(For help, go to Lessons 1-2 and 1-7.)
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
Simplify each expression.
1. 2n – 3n 2. –4 + 3b + 2 + 5b
3. 9(w – 5) 4. –10(b – 12)
5. 3(–x + 4) 6. 5(6 – w)
Evaluate each expression.
7. 28 – a + 4a for a = 5 8. 8 + x – 7x for x = –3
9. (8n + 1)3 for n = –2 10.–(17 + 3y) for y = 6
Solving Multi-Step Equations
2-3
Solutions
1. 2n – 3n = (2 – 3)n = –1n = –n
2. –4 + 3b + 2 + 5b = (3 + 5)b + (–4 + 2) = 8b – 2
3. 9(w – 5) = 9w – 9(5) = 9w – 45
4. –10(b – 12) = –10b – (–10)(12) = –10b + 120
5. 3(–x + 4) = 3(–x) + 3(4) = –3x + 12
6. 5(6 – w) = 5(6) – 5w = 30 – 5w
7. 28 – a + 4a for a = 5: 28 – 5 + 4(5) = 28 – 5 + 20 = 23 + 20 = 43
8. 8 + x – 7x for x = –3: 8 + (–3) – 7(–3) = 8 + (–3) + 21 = 5 + 21 = 26
9. (8n + 1)3 for n = –2: (8(–2) + 1)3 = (–16 + 1)3 = (–15)3 = –45
10. –(17 + 3y) for y = 6: –(17 + 3(6)) = –(17 + 18) = –35
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
Solve 3a + 6 + a = 90
4a + 6 = 90 Combine like terms.
4a + 6 – 6 = 90 – 6 Subtract 6 from each side.
4a = 84 Simplify.
3a + 6 + a = 90Check:
3(21) + 6 + 21 90 Substitute 21 for a.
63 + 6 + 21 90
90 = 90
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
= Divide each side by 4.
a = 21 Simplify.
4a4
844
You need to build a rectangular pen in your back yard for
your dog. One side of the pen will be against the house. Two sides of
the pen have a length of x ft and the width will be 25 ft. What is the
greatest length the pen can be if you have 63 ft of fencing?
Relate: length plus 25 ft plus length equals amount of side of side of fencing
Define: Let x = length of a side adjacent to the house.
Write: x + 25 + x = 63
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
x + 25 + x = 63
The pen can be 19 ft long.
2x + 25 = 63 Combine like terms.
2x + 25 – 25 = 63 – 25 Subtract 25 from each side.
2x = 38 Simplify.
x = 19
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
(continued)
= Divide each side by 2. 2x2
382
Solve 2(x – 3) = 8
2x – 6 = 8 Use the Distributive Property.
2x – 6 + 6 = 8 + 6 Add 6 to each side.
2x = 14 Simplify.
x = 7 Simplify.
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
= Divide each side by 2. 2x2
142
Solve + = 173x2
x5
x = 10 Simplify.
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
Method 1: Finding common denominators
+ = 173x2
x5
x + x = 17 Rewrite the equation.32
15
x = 17 Combine like terms. 1710
( x) = (17) Multiply each each by the reciprocalof , which is .
1710
1017
1017 17
101017
x + x = 17 A common denominator of and is 10.
1510
210
32
15
Solve + = 173x2
x5
15x + 2x = 170 Multiply.
17x = 170 Combine like terms.
x = 10 Simplify.
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
Method 2: Multiplying to clear fractions
+ = 173x2
x5
10( + ) = 10(17) Multiply each side by 10, a commonmultiple of 2 and 5.
3x2
x5
10( ) + 10( ) = 10(17) Use the Distributive Property.3x2
x5
= Divide each side by 17.17x17
17017
Solve 0.6a + 18.65 = 22.85.
100(0.6a + 18.65)
=
100(22.85)
The greatest of decimal places is two places. Multiply each side by 100.
100(0.6a) + 100(18.65)
=
100(22.85)
Use the Distributive Property.
60a + 1865
=
2285
Simplify.
