How do you solve #6+ \frac { 1} { x - 1} = - \frac { 6} { x - 2}#?

1 Answer
Nov 29, 2017

Ultimately, you're solving for #x#. Firstly, add all the terms and equate the equation to #0#. Then, using quadratic formula, solve for #x#.

Explanation:

Solving implies we determine the value of the variable. To do this, we must isolate and solve for #x#.

#6 + 1/(x-1) = 6/(x-2)#

Before we do that, let's simplify the left side. To do this, let's make the #6# have the same denominator as the other term.

#[6xx (x-1)/(x-1) ] + 1/(x-1) = 6/(x-2)#

#(6(x-1))/(x-1) + 1/(x-1) = 6/(x-2)#

#(6x-6)/(x-1) + 1/(x-1) = 6/(x-2)#

Now we can add the two terms.

#(6x-6+1)/(x-1) = 6/(x-2)#

#(6x-5)/(x-1) = 6/(x-2)#

Now we can cross-multiply. Cross-multiplying is done by:

Meritnation

#(6x-5)(x-2)=(6)(x-1)#

Now we simplify both sides. On the left side, we have to expand the brackets.

#6x^2-12x-5x+10=6x-6#

#6x^2-17x+10=6x-6#

Now, we are going to bring the terms to one side and simplify again. We are also going to equate the expression to #0#.

#6x^2-17x+10-6x+6=0#

#6x^2-23x+16=0#

Now we are going to factor the equation. To avoid any complications, we will just use the quadratic formula.

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#x = (-[-23]+-sqrt([-23]^2-4[6][16]))/(2[6])#

Simplify.

#x = (23+-sqrt(145))/(12)#

Now we solve for #x#. Remember there will be two answers because of the #+-#.

#x (+) ~~ 2.92#

#x (-) ~~ 0.91#

We can double check our work by graphing the function and finding the zeros.

Hope this helps :)