If AB=2 (x+1) BC=3x+1 and AC=4 (x+2) then find the value for x, AB, BC and AC

Givens

AB + BC = AC

AB = 2 (x + 1)

BC = 3x + 1

AC = 4 (x + 2)

Substitute and Solve

AB + BC = AC

2 (x + 1) + 3x + 1 = 4 (x + 2) Remove the brackets on the left

2x + 2 + 3x + 1 = 4 (x + 2) Collect the like terms on the left

5x + 3 = 4 (x + 2) Remove the brackets on the right.

5x + 3 = 4x + 8 Subtract 4x from both sides.

5x - 4x + 3 = 8

x + 3 = 8 Subtract 3 from both sides

x = 8 - 3

x = 5

Answers

AB=2 (5 + 1) = 2 * 6 = 12

BC = 3x + 1 = 3*5 + 1 = 15 + 1 = 16

AC = 4 (5 + 2) = 4*7 = 28