Warm-Up Exercises
Solve the system by substitution.
1.
y = – 2x
– 3x – y = 1
ANSWER
2.
(– 1, 2 )
x +y = 4
– 2x + 3y = 7
ANSWER
( 1, 3 )
Warm-Up Exercises
3. Three T-shirts plus 4 sweatshirts cost $96. A T-shirt
costs $3 less than a sweatshirt. How much does a
T-shirt cost?
ANSWER
$12
Example 1
Multiply One Equation
Solve the linear system using the linear combination
method.
2x – 3y = 6
Equation 1
4x – 5y = 8
Equation 2
SOLUTION
STEP 1 Multiply the first equation by – 2 so that the
coefficients of x differ only in sign.
2x – 3y = 6
– 4x + 6y = –12
4x – 5y = 8
4x – 5y = 8
y = –4
Example 1
Multiply One Equation
STEP 2 Add the revised equations
and solve for y.
y = –4
STEP 3 Substitute – 4 for y in one of the original
equations and solve for x.
2x – 3y = 6
2x – 3( – 4) = 6
2x + 12 = 6
2x = – 6
x = –3
Write Equation 1.
Substitute – 4 for y.
Simplify.
Subtract 12 from each side.
Solve for x.
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.
7x – 12y = – 22
Equation 1
– 5x + 8y = 14
Equation 2
SOLUTION
STEP 1 Multiply the first equation by 2 and the second
equation by 3.
7x – 12y = – 22
– 5x + 8y = 14
14x – 24y = – 44
–15x + 24y = 42
–x
= –2
Example 2
Multiply Both Equations
STEP 2 Add the revised equations
and solve for x.
–x
= –2
x = 2
STEP 3 Substitute 2 for x in one of the original equations
and solve for y.
– 5x + 8y = 14
Write Equation 2.
– 5( 2) + 8y = 14
Substitute 2 for x.
–10 + 8y = 14
y =3
Multiply.
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.
– 4x + 8y = –12
Equation 1
2x – 4y = 7
Equation 2
SOLUTION
Multiply the second equation by 2 so that the
coefficients of y differ only in sign.
– 4x + 8y = –12
2x – 4y = 7
Add the revised equations.
– 4x + 8y = –12
4x – 8y = 14
0=2
Example 3
A Linear System with No Solution
ANSWER
Because the statement 0 = 2 is false, there is no solution.
Checkpoint
Solve a Linear System
Solve the system using the linear combination method.
1. x – 4y = 5
ANSWER
(1, – 1)
ANSWER
infinitely many solutions
ANSWER
(4, 5 )
2x + y = 1
2.
2x – y = 4
4x – 2y = 8
3. 3x – 2y = 2
4x – 3y = 1
Checkpoint
Solve a Linear System
4. How can you tell when a system has no solution?
infinitely many solutions?
ANSWER
if you get a false equation; if you get a true equation
Example 4
Use a Linear System as a Model
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?
Example 4
Use a Linear System as a Model
SOLUTION
VERBAL
MODEL
Servings
per
Servings Pans of
Sandwich Servings
= needed
per pan • pasta + sandwich •
trays
tray
Money to
Price
Pans of
Price
Sandwich
•
= spend
+ per tray •
per pan
pasta
trays
on food
Example 4
LABELS
Use a Linear System as a Model
Servings per pan of pasta = 10
(servings)
Pans of pasta = p
(pans)
Servings per sandwich tray = 5
(servings)
Sandwich trays = s
(trays)
Servings needed = 30
(servings)
Price per pan of pasta = 40
(dollars)
Price per sandwich tray = 10
(dollars)
Money to spend on food = 80
(dollars)
Example 4
ALGEBRAIC
MODEL
Use a Linear System as a Model
10p + 5s = 30
Equation 1 (servings needed)
40p + 10s = 80
Equation 2 (money to spend
on food)
Multiply Equation 1 by – 2 so that the coefficients of s
differ only in sign.
10p + 5s = 30
– 20p – 10s = – 60
40p + 10s = 80
40p + 10s = 80
Add the revised equations and solve
for p.
20p
= 20
p=1
Example 4
Use a Linear System as a Model
Substitute 1 for p in one of the original equations and
solve for s.
10p + 5s = 30
Write Equation 1.
10( 1) + 5s = 30
Substitute 1 for p.
5s = 20
s=4
Subtract 10 from each side.
Solve for s.
ANSWER
The caterer should make 1 pan of pasta and 4 sandwich
trays.
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