Question

How do you solve and check for extraneous solutions in `(x-4) / (x-1)= 10/ (x+7)`?

Answer 1

`x = 9" "` or `" "x = -2`

Explanation:

Every time you're dealing with rational expressions, you need to make sure that you exclude from the solution set any value of `x` that can make the denominator(s) equal to zero.

In your case, you need

`x - 1 !=0 implies x !=1`

and

`x + 7 !=0 implies x != - 7`

The common denominator for these two fractions will be `(x-1)(x+7)`, so multiply each side by the appropriate expression to get

`( (x-4) * (x+7))/((x-1)(x+7)) = (10 * (x-1))/((x-1)(x+7))`

This is equivalent to

`(x-4)(x+7) = 10 (x-1)`

`x^2 + 3x - 28 = 10x - 10`

`x^2-7x - 18 = 0`

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

`x_(1,2) = (-(-7) +- sqrt((-7)^2 - 4 * 1 * (-18)))/(2 * 1)`

`x_(1,2) = (7 +- sqrt(121))/2`

`x_(1,2) = (7 +- 11)/2 = {(x_1 = (7 + 11)/2 = 9), (x_2 = (7 - 11)/2 = -2) :}`

Both `x_1` and `x_2` satisfy the conditions `x!=1` and `x !=-7`, which means that both will be valid solutions to the original equation.