lesson 19: rearranging formulas - mr....
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ALGEBRA I
M1 Hart Interactive β Algebra 1
Lesson 19
Lesson 19: Rearranging Formulas
Exploratory Challenge
Rearranging Familiar Formulas
1. The area π΄π΄ of a rectangle is 25 in2. The formula for area is π΄π΄ = ππππ.
A. If the width ππ is 10 inches, what is the length ππ?
B. If the width ππ is 15 inches, what is the length ππ?
2. A. Joey rearranged the area formula to solve for ππ. His beginning work is shown below. Finish his work to
isolate .
=
=
= _____
A lw
A lww w
l
B. Verify that the area formula, solved for ππ, will give the same results for ππ as having solved for ππ in the
original area formula. Use both w is 10 inches and w is 15 inches with an area of 25 in2.
π΄π΄ = 25 in2 ππ
ππ
Lesson 19: Rearranging Formulas
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This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org
This file derived from ALG I-M1-TE-1.3.0-07.2015
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
ALGEBRA I
M1 Hart Interactive β Algebra 1
Lesson 19
3. In the first column solve each equation for π₯π₯. Then follow the same steps to solve the βformulaβ for x in
the second column. Remember a variable symbol, like ππ, ππ, c, and ππ, represents a number.
Equation βFormulaβ
A. β =2 6 10x β =ax b c
B. β β = β3 3 12x β β = βax b c
C. β =9 4 21x β =a bx c
D. β
=3 1
102
x
β=
ax b dc
E. + =5 152
x + =
x b ca
Lesson 19: Rearranging Formulas
S.148
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org
This file derived from ALG I-M1-TE-1.3.0-07.2015
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
ALGEBRA I
M1 Hart Interactive β Algebra 1
Lesson 19
4. Solve the equation πππ₯π₯ β ππ = ππ for ππ. The variable symbols π₯π₯, ππ, and ππ, represent numbers.
5. Complete the chart below.
Formula Use the Given Values
and Solve Solve the Formula for
One Variable
Use the Given Values and the Equation from the Previous Column
then Solve
The perimeter formula
for a rectangle is
ππ = ππ(ππ + ππ), where
ππ represents the
perimeter,
ππ represents the length,
and ππ represents the
width.
Calculate ππ when ππ = 70
and ππ = 15.
Solve ππ = 2(ππ + ππ) for ππ. Calculate ππ when ππ = 70
and ππ = 15.
The area formula for a
triangle is π΄π΄ = 12 ππβ,
where
π΄π΄ represents the area,
ππ represents the length
of the base, and
β represents the height.
Calculate ππ when π΄π΄ =100 and β = 20.
Solve π΄π΄ = 12 ππβ for b. Calculate ππ when π΄π΄ =
100 and β = 20.
6. Rearrange each formula to solve for the specified variable. Assume no variable is equal to 0.
A. Given π΄π΄ = ππ(1 + ππππ), solve for ππ. B. Given π΄π΄ = ππ(1 + ππππ), solve for ππ.
C. Given πΎπΎ = 12πππ£π£
2, solve for ππ. D. Given πΎπΎ = 1
2πππ£π£2
, solve for π£π£.
Lesson 19: Rearranging Formulas
S.149
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org
This file derived from ALG I-M1-TE-1.3.0-07.2015
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
ALGEBRA I
M1 Hart Interactive β Algebra 1
Lesson 19
7. Comparing Equations with One Variable to Those with More Than One Variable
Solve for x in each equation. You may want to start with the equations on the right and then solve the
equations on the left.
Equation Containing More Than One Variable Related Equation Solve πππ₯π₯ + ππ = ππ β πππ₯π₯ for π₯π₯.
Solve 3π₯π₯ + 4 = 6 β 5π₯π₯ for π₯π₯.
Solve for π₯π₯.
πππ₯π₯ππ + πππ₯π₯
ππ = ππ
Solve for π₯π₯.
2π₯π₯5 + π₯π₯
7 = 3
Lesson 19: Rearranging Formulas
S.150
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org
This file derived from ALG I-M1-TE-1.3.0-07.2015
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
ALGEBRA I
M1 Hart Interactive β Algebra 1
Lesson 19
Lesson Summary
Homework Problem Set
For Problems 1β8, solve for π₯π₯.
1. πππ₯π₯ + 3ππ = 2ππ 2. πππ₯π₯ + β = π π π₯π₯ β ππ 3. 3πππ₯π₯ = 2ππ(ππ β 5π₯π₯) 4. π₯π₯+ππ4 = ππ
5. π₯π₯5 β 7 = 2ππ 6.
π₯π₯6 β
π₯π₯7 = ππππ 7.
π₯π₯ππ β π₯π₯
ππ = 1ππ 8.
3πππ₯π₯+2ππππ = 4ππ
The properties and reasoning used to solve equations apply regardless of how many variables appear in an equation or formula. Rearranging formulas to solve for a specific variable can be useful when solving applied problems.
Lesson 19: Rearranging Formulas
S.151
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org
This file derived from ALG I-M1-TE-1.3.0-07.2015
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
ALGEBRA I
M1 Hart Interactive β Algebra 1
Lesson 19
9. Solve for ππ.
ππ = πππ π ππ + ππ
10. Solve for π’π’.
1π’π’ + 1
π£π£ = 1ππ
11. Solve for π π .
π΄π΄ = π π 2
12. Solve for β.
ππ = ππππ2β
13. Solve for ππ.
ππ = 4βππ
14. Solve for ππ.
πΉπΉ = πΊπΊππππππ2
15. Solve for π¦π¦.
πππ₯π₯ + πππ¦π¦ = ππ
16. Solve for ππ1.
π΄π΄ = 12β(ππ1 + ππ2)
17. The science teacher wrote three equations on a board that relate velocity, π£π£, distance traveled, ππ, and
the time to travel the distance, ππ, on the board.
π£π£ = ππππ ππ = ππ
π£π£ ππ = π£π£ππ
Would you need to memorize all three equations? Explain your reasoning.
Lesson 19: Rearranging Formulas
S.152
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org
This file derived from ALG I-M1-TE-1.3.0-07.2015
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.