Question

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.

Answer 1

Answer:

Dimensions of the rectangle = 16 by 6

Step-by-step explanation:

Let the length and breadth represent L and B

Hence L= B+10=------eqn 1

LB = 96 -----eqn2

From eqn1 L=B+10, it's substituted into eqn2

B(B+10)=96

B²+10B-96=0

It's now solved quadratically

B²+16B-6B-96

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

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

B=6 and -16

Hence B=6 ft since it's the positive value

L= 96/B = 96/6 = 16 ft

Dimensions of the rectangle = 16 by 6

Answer 2

Answer: 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.