Question

The length of a rectangle plus its width is 15 cm. The area is 44 square cm. What are the length and width of the rectangle?

Answer 1

The length is 11 and the width is 4.

Answer 2

This problem requires a system of equations, which we can easily come up with.  "The length plus the width is 15" is L + W = 15.  That is the first of 2 equations.  The area is found by multiplying the length by the width, so the next statement, "The area is 44" is L*W=44.  Solve that first equation for L to get L=15-W.  Now sub that L value into the second equation in place of L to get an equation in terms of W only.  (15-W)W=44.  We will distribute to get $15w-w^2=44$.  Now we will bring the 44 ovr by subtraction and put this into standard form: $-w^2+15w-44=0$.  I personally do not like to try to factor polynomials with negative leading coefficients, so let's change all those signs (yes, that's legal): $w^2-15w+44=0$.  If you plug those values into the quadratic formula you will get w values of 11 and 4.  It just so happens (and NOT by coincidence) that 11+4 = 15, which fits our "the length of a rectangle plus its width is 15", and 11*4 = 44, which fits our "the area of the rectangle is 44".  So the length and width are 11 and 4, respectively.