60a + 1865 – 1865 =2285 – 1865Subtract 1865 from each side.60a =420 Simplify.
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
=Divide each side by 60.
60a60
42060
a = 7Simplify.
Solve each equation.
1. 4a + 3 – a = 24 2. –3(x – 5) = 66
3. + = 7 4. 0.05x + 24.65 = 27.5n3
n4
7 –17
12 57
Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3ALGEBRA 1 LESSON 2-3
2-3
(For help, go to Lessons 1-7 and 2-3.)
Simplify.
1. 6x – 2x 2. 2x – 6x 3. 5x – 5x 4.–5x + 5x
Solve each equation.
5. 4x + 3 = –5 6. –x + 7 = 12
7. 2t – 8t + 1 = 43 8. 0 = –7n + 4 – 5n
Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4ALGEBRA 1 LESSON 2-4
2-4
Solutions
1. 6x – 2x = (6 – 2)x = 4x 2. 2x – 6x = (2 – 6)x = –4x
3. 5x – 5x = (5 – 5)x = 0x = 0 4. –5x + 5x = (–5 + 5)x = 0x = 0
5. 4x + 3 = –5 6. –x + 7 = 124x = –8 –x = 5x = –2 x = –5
7. 2t – 8t + 1 = 43 8. 0 = –7n + 4 – 5n–6t + 1 = 43 0 = –12n + 4
–6t = 42 12n = 4 t = –7 n = 1
3
Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4ALGEBRA 1 LESSON 2-4
2-4
The measure of an angle is (5x – 3)°. Its vertical angle has a
measure of (2x + 12)°. Find the value of x.
5x – 3 = 2x + 12Vertical angles are congruent.5x – 3 – 2x = 2x + 12 – 2xSubtract 2x from each side.3x – 3 = 12Combine like terms.3x – 3 + 3 = 12 + 3Add 3 to each side.
3x = 15Simplify.
Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4ALGEBRA 1 LESSON 2-4
2-4
= Divide each side by 3.
3x3
153
x = 5 Simplify.
You can buy a skateboard for $60 from a friend and rent the
safety equipment for $1.50 per hour. Or you can rent all items you
need for $5.50 per hour. How many hours must you use a
skateboard to justify buying your friend’s skateboard?
Relate: cost of plus equipment equals skateboard and equipment friend’s rental rental skateboard
Define: let h = the number of hours you must skateboard
Write: 60 + 1.5 h = 5.5 h
Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4ALGEBRA 1 LESSON 2-4
2-4
60 + 1.5h = 5.5h
60 + 1.5h – 1.5h = 5.5h – 1.5hSubtract 1.5h from each side.60 = 4hCombine like terms.
You must use your skateboard for more than 15 hours to justify buying the skateboard.
Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4ALGEBRA 1 LESSON 2-4
2-4
(continued)
604
4h4 = Divide
each side by 4.15 = h Simplify.
Solve each equation.
a. –6z + 8 = z + 10 – 7z
–6z + 8 = z + 10 – 7z
–6z + 8 = –6z + 10Combine like terms.–6z + 8 + 6z = –6z + 10 + 6zAdd 6z to each side.8 = 10Not true for any value of z!This equation has no solution
Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4ALGEBRA 1 LESSON 2-4
b. 4 – 4y = –2(2y – 2)
The equation is true for every value of y, so the equation is an identity.
4 – 4y = –2(2y – 2)
4 – 4y = –4y + 4Use the Distributive Property.4 – 4y + 4y = –4y + 4 + 4yAdd 4y to each side.4 = 4 Always true!
2-4
Solve each equation.
1. 3 – 2t = 7t + 4 2. 4n = 2(n + 1) + 3(n –
1)
3. 3(1 – 2x) = 4 – 6x
4. You work for a delivery service. With Plan A, you can earn $5 per hour plus $.75 per delivery. With Plan B, you can earn $7 per hour plus $.25 per delivery. How many deliveries must you make per hour with Plan A to earn as much as with Plan B?
4 deliveries
1
no solution
Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4ALGEBRA 1 LESSON 2-4
2-4
– 19