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Over Lesson 6–4
Over Lesson 6–4
Applying Systems of
Linear Equations
Lesson 6-5
You solved systems of equations by using
substitution and elimination.
• Determine the best method for solving
systems of equations and apply systems of
equations to real-world problems.
Choose the Best Method
Determine the best method to solve the system of
equations. Then solve the system.
2x + 3y = 23
4x + 2y = 34
Understand
To determine the best method to solve the system of
equations, look closely at the coefficients of each term.
Plan
Since neither the coefficients of x nor the coefficients of
y are 1 or –1, you should not use the substitution method.
Since the coefficients are not the same for either
x or y, you will need to use elimination with multiplication.
Choose the Best Method
Solve
Multiply the first equation by –2 so the coefficients of the
x-terms are additive inverses. Then add the equations.
2x + 3y = 23
–4x – 6y = –46
4x + 2y = 34
(+) 4x + 2y = 34
–4y = –12
y=3
Multiply by –2.
Add the
equations.
Divide each side
by –4.
Simplify.
Choose the Best Method
Now substitute 3 for y in either equation to find the
value of x.
4x + 2y = 34
Second equation
4x + 2(3) = 34
4x + 6 = 34
4x + 6 – 6 = 34 – 6
4x = 28
y=3
Simplify.
Subtract 6 from each side.
Simplify.
Divide each side by 4.
x=7
Simplify.
Answer: The solution is (7, 3).
Choose the Best Method
Check
Substitute (7, 3) for (x, y) in the first equation.
2x + 3y = 23
?
2(7) + 3(3) = 23
23 = 23 
First equation
Substitute (7, 3) for (x, y).
Simplify.
POOL PARTY At the school pool party, Mr. Lewis
bought 1 adult ticket and 2 child tickets for $10.
Mrs. Vroom bought 2 adult tickets and 3 child tickets
for $17. The following system can be used to
represent this situation, where x is the number of
adult tickets and y is the number of child tickets.
Determine the best method to solve the system of
equations. Then solve the system.
x + 2y = 10
2x + 3y = 17
A. substitution; (4, 3)
B. substitution; (4, 4)
C. elimination; (3, 3)
D. elimination; (–4, –3)
Apply Systems of Linear
Equations
CAR RENTAL Ace Car Rental rents a car for $45
and $0.25 per mile. Star Car Rental rents a car for
$35 and $0.30 per mile. How many miles would a
driver need to drive before the cost of renting a
car at Ace Car Rental and renting a car at Star Car
Rental were the same?
Let x = number of miles and y = cost of renting a car.
y = 45 + 0.25x
y = 35 + 0.30x
Apply Systems of Linear
Equations
Subtract the equations to eliminate the y variable.
y = 45 + 0.25x
(–) y = 35 + 0.30x
0 = 10 – 0.05x
Write the equations
vertically and subtract.
–10 = –0.05x
Subtract 10 from each side.
200 = x
Divide each side by –0.05.
Apply Systems of Linear
Equations
Substitute 200 for x in one of the equations.
y = 45 + 0.25x
First equation
y = 45 + 0.25(200) Substitute 200 for x.
y = 45 + 50
Simplify.
y = 95
Add 45 and 50.
Answer: The solution is (200, 95). This means that
when the car has been driven 200 miles, the
cost of renting a car will be the same ($95) at
both rental companies.
Homework
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