Download 6.7 Literal Equations

Section 6.7 / Literal Equations
6.7
Objective A
327
Literal Equations
To solve a literal equation for one of the variables
A literal equation is an equation that contains more than
one variable. Examples of literal equations are shown at
the right.
Formulas are used to express a relationship among physical quantities. A
formula is a literal equation that states
a rule about measurements. Examples
of formulas are shown at the right.
2x 3y 6
4w 2x z 0
1
1
1
R1
R2
R
s a n 1d
A P Prt
(Physics)
(Mathematics)
(Business)
The Addition and Multiplication Properties can be used to solve a literal equation for one of the variables. The goal is to rewrite the equation so that the variable being solved for is alone on one side of the equation and all the other
numbers and variables are on the other side.
HOW TO
Video
Solve A P1 i for i.
The goal is to rewrite the equation so that i is on one side of the equation
and all other variables are on the other side.
A P1 i
A P Pi
A P P P Pi
A P Pi
AP
Pi
P
P
• Use the Distributive Property to remove parentheses.
• Subtract P from each side of the equation.
• Divide each side of the equation by P.
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AP
i
P
Example 1
You Try It 1
Solve 3x 4y 12 for y.
Solve 5x 2y 10 for y.
Solution
Your solution
3x 4y 12
3x 3x 4y 3x 12
4y 3x 12
4y
3x 12
4
4
3
y x3
4
• Subtract 3x.
• Divide by 4.
Solution on p. S17
328
Chapter 6 / Rational Expressions
Example 2
Solve I You Try It 2
E
for R.
Rr
Solve s Solution
AL
for L.
2
Your solution
E
I
Rr
R rI R r
E
Rr
RI rI E
RI rI rI E rI
RI E rI
RI
E rI
I
I
E rI
R
I
• Multiply by (R r).
• Subtract rI.
• Divide by I.
Example 3
You Try It 3
Solve L a1 ct for c.
Solve S a n 1d for n.
Solution
Your solution
L a1 ct
L a act
L a a a act
L a act
La
act
at
at
La
c
at
• Distributive Property
• Subtract a.
Example 4
You Try It 4
Solve S C rC for C.
Solve S rS C for S.
Solution
Your solution
S C rC
S 1 rC
S
1 rC
1r
1r
S
C
1r
• Factor.
• Divide by (1 r).
Solutions on p. S18
Copyright © Houghton Mifflin Company. All rights reserved.
• Divide by at.
Section 6.7 / Literal Equations
329
6.7 Exercises
Objective A
To solve a literal equation for one of the variables
For Exercises 1 to 15, solve for y.
1. 3x y 10
2. 2x y 5
3. 4x y 3
4. 5x y 7
5. 3x 2y 6
6. 2x 3y 9
7. 2x 5y 10
8. 5x 2y 4
9. 2x 7y 14
10. 6x 5y 10
11.
x 3y 6
12.
x 2y 8
14.
y 4 2x 3
15.
2
y 1 x 6
3
16. x 3y 6
17.
x 6y 10
18.
3x y 3
19. 2x y 6
20.
2x 5y 10
21.
4x 3y 12
22. x 2y 1 0
23.
x 4y 3 0
24.
5x 4y 20 0
13. y 2 3x 2
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For Exercises 16 to 24, solve for x.
For Exercises 25 to 40, solve the formula for the given variable.
25.
d rt; t
(Physics)
26.
E IR; R
(Physics)
27.
PV nRT; T
(Chemistry)
28.
A bh; h
(Geometry)
330
Chapter 6 / Rational Expressions
(Geometry)
30.
F
9
C 32; C
5
1
hb1 b2; b1
2
(Geometry)
32.
C
5
F 32; F
9
V
1
Ah; h
3
(Geometry)
34.
P R C; C
(Business)
35.
R
CS
;S
t
(Business)
36.
P
RC
;R
n
(Business)
37.
A P Prt; P
(Business)
38.
T fm gm; m
(Engineering)
39.
A Sw w; w
(Physics)
40.
a S Sr; S
(Mathematics)
29.
P 2l 2w; l
31.
A
33.
(Temperature
conversion)
(Temperature
conversion)
APPLYING THE CONCEPTS
Business Break-even analysis is a method used to determine the sales volume
required for a company to break even, or experience neither a profit nor a loss
on the sale of a product. The break-even point represents the number of units
that must be made and sold for income from sales to equal the cost of the prodF
,
SV
where
F is the fixed costs, S is the selling price per unit, and V is the variable costs per
unit. Use this information for Exercise 41.
41. a. Solve the formula B F
SV
for S.
b. Use your answer to part a to find the selling price per unit required for
a company to break even. The fixed costs are $20,000, the variable
costs per unit are $80, and the company plans to make and sell
200 desks.
c. Use your answer to part a to find the selling price per unit required for
a company to break even. The fixed costs are $15,000, the variable
costs per unit are $50, and the company plans to make and sell
600 cameras.
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uct. The break-even point can be calculated using the formula B