To the area of a triangle, the excess of the length over the width is added, giving 120; moreover, the sum of the length and width is 24. Find the dimensions of the rectangle.
[Hint: The problem can be put in the form of two equations xy+x-y=120, x+y=24; if the substitution y=z-2 is made, the system becomes x+z=26, xz=144.] I know x=18, y=6, but what steps are used to get there?

Question

To the area of a triangle, the excess of the length over the width is added, giving 120; moreover, the sum of the length and width is 24. Find the dimensions of the rectangle.
[Hint: The problem can be put in the form of two equations xy+x-y=120, x+y=24; if the substitution y=z-2 is made, the system becomes x+z=26, xz=144.] I know x=18, y=6, but what steps are used to get there?