and 7.

(iii) Add a number to the product of that number and 6.

(iv) Subtract a number from the product of that number and 8.

(v) Add a number to the product of that number and negative 5.

(i) Add 4 to the product of a number and 7 = 7x + 4

(ii) Subtract 4 from the product of a number and 7 = 7x – 4

(iii) Add a number to the product of that number and 6 = 6x + 6

(iv) Subtract a number from the product of that number and 8 = 8x – x

(v) Add a number to the product of that number and negative 5 = -5x + x

5. Simplify the expression: −2a(a + b) − 2a − (a + b)(−2a) − a − 2

−2a(a + b) − 2a − (a + b)(−2a) − a − 2

Now, expand the terms.

-2a 2 – 2ab – 2a – (-2a 2 – 2ab) – a – 2

= -2a 2 – 2ab – 2a + 2a 2 + 2ab – a – 2

Group the like terms and simplify.

= (-2a 2 + 2a 2 ) + (2ab – 2ab) + (-2a – a) – 2

= 0 + 0 – 3a – 2

= -3a – 2

Therefore, −2a(a + b) − 2a − (a + b)(−2a) − a − 2= -3a – 2.

6. Add the following algebraic expressions.

5x 3 + 7 + 6x – 5x 2 , 2x 2 – 8 – 9x, 4x – 2x 2 + 3x 3 , 3x 3 – 9x – x 2 and x – x 2 – x 3 – 4

(5x 3 + 7 + 6x – 5x 2 ) + (2x 2 – 8 – 9x) + (4x – 2x 2 + 3x 3 ) + (3x 3 – 9x – x 2 ) + (x – x 2 – x 3 – 4)

Combining the like terms, we get;

(5x 3 + 3x 3 + 3x 3 – x 3 ) + (– 5x 2 + 2x 2 – 2x 2 – x 2 – x 2 ) + (6x – 9x + 4x – 9x + x) + (7 – 8 – 4)

= 10x 3 – 7x 2 – 7x – 5

7. Subtract the sum of – 3x 3 y 2 + 2x 2 y 3 and – 3x 2 y 3 – 5y 4 from x 4 + x 3 y 2 + x 2 y 3 + y 4 .

Let us find the sum of – 3x 3 y 2 + 2x 2 y 3 and – 3x 2 y 3 – 5y 4 .

i.e., (– 3x 3 y 2 + 2x 2 y 3 ) + (– 3x 2 y 3 – 5y 4 )

= -3x 3 y 2 + (2x 2 y 3 – 3x 2 y 3 ) – 5y 4

= -3x 3 y 2 – x 2 y 3 – 5y 4

Now, subtract the above sum from x 4 + x 3 y 2 + x 2 y 3 + y 4 .

i.e., (x 4 + x 3 y 2 + x 2 y 3 + y 4 ) – (-3x 3 y 2 – x 2 y 3 – 5y 4 )

= x 4 + (x 3 y 2 + 3x 3 y 2 ) + (x 2 y 3 + x 2 y 3 )+ (y 4 + 5y 4 )

= x 4 + 4x 3 y 3 + 2x 2 y 3 + 6y 4

8. What should be added to 3pq + 5p 2 q 2 + p 3 to get p 3 + 2p 2 q 2 + 4pq?

In order to find the required expression, we should subtract 3pq + 5p 2 q 2 + p 3 from p 3 + 2p 2 q 2 + 4pq.

Thus, the required expression will be:

p 3 + 2p 2 q 2 + 4pq − (3pq + 5p 2 q 2 + p 3 )

= p 3 + 2p 2 q 2 + 4pq − 3pq − 5p 2 q 2 − p 3

= p 3 − p 3 + 2p 2 q 2 − 5p 2 q 2 + 4pq − 3pq

= −3p 2 q 2 + pq

Therefore, −3p 2 q 2 + pq to be added to 3pq + 5p 2 q 2 + p 3 , to get p 3 + 2p 2 q 2 + 4pq.

9. Subtract 4.5x 5 – 3.4x 2 + 5.7 from 5x 4 – 3.2x 2 – 7.3x.

Let us subtract 4.5x 5 – 3.4x 2 + 5.7 from 5x 4 – 3.2x 2 – 7.3x.

(5x 4 – 3.2x 2 – 7.3x) – (4.5x 5 – 3.4x 2 + 5.7)

= 5x 4 – 3.2x 2 – 7.3x – 4.5x 5 + 3.4x 2 – 5.7

= 5x 4 + (3.4x 2 – 3.2x 2 ) – 7.3x – 4.5x 5 – 5.7

= 5x 4 + 0.2x 2 – 7.3x – 4.5x 5 – 5.7

= -4.5x 5 + 5x 4 + 0.2x 2 – 7.3x – 5.7

10. Factorise the expression 10x 2 + 5x + 2xy + y.

10x 2 + 5x + 2xy + y

Take the common factors out.

