Step 4 Solvethe system of equations. Since equation (2), is solved
for L, we can substitute for Lin equation (1) and solve for W.
(1)
Let
Distributive property
Combine like terms.
Subtract 80.
Divide by 4.Let in the equation to find L.
L= 60 + 40 = 100
W= 60 L= W+ 40
W= 60
4 W= 240
4 W+ 80 = 320
2 W+ 80 + 2 W= 320
21 W+ 402 + 2 W= 320 L=W+40.
2 L+ 2 W= 320
W+ 40
L= W+40,
SECTION 4.3 Applications of Systems of Linear Equations 235
NOW TRY
EXERCISE 1
A rectangular parking lot has
a length that is 10 ft more
than twice its width. The
perimeter of the parking lot
is 620 ft. What are the dimen-
sions of the parking lot?
Step 5 State the answer.The length is 100 yd, and the width is 60 yd. Both
dimensions are within the ranges given in the problem.
Step 6 Check.Calculate the perimeter and the difference between the length and
the width.
The perimeter is 320 yd, as required.100 - 60 = 40
211002 + 21602 = 320
The answer is correct. NOW TRY
OBJECTIVE 2 Solve money problems by using two variables.
Solving a Problem about Ticket PricesFor the 2008 – 2009 National Hockey League and
National Basketball Association seasons, two hockey
tickets and one basketball ticket purchased at their
average prices would have cost $148.79. One hockey
ticket and two basketball tickets would have cost
$148.60. What were the average ticket prices for the
two sports? (Source:Team Marketing Report.)
Step 1 Read the problem again. There are two
unknowns.
Step 2 Assign variables.
Let the average price for a hockey ticket
and the average price for a basketball ticket.
Step 3 Write a system of equations. Because two hockey tickets and one
basketball ticket cost a total of $148.79, one equation for the system is
By similar reasoning, the second equation is
These two equations form a system of equations.
(1)h+ 2 b= 148.60 (2)
2 h+ b= 148.79
h+ 2 b=148.60.
2 h+b=148.79.
b=
h=
EXAMPLE 2
Be sure to use
parentheses
around W+40.Length is 40 yd more than width,
as required.NOW TRY ANSWER
- length: 210 ft; width: 100 ft
Don’t stop here.