How do you solve #8/(x-4) -3=1/(x-10)#?

1 Answer
Oct 21, 2015

Answer:

#x_(1,2) = (49 +- 7)/6#

Explanation:

The first thing to notice here is that you have two values of #x# for which the denominators are equal to zero.

This means that any possible solution set will not include thes values. In other words, you need

#x - 4 !=0 implies x != 4" "# and #" "x - 10 != 0 implies x != 10#

The next thing to do is use the common denominator of the two fractions, which is equal to #(x-4)(x-10)#, to rewrite the equation without denominators.

To do that, multiply the first fraction by #1 = (x-10)/(x-10)#, #3# by #1 = ((x-4)(x-10))/((x-4)(x-10))#, and the second fraction by #1 = (x-4)/(x-4)#.

This will get you

#8/(x-4) * (x-10)/(x-10) - 3 * ((x-4)(x-10))/((x-4)(x-10)) = 1/(x-10) * (x-4)/(x-4)#

#(8(x-10))/((x-4)(x-10)) - (3(x-4)(x-10))/((x-4)(x-10)) = (x-4)/((x-4)(x-10))#

This is of course equivalent to

#8x - 80 - 3(x^2 - 14x + 40) = x-4#

#7x - 76 - 3x^2 + 42x- 120 = 0#

#3x^2 - 49x +196 = 0#

Use the quadratic formula to find the two roots of this quadratic equation

#x_(1,2) = (-(-49) +- sqrt( (-49)^2 - 4 * 3 * 196))/(2 * 3)#

#x_(1,2) = (49 +- sqrt(49))/6 = (49 +- 7)/6#

Therefore, you have

#x_1 = (49 - 7)/6 = 7" "# and #x_2 = (49 + 7)/6 = 28/3#

Since both solutions satisfy the condtions #x !=4# and #x != 10#, both will be valid solutions to the original equation.