Writing and Solving One-Step
Linear Equations In One Variable

Writing and solving one-step linear equations in one variable sounds quite ominous. However, as we analyze what this means, you will see that it is a skill your students can acquire comfortably.

Equations are statements which say that two quantities are equal. Numerical equations, such as 7 + 9 = 16, include only numbers. Algebraic equations, such as 12 ÷ y = 2, include at least one variable. Solving an algebraic equation is like solving a mystery. You are seeking the value of the variable that will make the equation a true statement. For example, in the equation x + 2 = 7, the value of x = 5 makes the equation 5 + 2 = 7 a true statement. Notice that the variable x in the equation is only raised to the first power. When the variable in an equation is raised to the first power, the equation is linear. When the variable in an equation is raised to a power greater than one, the equation is not linear. For example, the equation 9 – s³ = 1 is not linear because the variable s is raised to the third power.

When solving a linear equation, there are three important ideas to keep in mind. The first is that you are trying to get the variable onto one side of the equation, by itself. To do this, employ the second idea, that is, use the inverse operation to undo what was done to the variable. Look at the examples below.

Type of Equation Equation	Example	Operation to Use
Addition Equation	x + 3 = 5	Subtract 3
Subtraction Equation	y – 4 = 3	Add 4
Multiplication Equation	2x = 12	Divide by 2
Division Equation	x ÷ 3 = 5	Multiply by 3

The third idea to keep in mind is that an equation is similar to a balance scale. You are given two quantities which are in balance. In order to solve the equation, you need to keep the equation balanced by doing the exact same thing to both sides of the equation. Therefore, if you add four to one side of the equation, you must add four to the other side of the equation to keep it in balance.

Let's apply these three ideas to some specific equations. Given the equation 3x = 12, you want to get the variable x all by itself. In order to do this, you must undo what was done to the variable x. Since x was multiplied by 3, use the inverse operation of multiplication and divide x by 3. To keep the equation in balance, you will then need to divide the other side by 3. Dividing both sides of the equation by 3 gives a value of x = 4. Applying these same ideas to the equation x – 7 = 13, you begin by getting x by itself. To do this, undo the operation of subtracting 7 by adding 7 to the variable. Then add 7 to the other side of the equation to keep the equation balanced. The value is x = 20.

Writing an algebraic equation involves translating a written statement into algebraic form. The variable in the equation represents the number you want to find.

It is very helpful to make use of the four-step process of problem solving when writing an equation: (1) understand, (2) decide, (3) solve, and (4) look back. This process is described in Unit 1, Lesson 12. Let's apply this four-step process to the following problem. Try to visualize what is happening as you read the problem. "Billy went to school with some change in his pocket. On the way to school he found 27¢. When Billy got to school, he counted his money and found that he had $1.14. How much money did Billy have in his pocket when he left for school?"

Understand: You want to know how much money Billy had in his pocket when he left for school.
He found 27¢ along the way, so that amount is being added to the amount he had when he left for school. When he got to school, he had $1.14, which is the sum of what he started with and what he found on the way to school.

Decide: You can write an equation to help you solve this problem. If you let m equal the amount of money Billy left for school with, the equation will be m + $0.27 = $1.14.

Solve: To solve m + $0.27 = $1.14, you need to subtract $0.27 from both sides to get the variable m by itself. The value is m = $0.87. Substituting $0.87 back in the equation for m, results in the statement $0.87 + $0.27 = $1.14, which is a true statement.

Look back: Does the answer of $0.87 make sense? Yes, because Billy must have left home with an amount close to $1.00 since he only found 27¢ and now has more than $1.00.

• Introducing the Concept
• Developing the Concept