P= 2L + 2w
A= length * width
PERIMETER EQUATION
P= 32
32= 2L + 2w
AREA EQUATION
A= 60
60= Lw
SOLVE FOR ONE VARIABLE
60= Lw
divide both sides by L
60/L= w
SUBSTITUTE
32= 2L + 2w
32= 2L + 2(60/L)
32= 2L + 120/L
Multiply everything by L to eliminate fraction
(32)(L)= (2L)(L) + (120/L)(L)
32L= 2L^2 + 120
subtract both sides by 32L
0= 2L^2 - 32L + 120
divide all by 2 to remove x^2 coefficient
0= L^2 - 16L + 60
factor
0= (L-6)(L-10)
0=L-6
6=L
0=L-10
10=L
FIND WIDTH
Substitute each possible length answer to find width.
If L= 6
32= 2L + 2w
32= 2(6) + 2w
32= 12 + 2w
20= 2w
10=w
If L=6
60= Lw
60= (6)w
10=w
If L=10
32= 2L + 2w
32= 2(10) + 2w
32= 20 + 2w
12= 2w
6= w
If L= 10
60= Lw
60= (10)w
6= w
ANSWER: There are two possible answers for the dimensions of the ticket. If L=6, then W=10. If L=10, then W= 6.
Hope this helps! :)
