GUIDED PRACTICE
for Example 1
1. Find the radius of a circle with a circumference of
25 feet.
SOLUTION
STEP 1 Solve the formula for r.
C=2πr
Write circumference formula.
C
Divide each side by 2π.
2π = r
STEP 2 Substitute the given value into the rewritten formula.
C 25
3.9 Substitute 25 for C and simplify.
r = 2π = 2π
ANSWER The radius of the circle is about 4 feet.
GUIDED PRACTICE
2.
for Example 1
The formula for the distance d between opposite
vertices of a regular hexagon is d = 2a where a is
3
the distance between opposite sides. Solve the
formula for a. Then find a when d = 10 centimeters.
SOLUTION
d 3
a= 2
When d = 10cm, a = 5 3
or
8.7cm
GUIDED PRACTICE
for Example 2
3. Solve the formula P = 2l + 2w for l. Then find the length
of a rectangle with a width of 7 inches and a perimeter
of 30 inches.
STEP 1 Solve the formula for p = 2l + 2w for l
P = 2l + 2w Write perimeter formula.
P – 2w = 2l
P – 2w = l
2
Subtract 2w from each side.
Divide each side by 2
GUIDED PRACTICE
for Example 2
STEP 2
Substitute the given values in.
Formula for
l = P – 2w
2
30 – 2 (7)
Multiply.
=
2
= 16
2
Subtract.
=8
Divide.
ANSWER
Length of rectangle is 8 in.
GUIDED PRACTICE
for Example 2
4. Solve the formula A = lw for w. Then find the width of a
rectangle with a length of 16 meters and an area of 40
square meters.
STEP 1 Solve the formula for w
A = lw
Write perimeter formula.
A =w
l
Divide each side by l
GUIDED PRACTICE
for Example 2
STEP 2
Substitute the given values into rewrite formula.
w = A Write perimeter formula.
l
w = 40 Subtract 40 from A and 16 for l.
16
w = 2.5 Divide
Write of rectangle is 2.5 m
GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the
given information to find the value of the variable.
1 bh
A
=
5.
2
Find h if b = 12 m
and A = 84 m2.
A=
1 bh
2
Write perimeter formula.
2A = bh
Multiply each side by 2.
2A = h
b
Divide each side by b
GUIDED PRACTICE
for Example 2
Find the value of h if b = 12m and A = 84m2.
2A = h
b
2A = h
b
2(84) = h
12
h = 14
h = 14m
Find h if b = 12 m
and A = 84 m2.
Write formula.
Substitute 84 for A and 12 for b.
Simplify
GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the
given information to find the value of the variable.
1 bh
A
=
6.
Find b if h = 3 cm
2
and A = 9 cm2.
A=
1 bh
2
Write perimeter formula.
2A = bh
Multiply each side by 2.
2A = b
h
Divide each side by h
GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the
given information to find the value of the variable.
1 bh
A
=
6.
Find b if h = 3 cm
2
and A = 9 cm2.
2A = b
h
2(9) = b
3
b=6
b = 6cm
Write formula.
Substitute 9 for A and 3 for b.
Simplify
GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the
given information to find the value of the variable.
1 (b + b )h
A
=
7.
2
Find h if b1 = 6 in.,
2 1
b2 = 8 in., and A = 70 in.2
A=
1 (b + b )h
2
2 1
2A = (b1 + b2)h
h=
2A
(b1 + b2)
Write perimeter formula.
Multiply each side by 2.
Divide by (b1 + b2)
GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the
given information to find the value of the variable.
1 (b + b )h
A
=
7.
2
Find h if b1 = 6 in.,
2 1
b2 = 8 in., and A = 70 in.2
h=
2A
(b1 + b2)
Write formula.
h=
2(70)
(6 + 8)
Substitute 70 for A and 6 for b1 and 8 for b2.
h = 10
h = 10 in.
Simplify
GUIDED PRACTICE
for Examples 3 and 4
Solve the equation for y. Then find the value of y when
x = 2.
8. y – 6x = 7
STEP 1 Solve the equation for y.
y – 6x = 7
y = 7 + 6x
STEP 2 Substitute the given value into the rewritten
equation.
y = 7 + 6 (2) Substitute 2 for n.
y = 7 + 12
Multiply.
y = 19
Add.
GUIDED PRACTICE
CHECK
y – 6x = 7
19 – 6 (2) =? 7
for Examples 3 and 4
Write original equation.
Substitute 2 for x and 19 for y.
7 = 7 Solution checks.
GUIDED PRACTICE
for Examples 3 and 4
Solve the equation for y. Then find the value of y when
x = 2.
9. 5y – x = 13
STEP 1 Solve the equation for y.
5y – x = 13
5y = 13 + x
Write original equation
Add x to each side
13 + x Divide each side by 5
y= 5
5
GUIDED PRACTICE
for Examples 3 and 4
STEP 2 Substitute the given value into the rewritten
equation.
13
2
y= 5 + 5
y=5
Substitute 2 for n.
Simplify.
CHECK
5y – x = 13
5(5) – 2 =? 7
3 = 3
Write original equation.
Substitute 2 for x and 19 for y.
Solution checks.
GUIDED PRACTICE
for Examples 3 and 4
Solve the equation for y. Then find the value of y when
x = 2.
10. 3x + 2y = 12
STEP 1 Solve the equation for y.
3x + 2y = 12
5y = 12 – 3x
Write original equation
Subtract 3x from each side
12 3x
y = 2 + 2 Divide each side by 2
= – 3x + 6
2
GUIDED PRACTICE
for Examples 3 and 4
STEP 2 Substitute the given value into the rewritten
equation.
(2)
Substitute 2 for n.
y = – 3 2 +6
–2
y = 2 +6
Multiply
y=3
Simplify
CHECK
3x – yx = 12
3(2) + 2(3) =? 12
12 = 12
Write original equation.
Substitute 3 for y and 2 for x.
Solution checks.
GUIDED PRACTICE
for Examples 3 and 4
Solve the equation for y. Then find the value of y when
x = 2.
11. 2x + 5y = –1
STEP 1 Solve the equation for y.
2x + 5y = –1
5y = –1 – 2x
Write original equation
Subtract 2x from each side
–1 2x
y = 5 – 5 Divide each side by 5
–1 2x
= 5 –5
GUIDED PRACTICE
for Examples 3 and 4
STEP 2 Substitute the given value into the rewritten
equation.
– 1 – 2(2)
y= 2
5
–1– 4
y= 5
5
y=–1
Substitute 2 for n.
Multiply
Simplify
CHECK
2x + 5y = –1
2(2) + 5(–1) =? –1
–1 = –1
Write original equation.
Substitute 2 for x and –1 for y.
Solution checks.
GUIDED PRACTICE
for Examples 3 and 4
Solve the equation for y. Then find the value of y when
x = 2.
12. 3 = 2xy – x
STEP 1 Solve the equation for y.
3 = 2xy – x
Write original equation
3 + x = 2xy
Add x to each side
3 +x = y
2x
Divide 2x to each side
GUIDED PRACTICE
for Examples 3 and 4
STEP 2 Substitute the given value into the rewritten
equation.
3+2 =y
2 (2)
5 =1 1
y= 4
4
Substitute 2 for n.
Simplify
CHECK
3= 2xy –x
3 =? 2(2)
3 = 3
Write original equation.
( 45 ) – (2) Substitute
2 for x and 5 1 for y.
Solution checks.
4
GUIDED PRACTICE
for Examples 3 and 4
Solve the equation for y. Then find the value of y when
x = 2.
13. 4y – xy = 28
STEP 1 Solve the equation for y.
4y – xy = 28
Write original equation
(4 – x)y = 28
Distributive property
28 = y
4–x
Divide each side by (4 – x)
GUIDED PRACTICE
for Examples 3 and 4
STEP 2 Substitute the given value into the rewritten
equation.
28
y= 4–2
Substitute 2 for x.
y = 14
Simplify
CHECK
4y –xy = 28
4(14) – (2) (14) =? 28
28 = 28
Write original equation.
Substitute 14 for y and 2 for y.
Solution checks.
EXAMPLE 5
Solve a multi-step problem
Movie Rental
A video store rents new movies for
one price and older movies for a
lower price, as shown at the right.
• Write an equation that represents
the store’s monthly revenue.
• Solve the revenue equation for
the variable representing the
number of new movies rented.
• The owner wants $12,000 in revenue per month. How
many new movies must be rented if the number of older
movies rented is 500? 1000?
EXAMPLE 5
Solve a multi-step problem
SOLUTION
STEP 1
Write a verbal model. Then write an equation.
An equation is R = 5n1 + 3n2.
STEP 2
Solve the equation for n1.
EXAMPLE 5
Solve a multi-step problem
R = 5n1 + 3n2 Write equation.
R – 3n2 = 5n1
Subtract 3n2 from each side.
R – 3n2
= n1
5
Divide each side by 5.
STEP 3 Calculate n1 for the given values of R and n2.
If n2 = 500, then n1 = 12,000 – 3 500 = 2100.
5
If n2 = 1000, then n1 = 12,000 – 3 1000 = 1800.
5
ANSWER
If 500 older movies are rented, then 2100 new movies
must be rented. If 1000 older movies are rented, then
1800 new movies must be rented.
GUIDED PRACTICE
for Example 5
14. What If? In Example 5, how many new movies must
be rented if the number of older movies rented is 1500?
SOLUTION
STEP 1
Write a verbal model. Then write an equation.
An equation is R = 5n1 + 3n2.
GUIDED PRACTICE
for Example 5
STEP 2
Solve the equation for n1.
R = 5n1 + 3n2 Write equation.
R – 3n2 = 5n1
R – 3n2
= n1
5
Subtract 3n2 from each side.
Divide each side by 5.
STEP 3 Calculate n1 for the given values of R and n2.
If n2 = 1500 = 12,000 – 3 1500 = 1500.
5
If 1500 older movies are rented, then 1500
new movies must be rented
GUIDED PRACTICE
for Example 5
15. What If? In Example 5, how many new movies
must be rented if customers rent no older movies at
all?
SOLUTION
STEP 1
Write a verbal model. Then write an equation.
An equation is R = 5n1 + 3n2.
GUIDED PRACTICE
for Example 5
STEP 2
Solve the equation for n1.
R = 5n1 + 3n2 Write equation.
R – 3n2 = 5n1
R – 3n2
= n1
5
Subtract 3n2 from each side.
Divide each side by 5.
STEP 3 Calculate n1 for the given values of R and n2.
If n2 = 0, then n1 = 12,000 – 3 0 = 2400.
5
If 0 older movies are rented, then 2400 new
movie must be rented
GUIDED PRACTICE
for Example 5
16. Solve the equation in Step 1 of Example 5 for n2.
Solve the equation for n1.
R = 5n1 + 3n2 Write equation.
R – 5n1 = 3n2
R – 5n1
= n2
3
Subtract 5n1 from each side.
Divide each side by 3.
Equation for n2 is R – 5n1
3