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

Question

1579 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-07T03:15:33

#x = 4#

Multiplying the numerators of the fractions by
the common denominator, #(x-2)(x-3)#,

#(x(x-3)-1(x-2))/((x-2)(x-3)) = 1#

Multiplying both sides by #(x-2)(x-3)#,

#implies (x^2 - 3x - x + 2) = x^2-5x+6#

#implies -4x + 5x = 6 - 2#

#implies x = 4#

Answer 2 · 2017-10-07T03:16:32

#x=4#

The first thing I'd do is combine the two fractions on the left:

#x/(x-2)(1)-1/(x-3)(1)=1#

#x/(x-2)((x-3)/(x-3))-1/(x-3)((x-2)/(x-2))=1#

#(x^2-3x)/((x-2)(x-3))-(x-2)/((x-2)(x-3))=1#

#(x^2-4x+2)/(x^2-5x+6)=1#

And now multiply through by the denominator of the fraction:

#x^2-4x+2=x^2-5x+6#

We can now simplify:

#x^2color(red)(-x^2)-4xcolor(red)(+5x)+2color(red)(-2)=x^2color(red)(-x^2)-5xcolor(red)(+5x)+6color(red)(-2)#

#x=4#