How do you solve #(x)/(2x+7)=(x-5)/(x-1)#?

1 Answer
Jun 17, 2017

Answer:

#x=-5# or #x=7#

Explanation:

Note : when you multiply something with a fraction, you multiply to the numerator.

First, let's simplify the LHS by multiplying #(x-5)/(x-1)# with #2x+7#

#(x-5*2x+7)/(x-1)#

#(2x^2-10x+7x-35)/(x-1)#

So we have

#x=(2x^2-3x-35)/(x-1)#

Then let's simplify the RHS by multiplying #x# with #x-1#.

#x/1=(2x^2-3x-35)/(x-1)# since #x=x/1#,

Then

#x*x-1=2x^2-3x-35#

Then #x^2-x=2x^2-3x-35#

#0=x^2-2x-35#

What we just did here is called cross multiplying.

Now, we have a quadratic equation.
We can factorise this to

#(x+5)(x-7) =0#

Set each factor equal to #0#, since if either #(x+5)# or #(x-7)# was 0, the equation would hold, because #0# multiplied by any number will always be 0.

#x+5=0#
#x=-5#

#x-7=0#
#x=7#

These are our two answers