SOLUTION: The width of a rectangle is 3 less than twice the length x. If the area of the rectangle is 43 square feet, which equation can be used to find the length, in feet?

Algebra ->  Rectangles -> SOLUTION: The width of a rectangle is 3 less than twice the length x. If the area of the rectangle is 43 square feet, which equation can be used to find the length, in feet?      Log On


   

Question 461475: The width of a rectangle is 3 less than twice the length x. If the area of the rectangle is 43 square feet, which equation can be used to find the length, in feet?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
The width of a rectangle y is 3 less than twice the length x.
y%2B3=2x.....->..y=2x-3

If the area of the rectangle is A=43ft%5E2,
which equation can be used to find the length, in feet?
xy=43ft%5E2
x%282x-3%29=43ft%5E2
2x%5E2-3x=43ft%5E2
2x%5E2-3x-43ft%5E2=0....use quadratic formula to find x
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-3%29+%2B-+sqrt%28+%28-3%29%5E2-4%2A2%2A%28-43%29+%29%29%2F%282%2A2%29+
x+=+%283+%2B-+sqrt%289%2B344+%29%29%2F4+
x+=+%283+%2B-+sqrt%28353+%29%29%2F4+
x+=+%283+%2B18.788294228055935999045204838699%29%2F4+
x+=+21.79%2F4+
x+=+5.45ft

y=2x-3
y=2%285.45%29-3
y=10.9-3
y=7.9ft

check:
5.45ft%2A7.9ft=43ft%5E2
43.055ft%5E2=43ft%5E2
43ft%5E2=43ft%5E2