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

1 Answer
Aug 30, 2015

Answer:

#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.