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 [tex]15w-w^2=44[/tex]. Now we will bring the 44 ovr by subtraction and put this into standard form: [tex]-w^2+15w-44=0[/tex]. 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): [tex]w^2-15w+44=0[/tex]. 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.
