The length and width of a rectangle are to each other as 4 is to 3. A second rectangle are 4 units longer and 2 units wider than the first one, and has twice as great an area as the first. Find the dimensions of the first rectangle.

Answer 1

let the length and width be 4x and 3x
old area = (4x)(3x) = 12x^2

new length = 4x+4
new width = 3x + 2
new area = (4x+4)(3x+2) = 12x^2 + 20x + 8

12x^2 + 20x + 8 = 2(12x^2) = 24x^2

12x^2 - 20x - 8 = 0
3x^2 - 5x - 2 = 0
(x-2)(3x + 1) = 0
x = 2 or x = -1/3, which is not admissible

the first triangle was 8 by 6

check:
8 by 6 gives an area of 48
new dimensions: 12 by 8
new area = 12(8) = 96
which is twice 48
All is good.