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Study Guides > College Algebra

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:
  1. 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.
  2. Apply the distributive property as needed: [latex]a\left(b+c\right)=ab+ac[/latex].
  3. Isolate the variable on one side of the equation.
  4. 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

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