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