The length of a rectangle is 3 feet greater than its width. The area of the rectangle is 18 square feet. What are the length and the width of the rectangle?

Question

Please help, and clearly explain how to solve it.

Answer 1

Let L=length of the rectangle. Let W=its width

For rectangles, area = length times width or LW

LW=18 L=3+W or W=L-3

substitute 3+W for L in the equation for area LW=18

then L(L-3)=18 or expanding L² - 3L = 18

subtract 18 from both sides: L² - 3L -18 = 0

You could use the quadratic formula but this equation is easily factored,

since 18=6 x 3 and 6-3=3

factors are (L-6) and (L+3) Finding the factors is a matter of trial and error

using factors of the constant term, which here is 18.

(L-6)(L+3)=0 Set each factor equal to zero and solve for L L-6=0 and L+3=0

L=-3 (you can ignore this negative length) and L=6

Since W=L-3 then W=6-3=3

Length = 6 Width = 3

Answer 2

Angela,

Let's start by declaring the variable(s). We don't know the width, so let's call the width w. We do know that the length is (3 + w). so now we have names for the two lengths. The area of a rectangle is A = l x w

Let's add our names and information from the question to that.

18 = (3+w)w

now let's expand the right side. (3+w)w = 3w + w^2 (w^2 means "w squared")

now let's move things around so we can solve the equation for w.

w^2 + 3w -18 = 0

we can factor that to be (w + 6)(w - 3) = 0

and we can further state that w + 6 = 0 and w - 3 = 0

solving we get

w = -6 and w = 3

even though w = -6 is an answer to the equation, there is no such thing as negative length so we'll not use that one.

But w = 3 makes sense so let's check it.

If the width is 3 and the length is 3 + 3 = 6, then the area would be 3 x 6 = 18!

Bingo!