2.3 Intermediate Linear Equations

Terrance Berg

Chapter 2: Linear Equations

2.3 Intermediate Linear Equations

When working with linear equations with parentheses, the first objective is to isolate the parentheses. Once isolated, the parentheses can be removed and then the variable solved.

Example 2.3.1

Solve for $x$ in the equation $4(2x-6) = 16.$

$\begin{array}{rrrl} \dfrac{4(2x-6)}{4}&=&\dfrac{16}{4}&\text{Divide both sides by 4} \\ \\ (2x - 6)&=&4&\text{Remove the parentheses} \\ 2x -6&=&4&\text{Add 6 to both sides to remove }-6 \\ +6&&+6& \\ \midrule \dfrac{2x}{2}&=&\dfrac{10}{2}&\text{Divide both sides by 2} \\ \\ x&=&5&\text{Solution} \end{array}$

Example 2.3.2

Solve for $x$ in the equation $3(2x - 4) + 9 = 15.$

$\begin{array}{rrrl} 3(2x - 4) + 9&=&15&\text{Subtract 9 from both sides} \\ -9&&-9& \\ \midrule \dfrac{3(2x - 4)}{3}&=& \dfrac{6}{3}&\text{Divide both sides by 3 and remove parentheses} \\ \\ 2x - 4&=&2&\text{Add 4 to both sides} \\ +4&&+4& \\ \midrule \dfrac{2x}{2}&=&\dfrac{6}{2}&\text{Divide both sides by 2} \\ \\ x&=&3&\text{Solution} \end{array}$

For some problems, it is too difficult to isolate the parentheses. In these problems, it is necessary to multiply or divide throughout the parentheses by whatever coefficient is in front of it.

Example 2.3.3

Solve for $x$ in the equation $3(4x - 5) - 4(2x + 1) = 5.$

$\begin{array}{rrrl} 3(4x - 5) - 4(2x + 1)&=&5&\text{Distribute} \\ 12x - 15 - 8x - 4&=&5&\text{Combine similar terms} \\ 4x-19&=&5&\text{Add 19 to both sides} \\ +19&&+19& \\ \midrule \dfrac{4x}{4}&=&\dfrac{24}{4}&\text{Divide both sides by 4} \\ \\ x&=&6& \end{array}$