Step-by-step explanation:
To solve this problem we can use a system of equations to help us determine the answer. Our first step is to change the information provided in the problem into mathematical equations.
The first sentence tells us, "The width of a rectangle is 14 units less than the length." We can represent this algebraically as W = L - 14 where w is the width of the rectangle and L is the length.
The second sentence tells us, "The area is 32 square units." We can represent this algebraically as L x W = 32 because area of a rectangle is length times width. Now we have the two equations needed for our system of equations.
W = L - 14
L x W = 32
To solve this we can substitute the first equation into the second equation and replace the W with what W equals in the first equation.
L x (L - 14) = 32
Next we multiply and distribute the L.
L2 - 14L = 32
To solve this quadratic we'll need to move 32 over to the other side of the equation.
L2 - 14L - 32 = 0
From here we can use the quadratic equation to solve for L or we can factor. Factoring gives us the following.
(L - 16)(L + 2) = 0
Now we can solve for L.
L = 16 and L = -2
Because this problem deals with area of a rectangle we can disregard the negative value for L.
The length is 16.
We can plug this value for L into our first equation from our system of equations to solve for W.
W = L - 14
W = 16 - 14
W = 2
The dimensions of the rectangle are length = 16 and width = 2. Please let me know if you have any questions or if I can be of further assistance in any way.
