1. Chapter 1 Solving Linear Equations
Section 1.5 Formulas Section 1.6 Absolute Value Equations

2. Section 1.5 Formulas
Objective: Solve linear formulas for a given variable.
When solving formulas for a variable we need to focus on the one variable we are trying to solve for, all the others are treated just like numbers.
For example:
Solve the formula 𝒂+𝟐𝒃= 𝒄 𝟑
for 𝑎 ) for 𝑏 ) for 𝑐

3. Solution Solve 𝒂+𝟐𝒃= 𝒄 𝟑 for 𝑎 𝑎= 𝑐 3 −2𝑏 2) for 𝑏
2𝑏= 𝑐 3 −𝑎 𝑏= 𝑐 3 −𝑎 2
3) for 𝑐
𝑎+2𝑏 ∙3=𝑐 3𝑎+6𝑏=𝑐
In your solution, the variable you solve for should be isolated on one side, and on the other side it doesn’t have this variable anymore. For example, in part 1), a is isolated on the left side, and no a on the right side.

4. Isolate the term containing the variable that we want to solve for
Example 1: Solve the following formula for 𝑥 𝑦 = 5𝑚𝑥−6𝑏 𝑥 =
Solution:
𝑦+6𝑏=5𝑚𝑥 𝑦+6𝑏 5𝑚 =𝑥

5. Example 2: Solve the formula for 𝑏. 𝐴 = 12 ℎ (𝑎 + 𝑏) 𝑏 =
Multiply out first and isolate the term containing the variable that we want to solve for
Solution:
𝐴=12ℎ𝑎+12ℎ𝑏 𝐴−12ℎ𝑎=12ℎ𝑏 𝐴−12ℎ𝑎 12ℎ =𝑏
Note: Cases are sensitive when you write variables. That means, you cannot enter upper case A as lower case a. They represent two different variables.
Example 2: Solve the formula for 𝑏. 𝐴 = 12 ℎ (𝑎 + 𝑏) 𝑏 =

6. When there are more than one term containing the variable that we want to solve for, collect those terms on one side and factor out the variable then solve.
Example 3: The formula, 𝐴 = 2𝑙𝑤 + 2𝑙ℎ + 2𝑤ℎ, gives the surface area 𝐴, of a rectangular solid with length, width, and height, 𝑙, 𝑤, and ℎ, respectively. Solve the formula for 𝑤. 𝐴 = 2𝑙𝑤 + 2𝑙ℎ + 2𝑤ℎ 𝑤 =
Solution:
𝐴−2𝑙ℎ=2𝑙𝑤+2𝑤ℎ 𝐴−2𝑙ℎ=𝑤 2𝑙+2ℎ 𝐴−2𝑙ℎ 2𝑙+2ℎ =𝑤

7. Solve the equation for h: 14(𝑎−ℎ)=ℎ𝑢+2𝑟
Multiply out first; Collect the terms containing the variable that we want to solve for; Factor out the variable and solve.
Example 4:
Solve the equation for h:
14(𝑎−ℎ)=ℎ𝑢+2𝑟
ℎ =
Solution:
14𝑎−14ℎ=ℎ𝑢+2𝑟 −14ℎ−ℎ𝑢=2𝑟−14𝑎 ℎ −14−𝑢 =2𝑟−14𝑎 ℎ= 2𝑟−14𝑎 −14−𝑢

8. Fractions---Multiply by LCD first
Example 5:
Solve the equation for p.
5 𝑠 = 8 𝑡 + 3𝑝
𝑝 =
Solution:
𝐿𝐶𝐷 𝑠, 𝑡 =𝑠𝑡 Multiply 𝑠𝑡 on both sides,
or by every term. 5 𝑠 𝑠𝑡 = 8 𝑡 𝑠𝑡 + 3𝑝 𝑠𝑡 5𝑡=8𝑠+3𝑝𝑠𝑡 5𝑡−8𝑠=3𝑝𝑠𝑡 5𝑡−8𝑠 3𝑠𝑡 =𝑝

