How do you solve #x/(2x-3) + 4/(x+1) = 1#?

1 Answer
Jul 9, 2016

Answer:

# x = 9 or x = 1#

Explanation:

Remember the value of a fraction does not change if we multiply the numerator and denominator by the same thing. We want to use this in order to get the fractions on the left hand side into one fraction, via a common denominator.

#(x)/(2x-3) * color(red)((x+1)/(x+1)) + 4/(x+1) * color(red)((2x-3)/(2x-3)) = 1#

#therefore (x(x+1) + 4(2x-3))/((2x-3)(x+1)) = 1#

#(x^2 + 9x -12)/(2x^2 - x - 3) = 1#

#x^2 + 9x -12 = 2x^2 - x - 3 implies x^2 - 10x + 9 = 0#

#therefore (x-9)(x-1) = 0#

#implies x = 9 or x = 1#