# Solving Linear Equations in One Variable

A**linear equation**is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form [latex]ax+b=0[/latex] and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An

**identity equation**is true for all values of the variable. Here is an example of an identity equation.

[latex]3x=2x+x[/latex]

The **solution set**consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for [latex]x[/latex] will make the equation true. A

**conditional equation**is true for only some values of the variable. For example, if we are to solve the equation [latex]5x+2=3x - 6[/latex], we have the following:

[latex]\begin{array}{l}5x+2\hfill&=3x - 6\hfill \\ 2x\hfill&=-8\hfill \\ x\hfill&=-4\hfill \end{array}[/latex]

The solution set consists of one number: [latex]\{-4\}[/latex]. It is the only solution and, therefore, we have solved a conditional equation.
An **inconsistent equation**results in a false statement. For example, if we are to solve [latex]5x - 15=5\left(x - 4\right)[/latex], we have the following:

[latex]\begin{array}{ll}5x - 15=5x - 20\hfill & \hfill \\ 5x - 15 - 5x=5x - 20 - 5x\hfill & \text{Subtract }5x\text{ from both sides}.\hfill \\ -15\ne -20 \hfill & \text{False statement}\hfill \end{array}[/latex]

Indeed, [latex]-15\ne -20[/latex]. There is no solution because this is an inconsistent equation.
Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.
### A General Note: Linear Equation in One Variable

A linear equation in one variable can be written in the form[latex]ax+b=0[/latex]

where *a*and

*b*are real numbers, [latex]a\ne 0[/latex].

### How To: Given a linear equation in one variable, use algebra to solve it.

The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x=_________, if*x*is the unknown. There is no set order, as the steps used depend on what is given:

- We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.
- Apply the distributive property as needed: [latex]a\left(b+c\right)=ab+ac[/latex].
- Isolate the variable on one side of the equation.
- When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.

### Example 1: Solving an Equation in One Variable

Solve the following equation: [latex]2x+7=19[/latex].### Solution

This equation can be written in the form [latex]ax+b=0[/latex] by subtracting [latex]19[/latex] from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.[latex]\begin{array}{ll}2x+7=19\hfill & \hfill \\ 2x=12\hfill & \text{Subtract 7 from both sides}.\hfill \\ x=6\hfill & \text{Multiply both sides by }\frac{1}{2}\text{ or divide by 2}.\hfill \end{array}[/latex]

The solution is [latex]x=6[/latex].
### Try It 1

Solve the linear equation in one variable: [latex]2x+1=-9[/latex]. Solution### Example 2: Solving an Equation Algebraically When the Variable Appears on Both Sides

Solve the following equation: [latex]4\left(x - 3\right)+12=15 - 5\left(x+6\right)[/latex].### Solution

Apply standard algebraic properties.[latex]\begin{array}{ll}4\left(x - 3\right)+12=15 - 5\left(x+6\right)\hfill & \hfill \\ 4x - 12+12=15 - 5x - 30 \hfill & \text{Apply the distributive property}.\hfill \\ 4x=-15 - 5x\hfill & \text{Combine like terms}.\hfill \\ 9x=-15 \hfill & \text{Place }x-\text{terms on one side and simplify}.\hfill \\ x=-\frac{15}{9}\hfill & \text{Multiply both sides by }\frac{1}{9}\text{, the reciprocal of 9}.\hfill \\ x=-\frac{5}{3}\hfill & \hfill \end{array}[/latex]

### Analysis of the Solution

This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, [latex]x=-\frac{5}{3}[/latex].### Try It 2

Solve the equation in one variable: [latex]-2\left(3x - 1\right)+x=14-x[/latex]. Solution## Licenses & Attributions

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