Question

The diagonals of rectangle nopq intersect at point r. if qr=3x-4 and np=5x+20, solve for x.

Answer 1

2(3x - 4) = 5x + 20
6x - 8 = 5x + 20
subtract 5x from both sides
x - 8 = 20
add 8 to both sides
x = 28

Answer 2

The value of "x" is 28.

Explanation:

As can see in the attach the length of both diagonals is the same. From the problem, you have the length of segment qr = 3x -4, which represents half of the diagonal. Then, length of segment np = 5x + 20, is the completa diagonal. So, you have to multiply the equation for qr by 2 and then equal it to equation of segment np.

2·(3x - 4) = 5x + 20 ⇒ 6x - 8 = 5x + 20 ⇒ 6x - 5x = 20 + 8 ⇒ x = 28.

Summarizing the value of "x" is 28.