Math assistance

Question

The area of a rectangle is 44 yd2, and the length of the rectangle is 3 yd less than double the width. Find the dimensions of the rectangle.

Answer 1

Length = L = (2W-3) yards

Width = W Yards

Area = L . W = W (2W-3) = 44

2W² -3W =44

2W² -3W - 44 = 0

 (W+4) (2W-11) = 0

W= -4 or W = 11/2 = 5.5

Distance cannot be negative. So reject x=-4

So Width = 5.5 yd

Length = 2W-3 = 2 x 5.5 -3 = 8 yards

Answer 2

Hi Kristina H.,

We know the area of a rectangle equals the length time the width, A = L*W.

We are given the area 44 yd² = to the area A.

We are also given the length equal to 3 yds less than double the width, L = 2W - 3.

So we have A = 44 and L = 2W - 3, we can rewrite A = L*W and solve for W. Then use W to solve for L.

Can you take it from here?

Let me know, Joe.

Answer 3

So this problem can be solved using a system of equations that will be derived directly from the question. First, to keep everything straight I will assign letters to the dimensions/measurements we are concerned with. Area of the rectangle will be "a", length of the rectangle will be "L", and width of the rectangle will be "w". The two equations you need to make the system will be that for area of a rectangle (a=L×w or 44=L×w) and that for the length of the rectangle, which we are told is 3 less than 2 times the width (L=2w-3). Now we need to do some substitution in order to get an equation with a single variable. There is more than one way to do this, however, the way I find the most straightforward is to plug our length equation into the area equation, giving us 44=(2w-3)×w which can be multiplied out to give 44=2w²-3w. We subtract 44 from both sides to get the equation into a familiar format (0=2w²-3w-44). From here we simply need to "plug-n-chug" using the quadratic formula ((-b ± √(b²-4×a×c))/2×a). Following this formula we narrow the answer down to either 5.5 or -4, and we can immediately rule out -4 as measurements cannot be negative numbers. In doing so we see that 5.5 is the width and 11 is the length.

I hope you find this helpful!