Question 1145118

Two rectangles A and B each have an area of 11 cm^2
The length of rectangle A is x cm
The length of rectangle B is (x+3) cm
Given that the width of rectangle A is 2 cm greater than the width of rectangle B form an equation in x and show that it simplifies to
2x^2 + 6x - 33 = 0
<pre>With the length and area of rectangle A being x and 11, respectively, the width of rectangle A becomes: {{{11/x}}}
With the length and area of rectangle B being x + 3 and 11, respectively, the width of rectangle B becomes: {{{11/(x + 3)}}}
As width of rectangle A is 2 cm GREATER than width of rectangle B, we get: {{{matrix(1,3, 11/x, "=", 11/(x + 3) + 2)}}}
11(x + 3) = 11x + 2x(x + 3) ------- Multiplying by LCD, x(x + 3)
{{{highlight_green(matrix(1,9, 11x + 33, "=", 11x + 2x^2 + 6x, "=====>", 0, "=", 2x^2 + 11x + 6x - 11x - 33, "======>", highlight(matrix(1,6, 0, "=", 2x^2 + 6x - 33, "<======", SIMPLIFIED, EQUATION))))}}}