= 5x(2x + 1) + y(2x + 1)

Again, take the common terms out.

= (2x + 1)(5x + y)

Therefore, 10x 2 + 5x + 2xy + y = (2x + 1)(5x + y).

Practice Questions on Algebraic Expressions

• Find the value of the expression a 2 + 3b 2 + 6ab for a = 1 and b = – 2.
• Find the number of terms of the expression 3x 2 y – 2y 2 z – z 2 x + + 4xy – 5.
• Simplify the expression 50x 3 – 21x + 107 + 41x 3 – x + 1 – 93 + 71x – 31x 3 .
• How much is y 4 – 12y 2 + y + 14 greater than 17y 3 + 34y 2 – 51y + 68?

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Solving Equations Worksheets

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About these 15 worksheets.

These worksheets help students practice and master the process of solving for variables in algebraic equations. These worksheets are designed with a variety of formats and problem types to engage students and reinforce their understanding of algebraic principles. The worksheets guide students through the fundamental skills needed to manipulate and simplify equations, ultimately solving for unknowns. They often incorporate creative elements, such as mazes, matching exercises, or themed designs, to keep the practice engaging and dynamic.

One common type of problem found on these worksheets is the straightforward linear equation. These problems focus on basic one-step or two-step equations, where students are required to isolate the variable by performing inverse operations. For example, students may need to add, subtract, multiply, or divide both sides of an equation to balance it and solve for the unknown. These types of problems are crucial because they form the foundation for more complex algebraic concepts. By repeatedly practicing these steps, students solidify their understanding of the equality principle and learn how to reverse operations to simplify equations.

In addition to traditional linear equations, many of these worksheets also feature problems that involve negative numbers, fractions, and decimals. This inclusion helps students develop fluency in handling a variety of number types, which is critical as they progress in their math studies. Solving equations that involve fractions requires students to multiply by the reciprocal or find common denominators, while working with decimals involves careful attention to place value. These challenges enhance a student’s precision and calculation skills, as even small errors in handling fractions or decimals can lead to incorrect solutions.

Another engaging format found on these worksheets is the maze or path-solving activity. In these exercises, students must follow a sequence of correct answers to navigate through a maze. Each correct solution leads them to the next step in the maze, while incorrect answers might lead them astray. This format adds a layer of fun to equation solving while reinforcing accuracy and critical thinking. Students are motivated to double-check their answers to ensure they’re on the right path, which encourages attention to detail. These mazes are an excellent way to build both speed and precision in solving equations.

Many worksheets also incorporate matching or fill-in-the-blank exercises where students solve an equation and then match their solution with a corresponding answer. This type of exercise can be particularly useful for self-checking, as it provides immediate feedback. If a student’s solution doesn’t match any of the available options, they know to revisit their work and look for errors. This format fosters independence and allows students to take ownership of their learning process, developing problem-solving skills that extend beyond the classroom.

These worksheets are designed with a theme or a creative twist, such as solving for variables in a secret message. In these problems, each correct answer corresponds to a letter, and when all the problems are solved, the letters spell out a hidden word or phrase. This type of worksheet adds an element of surprise and fun, keeping students engaged and motivated to complete the task. The process of solving each equation remains the same, but the added challenge of uncovering a hidden message makes the practice more enjoyable and rewarding.

Another important aspect of these worksheets is that they often include a mix of equation types to ensure students are exposed to a broad range of problem-solving scenarios. For instance, students might encounter equations with variables on both sides, which require them to combine like terms and move all variable terms to one side of the equation. These problems teach students how to simplify more complex algebraic expressions and develop a deeper understanding of how to manipulate equations to find solutions.

Equations involving parentheses and the distributive property are also commonly featured. These problems require students to expand expressions by applying the distributive property before solving the equation. This introduces another layer of complexity, as students must remember to correctly distribute terms across all elements within the parentheses. These exercises reinforce the importance of following the correct order of operations and provide practice in breaking down more complex expressions into manageable steps.

The skills taught through these worksheets are foundational for success in higher-level math. By practicing with these worksheets, students develop a strong understanding of how to manipulate algebraic expressions and solve for unknowns, a skill that is critical for more advanced math topics such as systems of equations, quadratic equations, and functions. The variety of problems and formats ensures that students are not only practicing routine calculations but also developing flexibility in their problem-solving approaches.

These worksheets promote a growth mindset by encouraging students to view mistakes as learning opportunities. With immediate feedback from matching or maze activities, students can identify errors and correct their understanding, reinforcing the idea that persistence and practice lead to improvement. This mindset is crucial for building confidence in math, as it teaches students that challenges are an integral part of the learning process.

Fun teaching resources & tips to help you teach math with confidence

FREE Writing Expressions Activity: Partner Challenge – Grades 5-7

Do your students struggle to translate words and phrases into numerical or algebraic expressions? This simple, low-prep writing expressions activity will get kids thinking and talking.

