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

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 2) for 𝑏 3) for 𝑐

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.

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

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 =

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

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

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

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

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

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

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

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.

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

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

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

Consider the expression inside the absolute value as a whole, write into two equations and then solve for the variable. Solution: 5𝑥+1=20 5𝑥=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.

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.

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

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= 10 5 𝑥=2 Or 5𝑥−4=−6 5𝑥=−6+4 5𝑥=−2 𝑥=− 2 5

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

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

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

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