Solving Linear Equations with One Variable

Site:	Saylor Academy
Course:	MA007: Algebra
Book:	Solving Linear Equations with One Variable

Printed by:	Guest user
Date:	Tuesday, 15 July 2025, 7:42 AM

Description

This is a book resource with multiple pages. Navigate between the pages using the buttons.

Linear Literal Equations
Exercises
Answers

Linear Literal Equations

Solving equations for a particular variable is useful, especially when we have some pieces of information, and we need to repeatedly use a formula to find others. For example, you may know that distance = rate times time. or D = rt. If I had to do a lot of calculations where I knew the rates and the distances but needed to find the times, it would have been easier if the equation had been written another way. In this section, you will learn how to change equations to make them easier to work with in particular situations. In our example, we can take D = rt and change it to t = D/r. This skill will be very helpful in Unit 5.

Literal equations, or formulas, usually have more than one variable. Since the letters are placeholders for values, the steps for solving them are the same. Use the properties of equality to isolate the indicated variable.

Solve for \(a\): \(P=2 a+b\)

\(\begin{aligned} P &=2 a+b \\ P-b &=2 a+b-b \quad \text { Subtract } b \text { on both sides. } \\ P-b &=2 a \\ \frac{P-b}{2} &=\frac{2 a}{2} \qquad \qquad \text { Divide both sides by } 2 . \\ \frac{P-b}{2} &=a \end{aligned}\)

Solution: \(a=\frac{P-b}{2}\)

Solve for \(x\): \(z=\frac{x+y}{2}\)

\( \begin{aligned} z&\frac{x+y}{2}\\ 2 \cdot z&=2 \cdot \frac{x+y}{2} \quad \text { Multiply both sides by 2. }\\ 2 z &=x+y \\ 2 z-y &=x+y-y \quad \text { Subtract } y \text { on both sides. } \\ 2 z-y &=\quad x \end{aligned} \)

Solution \(x=2z−y\)

Source: Ann Simao, https://cnx.org/contents/F9lGcov0@1/Elementary-Algebra-Solving-Linear-Equations-in-One-Variable#element-515
This work is licensed under a Creative Commons Attribution 4.0 License.

Exercises

Literal Equations

Solve for w: \(P=2 l+2 w\)
Solve for \(\mathrm{b}: P=a+b+c\)
Solve for \(C\): \(F=\frac{9}{5} C+32\)
Solve for \(r\): \(C=2 \pi r\)
Solve for \(y\): \(z=\frac{x-y}{5}\)

Answers

\(w=\frac{P-2 l}{2}\)
\(b=P-a-c\)
\(C=\frac{5 F-160}{9}\)
\(r=\frac{C}{2 \pi}\)
\(y=-5 z+x\)