chapter 4: solving literal equations
TRANSCRIPT
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Chapter 4:
Solving Literal
Equations
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Day 1: The Basics of Literal Equations
A-REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Warm-Up
What is a literal equation?
A literal equation is an equation with two or more variables. Instead of solving for a numerical value, we solve for one variable in terms of another. A formula is one type of literal equation that has special applications in
math or science.
Observe the similarities between the linear equation (left) and the literal equation (right):
One-Step Linear Equation: One-Step Literal Equation: 1) 5510 y
2) 55 xy solve for y
3) 8540 s solve for s
4) 85 xs solve for s
Two-Step Linear Equation: Two-Step Literal Equation:
5) 87132 a solve for a
6) cba 2 solve for a
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Solving for a variable using division:
7) 453 x 8) yx 3 solve for x
Quick Check for Understanding
9) 92 yx solve for y
10) dcb 3 solve for b
11) P=mv solve for m
Application
12) The formula rtd relates the distance an object travels, d, to its average rate of speed r, and amount of
time t that it travels.
a) Solve the formula rtd for t.
b) How many hours would it take for a car to travel 150 miles at an average rate of 50 miles per hour?
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Independent Practice Solve for the variable indicated.
1) π = ππ‘ ππππ£π πππ π 2) π = π + π ππππ£π πππ π
3) π¦ = ππ₯ + π ππππ£π πππ π₯ 4) π = π β π ππππ£π πππ π
5) π΄π₯ + π΅ = πΆ ππππ£π πππ π₯ 6) π΄π₯ + π΅π¦ = πΆ ππππ£π πππ π¦
7) πΌ = πππ‘ ππππ£π πππ π 8) πΆ = ππ ππππ£π πππ π
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9) π = ππ ππππ£π πππ π 10) πΈ = πΌπ ππππ£π πππ π
11) πΈ = ππ2 ππππ£π πππ π2 12) π = 2π + 2π€ ππππ£π πππ π¦
13) 5π + π = 10 ππππ£π πππ π 14) 5 β π = 2π‘ ππππ£π πππ π‘
15) ππ = ππ π ππππ£π πππ π 16) π¦ = 3π₯ β 1 ππππ£π πππ π₯
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17) The volume of a prism is π = ππ€β.
a) Solve this formula for β.
b) If the volume of a prism is 64, its length 4, and its width 2, what is its height?
Summary
Homework
Chapter 4- Day 1 -Textbook pp. 109-110 #2, 5, 8, 9, 10, 11, 14, 20, 23, 26, 36-37
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Day 2: Solving Literal Equations with Proportions
A-REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Warm-Up
Model Problems Solving Proportions
Solve for x in each equation.
Linear Equations: Literal Equation:
1) π₯
3= 9 2)
π₯
3= π¦
3) 2π₯ β 1 =3
2 4)
π₯
2= π β 6
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5) 10 = 2
3(x β 4) 6) D=
11
5(π₯ β 15)
Application
7) The formula to convert Celsius to Fahrenheit is given by C = 5
9(πΉ β 32).
a) Solve this formula for F.
b) The boiling point of water is 100β. What is the Fahrenheit equivalent of this temperature?
Reminder: Donβt distribute a
coefficient unless absolutely necessary!
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8) Check for Understanding Solve for the given variable.
d) The formula for the mean (average) π΄ of two numbers y and z is one-half their sum, or π΄ =1
2(π¦ + π§).
If the average of two numbers is 7 and one of the numbers is 4, find the other number.
Cumulative Independent Practice Days 1-2 Solve for the value of the variable.
1) π
π= π₯ πππ π 2) π =
1
3π΄β πππ π΄
a)
π =π
π solve for n
b)
π΄ =π+π
2 π πππ£π πππ π
c)
πΉ =πΊπ1π2
π2 solve for π1
10
3) π =1
2ππ‘2 πππ π 4) π =
π€β10π
π πππ π€
5) π + π =π
5 πππ π 6) π + π =
π
5 πππ π
7) π₯
7β π¦ = π‘ πππ π₯ 8)
π₯βπ¦
7= π‘ πππ π₯
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9) π₯βπ¦
7= π‘ πππ π¦ 10) π = π β πΆ πππ πΆ
11) π =πΆβπ
π‘ πππ πΆ 12) 2π₯ + 7π¦ = 14 πππ π¦
13) π =π¦2βπ¦1
π₯2βπ₯1 πππ π¦2 14) π =
2
3(π₯ + 2π¦) for x
15) The formula π =1
3ππ2β is the formula for the volume of a cylinder. To the nearest tenth, what is the height
of a cylinder with volume 100 cm3 and radius 2 cm?
HW Chapter 4-Day 2 Textbook pp. 109-110 #6, 7, 12, 13, 15, 18, 24, 30, 45
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Day 3: Using the Distributive Property and Rational Equations
A-REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Warm-Up
The formula π = 2(πΏ + π) is the formula for the perimeter of a rectangle. Solve this formula for L. What is the length of a rectangle whose perimeter is 48 and whose width is 6?
