Let be the sides of the rectangle.
We know that and
Now, solve.
Let L = length of rectangle, in inches
And, w = width, in inches
Perimeter, P = 2L +2w
32 = 2L +2w
16 = L +w -------------(1)
Area, A = w*L
60 = wL ----------------(2)
There are your two simultaneous equations.
One way to continue is by substitution,
there are many ways to do this substitution itself,
one way is we use (1) to get the value of L in terms of w,
L = 16-w --------------(1a)
Substitute that into (2),
60 = w(16-w)
60 = 16w -w^2
w^2 -16w +60 = 0
Factor that,
(w-6)(w-10) = 0
w = 6 or 10
When w = 6 inches,
L = 16-w --------------(1a)
L = 16-6
L = 10 inches
When w = 10 inches,
L = 16-w --------------(1a)
L = 16-10
L = 6 inches
What? L is shorter than w?
Ignore this.
Therefore, length = 10 inches and width = 6 inches. ----------answer.