We have been solving problems involving ratios, also called proportions. These model situations where the ratio/proportion between items never changes. We also call these linear equations.
Subsection2.6.1Solving Linear Equations
Before reading farther solve the equation \(5x-7=12\text{.}\) What steps did you use? Why do they work? Example 2.6.1 is an example of solving another linear equation.
Note we added three because it eliminates the -3 (undoes subtraction of 3). We divided by negative eight because it eliminates the -8 (undoes the multiplication by -8).
Checkpoint2.6.2.
What is the solution to \({4}x+{2}={14}\text{.}\)\(x=\)
Answer.
\(3\)
Solution.
\begin{equation*}
\begin{aligned}
{4}x+{2} \amp = {14}\\
{4}x \amp = {14}-{2}\\
x \amp = \frac{{14}-{2}}{{4}}\\
x \amp = {3}.
\end{aligned}
\end{equation*}
Some linear equations need one more technique. What would you need to solve \(17-4y = 14-y\text{?}\)Example 2.6.3 is an example of solving a similar linear equation.
This is referred to as clearing denominators. We are once again eliminating division by multiplying. Always remember to distribute. Note, we could multiply once if we figured out the correct number, but there are no prizes for doing this fast, so you can do this either way.
Subsection2.6.2Solving with Multiple Variables
When we are using models (also called formulas) they often include more than one variable (also known as named constants). The process for solving such models is the same.
Example2.6.6.
Solve the equation \(V=IR\) for \(R\text{.}\) Note, this formula is explained in Example 1.3.1.
Solution.
\begin{align*}
V & = IR.\\
\frac{V}{I} & = \frac{IR}{I} & \text{divide to undo multiplication}.\\
\frac{V}{I} & = R.
\end{align*}
Example2.6.7.
Solve the lift equation \(L=\frac{1}{2}\rho S C_L v^2\) for \(S\text{.}\)
Solution.
\begin{align*}
L \amp = \frac{1}{2}\rho S C_L v^2.\\
2L \amp = 2\frac{1}{2}\rho S C_L v^2.\\
2L \amp = \rho S C_L v^2.\\
\frac{2L}{\rho} \amp = \frac{\rho S C_L v^2}{\rho}.\\
\frac{2L}{\rho} \amp = S C_L v^2.\\
\frac{2L}{\rho C_L} \amp = \frac{S C_L v^2}{C_L}.\\
\frac{2L}{\rho C_L} \amp = S v^2.\\
\frac{2L}{\rho C_L v^2} \amp = \frac{S v^2}{v^2}.\\
\frac{2L}{\rho C_L v^2} \amp = S.
\end{align*}
Example2.6.8.
Solve the lift equation \(L=\frac{1}{2}\rho S C_L v^2\) for \(S\) when \(L=255\bar{0}\text{,}\)\(\rho=0.002378\text{,}\) and \(C_L=1.430\text{.}\)
Solution.
Because we know some of the values, we can first insert them before solving.
\begin{align*}
255\bar{0} \amp = \frac{1}{2}(0.002378)S(1.430)v^2.\\
255\bar{0} \amp = 0.001700 S v^2.\\
\frac{255\bar{0}}{0.001700} \amp = \frac{0.001700 S v^2}{0.001700}.\\
150\bar{0}000 \amp = S v^2.\\
\frac{150\bar{0}000}{v^2} \amp = \frac{S v^2}{v^2}.\\
\frac{150\bar{0}000}{v^2} \amp = S.
\end{align*}
Checkpoint2.6.9.
Solve the lift equation \(L=\frac{1}{2}\rho S C_L v^2\) for \(\rho\text{.}\)
Subsection2.6.3Identifying Linear Expressions
All of the equations with which we just worked are linear. What do you use to identify linear expressions or linear equations? Table 2.6.10 shows examples of linear expressions and non-linear expressions.
Table2.6.10.Linear and Non-linear
Linear
Non-linear
\(5x+3\)
\(5x^2-x+3\)
\(y=11-\frac{7}{13}x\)
\(y=\frac{17}{x}\)
\(7x-9y=8\)
\(3-2xy=12\)
Some equations that may not appear to be linear can be solved using the same methods.
