Appendix C Review: Linear Equations
A linear equation is a special type of equation in which the only power on the variable is 1. It is called linear because its graph forms a straight line.
How to Solve a Linear Equation.
Simplify the expressions on both sides of the equation, if needed, using the distributive property and/or combining like terms.
Add or subtract terms from both sides of the equation until only one term has a variable attached, and that term is isolated on one side of the equation.
Multiply or divide both sides of the equation by the same number to solve for the variable.
Check that the solution is correct.
Example C.0.1.
Solve \(2x-15=7\)
Solution.
\begin{align*}
2x-15\amp =7\\
2x-15+15\amp =7+15\\
2x\amp=22 \\
\frac{2x}{2}\amp= \frac{22}{2}\\
x\amp=11
\end{align*}
To check that the solution is correct, we can substitute \(x=11\) into the original equation and ensure that it results in a true statement:
\begin{align*}
2x-15\amp=7 \\
2(11)-15\amp=7 \\
22-15\amp=7 \\
7 \amp=7
\end{align*}
Since 7=7 is true, we know that the solution \(x=11\) is correct.
We also could have checked the solution using graphing technology by verifying that 11 is the x-coordinate of the intersection point of the graphs of \(y=2x-15\) and \(y=7\text{.}\)
Example C.0.2.
Solve for \(t\text{:}\)
\begin{equation*}
17(t-2)+15=11t-10
\end{equation*}
Solution.
Simplify:
\begin{align*}
17(t-2)+15\amp =11t-10\\
17t-34+15\amp =11t-10\\
17t-19\amp=11t-10
\end{align*}
Subtract \(11t\) from both sides and add \(19\) to both sides to get the variable term isolated on the left side of the equation:
\begin{align*}
17t-11t-19+19\amp=11t-11t-10+19 \\
6t\amp=9
\end{align*}
Divide by 6 to solve:
\begin{align*}
\frac{6t}{6}\amp =\frac{9}{6}\\
t\amp =\frac{3}{2}
\end{align*}
We can express the answer as \(t=\frac{3}{2}\text{,}\) \(t=1 \frac{1}{2}\text{,}\) or \(t=1.5\text{.}\)
Example C.0.3.
Solve for \(y\) and round your answer to four decimal places:
\begin{equation*}
10-2.79(y+6)=1.4y+1
\end{equation*}
Solution.
\begin{align*}
10-2.79(y+6)\amp=1.4y+1 \\
10-2.79y-16.74\amp=4.4y+1 \\
10-1-2.79y+2.79y-16.74\amp =4.4y+2.79y+1-1\\
-7.74\amp =7.79y\\
\frac{-7.74}{7.79}\amp=\frac{7.79y}{7.79} \\
0.9936\amp\approx y
\end{align*}
