Algebra problem

Question

The length of a rectangle is two feet more than five times the width. The perimeter is 40 feet. Find the dimensions of the rectangle

Answer 1

Start by changing the written explanation into equations!

Length (l) = 2 more (add) than 5 times (multiply) the width (w)

So, l = 2 + 5w

Perimeter (P) = 40

Next, remember formula for perimeter of a rectangle:

P = 2(l) + 2(w)

Now, fill in the perimeter formula with what you know:

40 = 2(2 + 5w) + 2w

Solve for w:

40 = 2(2 +5w) + 2w

40 = 4 + 10w + 2w

40 = 4 + 12w

36 = 12w

3 = w

Now plug in your answer to solve for l:

l = 2 + 5(3)

l = 2 +15

l = 17

Check your work by re-solving for the perimeter

P = 2(17) + 2(3)

P = 34 + 6

P = 40

Answer 2

Let 'l' be the length of the rectangle and let 'w' be the width.

"The length of a rectangle is two feet more than five times the width" translates to:

l = 5w+2.

The perimeter of a rectangle is given by P = 2l + 2w.

So we know that P = 2(5w+2) + 2w = 10w+4+2w = 12w+4

And since we are told that P = 40 we can solve for w:

40 = 12w+4

36 = 12w by subtracting 4 from both sides

w = 3 by dividing 12 on both sides

Plug w back into l: l = 5(3) + 2 = 17.

So the dimensions of the rectangle are l = 17 and w = 3.