How do you solve #\frac { x + 2} { x } = \frac ( x - 1} { x } - \frac { 4x + 2} { x ^ { 2} - 3x }#?

1 Answer
Jan 25, 2018

Answer:

#x=1#

(the solution #x=0# is extraneous)

Explanation:

First, move the #(x-1)/x# to the left side.

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

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

Now you can combine the numerators on the left side since the denominators are the same.

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

#3/x = (-(4x+2))/(x^2-3x)#

Now we can just cross multiply and solve it like a quadratic:

#3(x^2-3x) = -x(4x+2)#

#3x^2 - 9x = -4x^2 - 2x#

#7x^2 - 7x = 0#

#x^2 - x = 0#

#x(x-1) = 0#

#therefore " " x = 0 " " or " " x=1#

So these should be our two solutions. But there's one last step - we need to plug these solutions back into the original equation to make sure they aren't extraneous (i.e. they produce an indeterminate form like #0/0# instead of a real number).

Here's the original equation:

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

Let's check #x = color(red)0#

#(color(red)0+2)/color(red)0 stackrel(?color(white)"=")(=) (color(red)0 - 1)/color(red)0 - (4(color(red)0)+2)/(color(red)0^2-3(color(red)0))#

This already has a lot of dividing by zero, so unfortunately it is NOT a real solution since it will produce undefined values.

Now, let's check #x=color(blue)1#

#(color(blue)1+2)/color(blue)1 stackrel(?color(white)"=")(=) (color(blue)1 - 1)/color(blue)1 - (4(color(blue)1) + 2)/(color(blue)1^2-3(color(blue)1))#

#3/1 stackrel(?color(white)"=")(=) 0/1 - 6/(-2)#

#3 stackrel(?color(white)"=")(=) -(-3)#

#3 stackrel(color(limegreen)sqrt"" color(white)(=))(=) 3#

So our only solution is #x=1#

Final Answer