The length of a rectangle is 10 feet more than the width. The area of the rectangle is 96 square feet. Find the dimensions of the rectangle. Solve algebraically.

The dimensions of the rectangle are 6ft and 16ft.

Step-by-step explanation:

Let the width of the rectangle be y

Let the length of the rectangle be y+10

Area= 96ft

Since area of a rectangle is length multiplied by width

y(y+10) = 96

y^2 +10y =96

y^2 +10y -96= 0

Solving algebraically,

y^2 +16y - 6y - 96= 0

y(y+16) - 6(y+16)= 0

(y-6)(y+16)= 0

y-6= 0

y =6ft

Recall that the width was denoted as y. That means the width is 6ft.

Since length is y+10= 6+10 = 16ft

Length is 16ft, width is 6ft.