9. More Practice Solve the following formula for 𝑚 𝑐=𝑎𝑚𝑡 𝑚=
Solve the formula for the specified variable. 𝑆=2𝜋𝑡ℎ+2𝜋𝑟ℎ ℎ=

10. More Practice Solve for 𝑧 8𝑧−8𝑏=𝑎−6 𝑧 = Solve the formula for 𝑎.
𝐴=12ℎ(𝑎+𝑏) 𝑎=

11. More Practice Solve the formula for 𝑠. 𝑠𝑡+𝑤𝑓=𝑠𝑗+𝑒 𝑠= Solve for 𝑝
𝑤=45(𝑦𝑝+𝑚)
𝑝 =

12. More Practice Solve the formula for 𝑚 6𝑚+𝑚𝑝 𝑝 =𝑟 𝑚=
Solve the equation for 𝑘:
1 4 (𝑦−𝑘)=𝑘𝑛+4𝑠 𝑘=

13. Section 1.6 Absolute Value Equations
Objective: Solve linear absolute value equations.
The absolute value is the distance between a number and zero on a number line. So an absolute value should not be a negative number. It doesn’t make sense to say a distance is negative.
If 𝒙 =𝟒,
then 𝑥=4 𝑜𝑟 −4.

14. Example 1:
Solve the equation. If there is more than one solution, separate them with a comma. |𝑥|=6 𝑥=6 𝑜𝑟 −6

15. You should isolate the absolute value part first.
Example 2: Solve the equation. If there is more than one solution, separate them with a comma. 8−|𝑥|=4 𝑥=
Solution:
− 𝑥 =4−8 − 𝑥 =−4 𝑥 = −4 −1 𝑥 =4 𝑥=4 𝑜𝑟 −4

16. More Practice
Solve the equation. If there is more than one solution, separate them with a comma. |𝑥|=2
Solve the equation. If there is more than one solution, separate them with a comma. 4−|𝑥|=2

17. Consider the expression inside the absolute value as a whole, write into two equations and then solve for the variable.
Solution:
5𝑥+1= 𝑥=20−1 5𝑥=19 𝑥= 19 5
Example 3: Solve the equation. If there is more than one solution, separate them with a comma. |5𝑥+1|=20 𝑥=
𝑜𝑟 5𝑥+1=−20 5𝑥=−20−1 5𝑥=−21 𝑥=− 21 5
There are two different solutions.
They are not just opposite to each other.

18. Example:
Solve the equation. If there is more than one solution, separate them with a comma. 4𝑥+3 =−11 𝑥=
Solution:
Since an absolute value should not be a negative number. This problem has no solution.

19. More Practice Solve the equation |5𝑥−1|=19 The solutions are:
|2𝑥−1|=18
The solutions are:

20. Remember you should isolate the absolute value part first, before you apply the definition and write into two equations.
Example 5: Solve the equation. If there is more than one solution, separate them with a comma. 2−4|5𝑥−4|=−22 𝑥=
Solution:
−4 5𝑥−4 =−22−2 −4 5𝑥−4 =−24 5𝑥−4 = −24 −4 5𝑥−4 =6
5𝑥−4=6 5𝑥=6+4 5𝑥=10 𝑥= 𝑥=2
Or 5𝑥−4=−6 5𝑥=−6+4 5𝑥=−2 𝑥=− 2 5

21. More Practice Solve the equation |4𝑥+4|=13 The solutions are:
−5+4|4𝑥−1|=11
The solutions are:

22. More Practice Solve the equation 5−2|5𝑥+3|=−11 The solutions are:
5−4|4𝑥+3|=−3
The solutions are:

23. Example 6: Solve the equation
Example 6: Solve the equation. If there is more than one solution, separate them with a comma. |4𝑥−1|=|7𝑥−4| 𝑥=
Solution:
Both sides contain only absolute values, we can start writing as two equations, and solve each equation for x.
4𝑥−1=7𝑥−4 4𝑥−7𝑥=−4+1 −3𝑥=−3 𝑥=1
Or 4𝑥−1=− 7𝑥−4 4𝑥−1=−7𝑥+4 4𝑥+7𝑥=4+1 11𝑥=5 𝑥= 5 11

24. More Practice Solve the equation |2𝑥+1|=|4𝑥−2| The solutions are:
|2𝑥+3|=|5𝑥−1|
The solutions are: