Pre-Algebra Solve 5v – 12 = 8. Solving Two-Step Equations Lesson 7-1 5v – 12 = 8 5v – 12 + 12 = 8 + 12Add 12 to each side. 5v = 20Simplify. v = 4Simplify

Pre-AlgebraPre-Algebra

Solve 5v – 12 = 8.

Solving Two-Step EquationsSolving Two-Step Equations

Lesson 7-1

5v – 12 = 8

5v – 12 + 12 = 8 + 12 Add 12 to each side.

5v = 20 Simplify.

v = 4 Simplify.

= Divide each side by 5.5v5

205

Check: 5v – 12 = 8

5(4) – 12 8 Replace v with 4.

20 – 12 8 Multiply.

8 = 8 Simplify.

Additional Examples


Solve 7 – 3b = 1.


Lesson 7-1

7 – 3b = 1

–7 + 7 – 3b = –7 + 1 Add –7 to each side.

0 – 3b = –6 Simplify.

–3b = –6 0 – 3b = –3b.

b = 2 Simplify.

= Divide each side by –3.–3b–3

–6–3

Additional Examples


You borrow $350 to buy a bicycle. You agree to

pay $100 the first week, and then $25 each week until the

balance is paid off. To find how many weeks w it will take

you to pay for the bicycle, solve 100 + 25w = 350.


Lesson 7-1

It will take you 10 weeks to pay for the bicycle.

100 + 25w = 350

100 + 25w – 100 = 350 – 100 Subtract 100 from each side.

25w = 250 Simplify.

w = 10 Simplify.

= Divide each side by 25.25w25

25025

Additional Examples


Solving Multi-Step EquationsSolving Multi-Step Equations

In his stamp collection, Jorge has five more than

three times as many stamps as Helen. Together they have

41 stamps. Solve the equation s + 3s + 5 = 41. Find the

number of stamps each one has.

Lesson 7-2

s + 3s + 5 = 41

4s + 5 = 41 Combine like terms.

4s + 5 – 5 = 41 – 5 Subtract 5 from each side.

4s = 36 Simplify.

s = 9 Simplify.

= Divide each side by 4.4s4

364

Additional Examples



(continued)

Lesson 7-2

Check: Is the solution reasonable? Helen and Jorge have a total of 41 stamps. Since 9 + 32 = 41, the solution is reasonable.

Helen has 9 stamps. Jorge has 3(9) + 5 = 32 stamps.

Additional Examples



The sum of three consecutive integers is 42. Find

the integers.

Lesson 7-2

sum of three consecutive integers 42isWords

Let = the least integer.n

Then = the second integer,n + 1

and = the third integer.n + 2

+ +n n + 1 n + 2Equation 42=

Additional Examples



(continued)

Lesson 7-2

n + (n + 1) + (n + 2) = 42

(n + n + n) + (1 + 2) = 42 Use the Commutative and Associative Properties of Addition to group like terms together.

3n + 3 = 42 Combine like terms.

3n + 3 – 3 = 42 – 3 Subtract 3 from each side.

3n = 39 Simplify.

n = 13 Simplify.

= Divide each side by 3.3n3

393

Additional Examples



(continued)

Lesson 7-2

If n = 13, then n + 1 = 14, and n + 2 = 15. The three integers are 13, 14, and 15.

Check: Is the solution reasonable? Yes, because 13 + 14 + 15 = 42.

Additional Examples



Solve each equation.

Lesson 7-2

a. 4(2q – 7) = –4

4(2q – 7) = –4

8q – 28 = –4 Use the Distributive Property.

8q – 28 + 28 = –4 + 28 Add 28 to each side.

8q = 24 Simplify.

q = 3 Simplify.

Divide each side by 8.=8q8

248

Additional Examples



(continued)

Lesson 7-2

b. 44 = –5(r – 4) – r

44 = –5(r – 4) – r

44 = –5r + 20 – r Use the Distributive Property.

44 = –6r + 20 Combine like terms.

44 – 20 = –6r + 20 – 20 Subtract 20 from each side.

24 = –6r Simplify.

–4 = r Simplify.

Divide each side by –6.=24–6

–6r–6

Additional Examples


Multi-Step Equations With Fractions and DecimalsMulti-Step Equations With Fractions and Decimals

Solve p – 7 = 11.

Lesson 7-3

p – 7 = 1134

Add 7 to each side.p – 7 + 7 = 11 + 734

Simplify.p = 1834

p =34

43 •

43 • 18 Multiply each side by , the reciprocal of .

34

43

1p =4 • 18

3 1

6Divide common factors.

p = 24 Simplify.

34

Additional Examples



(continued)

Lesson 7-3

Check: p – 7 = 11

(24) – 7 11 Replace p with 24.

Simplify.18 – 7 11

11 = 11

34

34

– 7 11 Divide common factors.3 • 24

4 1

6

Additional Examples



Solve y + 3 = .

Lesson 7-3

y + 3 = 12

23

Multiply each side by 6, the LCM of 2 and 3.6 y + 3 = 12 6

23

3y + 18 – 18 = 4 – 18 Subtract 18 from each side.

3y = –14 Simplify.

3y + 18 = 4 Simplify.

Use the Distributive Property.6 • y + 6 • 3 =12 6

23

Divide each side by 3.3y3

–143=

Simplify.y = –423

23

12

Additional Examples



Lesson 7-3

Suppose your cell phone plan is $30 per month plus

$.05 per minute. Your bill is $36.75. Use the equation 30 +

0.05x = 36.75 to find the number of minutes on your bill.

There are 135 minutes on your bill.

30 + 0.05x = 36.75

x = 135 Simplify.

30 – 30 + 0.05x = 36.75 – 30 Subtract 30 from each side.

0.05x = 6.75 Simplify.

Divide each side by 0.05.=6.750.05

0.05x0.05

Additional Examples


Problem Solving Strategy: Write an EquationProblem Solving Strategy: Write an Equation

A moving van rents for $29.95 a day plus $.12 a

mile. Mr. Reynolds’s bill was $137.80 and he drove the van

150 mi. For how many days did he have the van?

Lesson 7-4

Words number of days • $29.95/d + $.12/mi • 150 mi $137.80=

Let d = number of days Mr. Reynolds had the van.

Equation 137.80d • 29.95 + 0.12 • 150 =

Additional Examples


Problem Solving Strategy: Write an EquationProblem Solving Strategy: Write an Equation

(continued)

Lesson 7-4

Mr. Reynolds had the van for 4 days.

d • 29.95 + 0.12 • 150 = 137.80

29.95d + 18 = 137.80 Multiply 0.12 and 150.

29.95d + 18 – 18 = 137.80 – 18 Subtract 18 from each side.

29.95d = 119.80 Simplify.

d = 4 Simplify.

Divide each side by 29.95.=29.95d29.95

119.8029.95

Additional Examples


Solving Equations With Variables on Both SidesSolving Equations With Variables on Both Sides

Solve 4c + 3 = 15 – 2c.

Lesson 7-5

4c + 3 = 15 – 2c

4c + 2c + 3 = 15 – 2c + 2c Add 2c to each side.

6c + 3 = 15 Combine like terms.

6c + 3 – 3 = 15 – 3 Subtract 3 from each side.

6c = 12 Simplify.

c = 2 Simplify.

Divide each side by 6.=6c6

126

Check: 4c + 3 = 15 – 2c4(2) + 3 15 – 2(2) 8 + 3 15 – 4 11 = 11

Substitute 2 for c.Multiply.

Additional Examples



Lesson 7-5

Steve types at a rate of 15 words/min and Jenny types at a

rate of 20 words/min. Steve and Jenny are both typing the same

document, and Steve starts 5 min before Jenny. How long will it take

Jenny to catch up with Steve?

words Jenny types = words Steve types

Words 20 words/min • Jenny’s time = 15 words/min • Steve’s time

Let = Jenny’s time.x

Then x + 5 = Steve’s time.

Equation 20 • x = 15 • (x + 5)

Additional Examples



Lesson 7-5

(continued)

20x = 15(x + 5)

Jenny will catch up with Steve in 15 min.

20x = 15x + 75 Use the Distributive Property.

20x – 15x = 15x – 15x + 75 Subtract 15x from each side.

5x = 75 Combine like terms.

x = 15 Simplify.

Divide each side by 5.=5x5

755

Additional Examples



Lesson 7-5

(continued)

Check: Test the result.At 20 words/min for 15 min, Jenny types 300 words.Steve’s time is five min longer. He types for 20 min.At 15 words/min for 20 min, Steve types 300 words.Since Jenny and Steve each type 300 words, the answer checks.

Additional Examples


Solving Two-Step InequalitiesSolving Two-Step Inequalities

Solve and graph 7g + 11 > 67.

Lesson 7-6

7g + 11 > 67

7g + 11 – 11 > 67 – 11 Subtract 11 from each side.

7g > 56 Simplify.

g > 8 Simplify.

Divide each side by 7.>7g7

567

Additional Examples



Solve 6 – r – 6.

Lesson 7-6

< 23

6 – r – 6<23

6 + 6 – r – 6 + 6< 23

Add 6 to each side.

Simplify.12 – r< 23

Simplify.>–18 r, or r –18<

32– (12)

32–

23– r Multiply each side by . Reverse the direction of

the inequality symbol.

32

–>

Additional Examples



Dale has $25 to spend at a carnival. If the admission to

the carnival is $4 and the rides cost $1.50 each, what is the

greatest number of rides Dale can go on?

Lesson 7-6

25Inequality 4 + •1.5 r <

Let = number of rides Dale goes on.

Words $4 admission + $1.50/ridenumberof rides

is less thanor equal to

$25•

r

Additional Examples



(continued)

Lesson 7-6

The greatest number of rides Dale can go on is 14.

4 + 1.5r 25<

Subtract 4 from each side.4 + 1.5r – 4 25 – 4<

Simplify.1.5r 21<

Divide each side by 1.5.1.5r1.5

211.5<

Simplify.r 14<

Additional Examples


Solve the circumference formula C = 2 r for r.

Transforming FormulasTransforming Formulas

Lesson 7-7

C = 2 r

Use the Division Property of Equality.C

2 2 r2 =

Simplify.C

2 = r, or r =C

2

Additional Examples



Solve the perimeter formula P = 2 + 2w for w.

Lesson 7-7

P = 2 + 2w

Simplify.P – 2 = 2w

Subtract 2 from each side.P – 2 = 2 + 2w – 2

Multiply each side by .12

12 (P – 2 ) = (2w)

12

12 P – = w Use the Distributive Property and simplify.

Additional Examples



You plan a 600-mi trip to New York City. You estimate

your trip will take about 10 hours. To estimate your average

speed, solve the distance formula d = rt for r. Then substitute to

find the average speed.

Lesson 7-7

d = rt

Your average speed will be about 60 mi/h.

= 60 Simplify.

Divide each side by t.=dt

rtt

Simplify.= r, or r =dt

dt

Replace d with 600 and t with 10. r =60010

Additional Examples



The high temperature one day in San Diego was 32°C.

Solve C = (F – 32) for F. Then substitute to find the

temperature in degrees Fahrenheit.

Lesson 7-7

59

C = (F – 32)59

(C) = (F – 32)95

95

59

Multiply each side by .95

Simplify.C = F – 3295

Additional Examples



(continued)

Lesson 7-7

32°C is 89.6°F.

Add 32 to each side.C + 32 = F – 32 + 3295

Simplify and rewrite.C + 32 = F, or F = C + 3295

95

Replace C with 32. Simplify.F = (32) + 32 = 89.695

Additional Examples


Simple and Compound InterestSimple and Compound Interest

Suppose you deposit $1,000 in a savings account

that earns 6% in interest per year.

Lesson 7-8

a. Find the interest earned in two years. Find the total of principal plus interest.

The account will earn $120 in two years. The total of principal plus interest will be $1,120.

I = prt Use the simple interest formula.

I = 1,000 • 0.06 • 2 Replace p with 1,000, r with 0.06, and t with 2.

I = 120 Simplify.

total = 1,000 + 120 = 1,120 Find the total.

Additional Examples



(continued)

Lesson 7-8

b. Find the interest earned in six months. Find the total of principal plus interest.

The account will earn $30 in six months. The total of principal plus interest will be $1,030.

I = prt Use the simple-interest formula.

I = 1,000 • 0.06 • 0.5 Replace p with 1,000, r with 0.06, and t with 0.5.

I = 30 Simplify.

Total = 1,000 + 30 = 1,030 Find the total.

Write the months as part of a year.t = = = 0.512

612

Additional Examples


Year 5 : $486.20 486.20 • 0.05 = 24.31 486.20 + 24.31 = 510.51

Year 6 : $510.51 510.51 • 0.05 25.53 510.51 + 25.53 = 536.04

Year 7 : $536.04 536.04 • 0.05 26.80 536.04 + 26.80 = 562.84

Year 8 : $562.84 562.84 • 0.05 28.14 562.84 + 28.14 = 590.98


You deposit $400 in an account that earns 5% interest compounded annually (once per year). The balance after the first four years is $486.20. What is the balance in your account after another 4 years, a total of 8 years? Round to the nearest cent.

Lesson 7-8

After the next four years, for a total of 8 years, the balance is $590.98.

Interest BalancePrincipal at

Beginning of Year

Additional Examples



Find the balance on a deposit of $2,500 that earns 3%

interest compounded semiannually for 4 years.

Lesson 7-8

The interest rate r for compounding semiannually is 0.03 ÷ 2, or 0.015.

The number of payment periods n is 4 years 2 interest periods per year, or 8.

The balance is $2,816.23.

B = p(1 + r)n Use the compound interest formula.

B = 2,500(1 + 0.015)8 Replace p with 2,500, r with 0.015, and n with 8.

Use a calculator. Round to the nearest cent.B 2,816.23

Additional Examples

Documents

Pre-Algebra Solve 5v – 12 = 8. Solving Two-Step Equations Lesson 7-1 5v – 12 = 8 5v – 12 + 12 = 8 + 12Add 12 to each side. 5v = 20Simplify. v = 4Simplify