Question

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

Answer 1

`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.