Warm-Up Exercises

Solve the system by substitution.

2xy = –1.

3x – y– = 1

4x + y =2.

72x + 3y =–

ANSWER ( )1, 2–

ANSWER ( )1, 3

Warm-Up Exercises

Three T-shirts plus 4 sweatshirts cost $96. A T-shirt costs $3 less than a sweatshirt. How much does a T-shirt cost?

3.

ANSWER $12

Example 1 Multiply One Equation

Solve the linear system using the linear combination method.

Equation 163y2x =–

Equation 285y4x =–

SOLUTION

STEP 1 Multiply the first equation by 2 so that the coefficients of x differ only in sign.

–

63y2x =–

85y4x =– 85y4x =–

126y4x =+– –

4y = –

Example 1 Multiply One Equation

STEP 2 Add the revised equations and solve for y.

STEP 3 Substitute 4 for y in one of the original equations and solve for x.

–

Write Equation 1.63y2x =–

Substitute 4 for y.62x =– ( )4–3 –

Simplify.6122x =+

Subtract 12 from each side.62x = –

Solve for x.3x = –

4y = –

Example 1 Multiply One Equation

STEP 4 Check by substituting 3 for x and 4 for y in the original equations.

– –

ANSWER The solution is .( )3,– –4

Example 2 Multiply Both Equations

Solve the system using the linear combination method.

Equation 1

Equation 2

SOLUTION

STEP 1

Multiply the first equation by 2 and the second equation by 3.

2212y7x =– –

148y5x =+–

2212y7x =– –

148y5x =+–

4424y14x =– –

4224y15x =+–

2x = ––

Example 2 Multiply Both Equations

STEP 2 Add the revised equations and solve for x.

2x =

2x = ––

STEP 3 Substitute 2 for x in one of the original equations and solve for y.

148y5x =+– Write Equation 2.

148y =+ Substitute 2 for x.– ( )25

148y10 =+– Multiply.

3y = Solve for y.

Example 2 Multiply Both Equations

STEP 4 Check by substituting 2 for x and 3 for y in the original equations.

ANSWER The solution is (2, 3).

Example 3 A Linear System with No Solution

Solve the system using the linear combination method.

Equation 1

Equation 274y2x =–

128y4x =+– –

SOLUTION

Multiply the second equation by 2 so that the coefficients of y differ only in sign.

8y

–

124x =+

148y4x =

– –

74y2x =–

128y4x =+– –

20 =Add the revised equations.

Example 3 A Linear System with No Solution

ANSWER

Because the statement 0 2 is false, there is no solution.=

Checkpoint

ANSWER infinitely many solutions

1.

Solve the system using the linear combination method.

Solve a Linear System

54yx =–

1y2x =+ANSWER (1, 1)–

2. 4y2x =–

82y4x =–

3. 22y3x =–

13y4x =–ANSWER (4, 5 )

Checkpoint

ANSWER

if you get a false equation; if you get a true equation

4. How can you tell when a system has no solution? infinitely many solutions?

Solve a Linear System

Use a Linear System as a ModelExample 4

Catering A customer hires a caterer to prepare food for a party of 30 people. The customer has $80 to spend on food and would like there to be a choice of sandwiches and pasta. A $40 pan of pasta contains 10 servings, and a $10 sandwich tray contains 5 servings. The caterer must prepare enough food so that each person receives one serving of either food. How many pans of pasta and how many sandwich trays should the caterer prepare?

Use a Linear System as a ModelExample 4

SOLUTION

VERBALMODEL

•Servingsper pan

Pans of pasta

Sandwich trays =+

Servings per

sandwich tray

Servings needed•

•Price

per panPans of pasta

Sandwich trays

=+Price

per tray

Money to spend

on food•

Use a Linear System as a ModelExample 4

LABELS Servings per pan of pasta 10 = (servings)

Pans of pasta p = (pans)

Servings per sandwich tray 5 =

(trays) Sandwich trays s=

Servings needed 30 =

Price per pan of pasta 40 = (dollars)

Price per sandwich tray 10 = (dollars)

Money to spend on food 80 = (dollars)

(servings)

(servings)

Use a Linear System as a ModelExample 4

ALGEBRAICMODEL

Equation 1 (servings needed) 30=10p + 5s

Equation 2 (money to spend on food)

80=40p + 10s

Multiply Equation 1 by 2 so that the coefficients of s differ only in sign.

–

30=10p + 5s

80=40p + 10s

=20p 10s– – 60–

80=40p + 10s

20=20pAdd the revised equations and solve for p.

1=p

Use a Linear System as a ModelExample 4

ANSWER

The caterer should make 1 pan of pasta and 4 sandwich trays.

Substitute 1 for p in one of the original equations and solve for s.

Write Equation 1. 30=10p + 5s

Substitute 1 for p. 30=10 + 5s( )1

Subtract 10 from each side. 20=5s

4=s Solve for s.

Checkpoint Solve a Linear System

5. Another customer asks the caterer in Example 4 to plan a party for 40 people. This customer also wants both sandwiches and pasta and has $120 to spend. How many pans of pasta and how many sandwich trays should the caterer prepare?

ANSWER

2 pans of pasta and 4 sandwich trays