How do you solve \frac{3}{2x-4}-\frac{2}{x^{2}-x-2}=\frac{9}{x+1}32x42x2x2=9x+1?

2 Answers
Apr 19, 2018

x=7/3x=73

Explanation:

First we factor the denominators, then find a common denominator so we can subtract, then cross multiply and solve the resulting polynomial equation.

frac{3}{2x-4} - frac{2}{x^2 - x -2 } = frac{9}{x+1}32x42x2x2=9x+1

frac{3}{2(x-2)} - frac{2}{(x+1)(x-2)} = frac{9}{x+1}32(x2)2(x+1)(x2)=9x+1

frac{3(x+1) - 2(2)}{2(x+1)(x-2)} = frac{9}{x+1}3(x+1)2(2)2(x+1)(x2)=9x+1

frac{3x-1}{2(x+1)(x-2)} = frac{9}{x+1}3x12(x+1)(x2)=9x+1

(3x-1)(x+1) = 18(x+1)(x-2) (3x1)(x+1)=18(x+1)(x2)

(x+1)( (3x - 1) - 18(x-2) ) = 0(x+1)((3x1)18(x2))=0

(x+1)(-15 x + 35) = 0 (x+1)(15x+35)=0

x = -1 x=1 or x = 35/15 = 7/3 x=3515=73

Apr 19, 2018

x=7/3x=73

Explanation:

3/(2x-4)-2/(x^2-x-2)=9/(x+1)32x42x2x2=9x+1

Factorising

2x-4=2(x-2)2x4=2(x2)

x^2-x-2=x^2-2x+x-2x2x2=x22x+x2

=x(x-2)+(x-2)=x(x2)+(x2)

=(x+1)(x-2)=(x+1)(x2)

3/(2x-4)-2/(x^2-x-2)=3/(2(x-2))-2/((x+1)(x-2))32x42x2x2=32(x2)2(x+1)(x2)

3/(2(x-2))-2/((x+1)(x-2))=9/(x+1)32(x2)2(x+1)(x2)=9x+1

3/(2(x-2))-9/(x+1)=2/((x+1)(x-2))32(x2)9x+1=2(x+1)(x2)

Multiplying throughout by (x+1)(x-2)(x+1)(x2)

3/2(x+1)-9(x-2)=232(x+1)9(x2)=2

3/2x+3/2-9x+18=232x+329x+18=2

(3/2-9)x+(3/2+18-2)=0(329)x+(32+182)=0

-15/2x+35/2=0152x+352=0

-15x+35=015x+35=0

15x=3515x=35

x=35/15x=3515

x=7/3x=73

Check:

lhs=
3/(2x-4)-2/(x^2-x-2)=3/(2xx7/3-4)-2/((7/3)^2-7/3-2)32x42x2x2=32×7342(73)2732

=3/(14/3-4)-2/(49/9-7/3-2)=9/(14-12)-18/(49-21-18)=314342499732=9141218492118

=9/2-18/10=45/10-18/10=(45-18)/10=27/10=921810=45101810=451810=2710

rhs=9/(x+1)=9/(7/3+1)=27/(7+3)=27/10rhs=9x+1=973+1=277+3=2710

lhs=rhslhs=rhs