Translating words and phrases into “math language” is an important skill because it will help them prepare for algebra and higher levels of math. But getting kids comfortable and confident with that takes time and practice. Whether you’ve got 5th graders working on numerical expressions or 6th and 7th graders writing algebraic expressions with variables, this low-prep and engaging partner challenge is a great warm-up!

What are Expressions?

Before jumping into this activity with your students, be sure you (and they) understand what expressions are.

Here’s a short, helpful definition of a numerical expression :

Numerical expressions are number sentences involving one or more operations.

Some examples include: 5 + 7 or 15/8 + 12 or (16 + 3) x 18

Numerical expressions contain NUMBERS and math OPERATIONS.

What about algebraic expressions? What’s the difference?

Algebraic expressions are similar to numerical expressions in that they include numbers and operations, but they also include at least one VARIABLE, which represents an unknown number.

Some examples of algebraic expressions are: 3x or 9 – 12y or (4 + p) – 15r

The key thing to note is that expressions include numbers and math operation symbols but NO EQUALS SIGN .

When you have two expressions that are equal to each other, it forms an equation . But that’s a post for another day. If you’d like to move on to finding equivalent expressions try this cut and paste equivalent expressions activity .

Introducing Expressions to Students

Although I shared some definitions and examples for you above, I would encourage you to NOT start there with your students.

Instead, give them the chance to think about their own way of translating situations, words and phrases into expressions using this partner challenge.

Walk around the room and listen in on student conversations to hear how they think about the wording and the math . Listen to student disagreements and how they justify their answers.

This will give you insights into what students already know , what they may need help with , and any misconceptions that you need to clear up after they finish the activity.

Writing Expressions Activity: Which Version Will You Use?

To use this in your class, all you need to do is choose which version is best (numeric expressions or algebraic expressions) then print a set for each pair of students .

But which version do you need for your students? Option one includes numerical expressions only . This is best for 5th grade students who are not yet ready to work with or think about variables.

But you might also use this version with older students who struggle with word problems , need further clarification on different math operations or who need a refresher before including variables.

Option two is to help introduce students to writing algebraic expressions . This is a fun warm-up for 6th or 7th grade students before you get into your expressions chapter or unit, or as a quick refresher before working on more challenging expressions and word problems.

If you enjoy this activity, become a Math Geek Mama+ member and gain access to the entire ad-free library of engaging math activities like this one, hundreds of math games and low-prep practice worksheets for grades 5-8 !

Learn more about Math Geek Mama+ right HERE .

How to Use the Writing Expressions Partner Activity

To begin, pair students with a partner and print a worksheet for each partner . One student should have ‘Partner A’ and the other student should have ‘Partner B.’

Using Option One: Writing Numerical Expressions

Students take turns reading one of the verbal expressions aloud , while their partner writes a numerical expression to model it on their white board. For example, if partner A reads the phrase, “the sum of 9 & 12,” partner B would write “9 + 12” on their white board.

They should discuss their expressions together as they work. Do they agree? Why or why not? Is there another way to write the expression?

Once they have each written a variety of expressions (or when you tell them time is up), students take a moment to answer they ‘think about it’ questions and you can discuss the activity as a whole class.

This will help discuss properties such as the commutative & associative property, as well as the importance of order of operations.

Using Option Two: Writing Algebraic Expressions

The second option (also labeled ‘Partner A’ & ’Partner B’) includes some of the same numerical expressions, but there are also some expressions that require variables .

This option is completed the same way, with partners taking turns reading expressions & writing expressions on their white board.

This is meant to introduce students to the idea of using variables in their expressions when there is a value that is unknown .

Working through a variety of expressions with a partner can give them an opportunity to talk together, to notice & wonder when they come across expressions that stump them & to think about how they might represent unknowns in their expression.

The goal of the activity is not that students will correctly interpret & write each expression, it is simply that they think & talk meaningfully about different examples with their partner.

When time is up, allow them a few minutes to write their reflections to the ‘Think About it” questions & then discuss as a whole class.

Ultimately, this short and simple activity is designed to get students talking and thinking about words and math and how to write expressions that accurately model situations .

And hopefully, it provides a great springboard for math talk in your classroom, spurring students on to other questions and examples and greater understanding.

Want to give this a try with your students? Just use the link below to get it free in my shop!

{Click HERE to grab the FREE Writing Expressions Activity: Partner Practice from my shop!}

Looking for more writing expressions resources check out the links below..

• Writing Algebraic Expressions: FREE Practice Pages
• Simplify Expressions: Equivalent Expressions Cut & Paste {FREE}
• Simplify Expressions Visually with Algebra Tiles – DIGITAL Activity

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Math Time Doesn't Have to End in Tears

Join 165,000+ parents and teachers who learn new tips and strategies, as well as receive engaging resources to make math fun. Plus, receive my guide, "5 Games You Can Play Today to Make Math Fun," as my free gift to get you started!