Question

the length of a rectangle exceeds its width by 6 inches and the area is 40 square inches. What are the length and width of the rectangle?

Answer 1

Classic Algebra and its unnecessarily complicated sentence structure. As you may have probably known, Algebra has its own "vocabulary set".

"the length of a rectangle exceeds its width by 6 inches" -> length is 6 in. longer than width -> l= w + 6

Since we're solving for the length and width, let's give them each variables.

length = l = w+6

width = w

The next bit of information is "the area is 40 square inches"

Applying the formula for the area of a rectangle we can set up:

l x w = 40

replace "l", or length, with it's alternate value.

(w+6) x w = 40

distribute

$w^{2}$ + 6w = 40

subtract 40 from both sides

$w^{2}$ + 6w - 40 = 0

factor

(w - 4)(w + 10) = 0

solve for w

w= 4, or -10

So great, we have 2 values; which one do we choose? Since this problem is referring to lengths and inches, we will have to choose the positive value. There is not such thing as a negative distance in the real world.

We now have half of the problem solved: width. Now we just need to find the length which we can do but substituting it back into the original alternate value of l.

l = w + 6

w=4

l = 4 + 6 = 10

The length is 10 in. and the width is 4 in. Hope this helps!