Distribution and Reverse distribution
1) When there is a common factor in all terms of an expression, we can use the distributive property in reverse
to write it in factored form.
Simplest form Factored Form
a) 2πΏ + 2π 2(πΏ + π)
b) 3π β 3π
c) 2ππ€ + 2π
d) ππ + ππ
e) 2ππβ + 2ππ2
2) Model Problem Using the Distributive Property in Reverse
Solve for c in terms of a and b: ππ + ππ = ππ
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3) Practice
a) b)
Using Rational Equations
4) Model Problem
The formula 1
π+
1
π=
1
π relates an objectβs distance, π, and its imageβs distance, π, to the focal length of the
lens, π. Solve this formula for π.
5) Practice
The total resistance in a circuit is given by the formula 1
π =
1
π 1+
1
π 2. Solve this formula for π 1.
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Unit Summary So Far:
Look for STRUCTURE in equations:
One- or Two-Step Equations
π΄π₯ + π΅ = πΆ π₯ + π = π
Proportions
π· =π
π πΎ =
1
2ππ£ 2 οΏ½Μ οΏ½ =
π₯1+π₯2
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Reverse Distribution (Common Factor)
S = 2ππ2 + 2ππβ
Rational Equations (Sums and Differences)
1
πΆ=
1
πΆ1+
1
πΆ2
Cumulative Practice/Homework Chapter 4 β Day 3 Solve for the requested variable.
1) π = ππ for n 2) π =1
3π΅β πππ π΅
3) π = 2π₯+π‘
π πππ π₯ 4) π£ = π£0 + ππ‘ for π£0
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5) π½ = ππ£π β ππ£π for m 6) πΈ = πΌπ πππ πΌ
7) π¦ β π¦1 = π(π₯ β π₯1)πππ π₯ 8) π = ππβ πππ π
9) π =ππ
π΄ πππ π΄ 10) πΉ + π β πΈ = 2 πππ π
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11) π =1
2ππ πππ π 12) π§ + π¦ = π₯ + π₯π¦2 πππ π₯
13) πΊ = π» β ππ πππ π» 14) πΉπΆ =
ππ£2
π for m
15) π = π0 + ππβ for h 16) π =ππ»βππΆ
ππ» πππ ππ»
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Multiple Choice Practice.
17)
18)
Real-World Application.
19)
If Patty paid 0.93 to mail her letter, how many
ounces was it? (C = cost, z = ounces)
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Day 4: Square Roots
A-REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Warm-Up
Mini-Lesson: Using Square Roots
To solve for a squared variable, take its square root.
Linear Equation: 64 = 16π₯2 Literal Equation: π΄ = ππ2 solve for r
Check for Understanding Solve for the indicated variable.
1) The formula for kinetic energy is πΎ =1
2ππ£ 2.
Write an expression for π£ in terms of K and π.
2) The gravitational force F that two planetary bodies
exert on one another is given by πΉ = πΊπ1π2
π2.
Solve this formula for π.
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Cumulative Practice/Homework Solve for the value of the indicated variable.
1) π = ππ€β Solve for h
2) π =1
2ππ‘2 solve for t
3) )(2
121
bbhA Solve for h
4) ππππ£π πππ π: π+π
3= π + 5
5) Solve π = π+3π€
2 for w 6) Solve ππ₯ + ππ¦ + π = 0 πππ π¦
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7) π΄ = 2ππβ + 2ππ2 Solve for Ο 8) Rewrite πΎ = 3
2ππ solved for T in terms of k and T.
9) π β 3π = 2 ππππ£π πππ π 10) In π + ππ₯ = π, what is a in terms of x and b?
11) 5βπ
6= π β 7 Solve for c 12) ππ =
π£2
π Solve for π£
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Regents Practice.
13) 14)
15) 16)
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Day 5: Review
A-REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Look for STRUCTURE in equations:
One-Step Equations
1) πΌ = πππ‘ Solve for π.
2) π = π β π Solve for M.
Two-Step Equations
3) 5π‘ β 2π = 25 Solve for t.
4) π£π‘ β 16π‘2 Solve for v.
Proportions
5) πΉ =ππ‘
π Solve for l.
6) π =144π
π¦ Solve for p.
7) π΄ =1
2β(π + π) Solve for a. 8) π =
π¦2βπ¦1
π₯2βπ₯1 Solve for π¦2
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Reverse Distribution
9) π = π β ππ Solve for R. 10) ππ₯ = ππ₯ + π Solve for x.
Rational Equations
11) 1
πΆ=
1
πΆ1+
1
πΆ2 Solve for πΆ1 12)
π₯
3+
π₯
4= π Solve for x.
Square Roots
13) πΎ =1
2ππ£2 Solve for v.
14) π =1
3ππ2β Solve for r.
Applications. The surface area of a sphere is given by the formula π = 4ππ2. Solve this formula for r. What
is the radius of a sphere whose surface area is 201 ππ2?