Question

A student solves the following equation for all possible values of x:

Answer 1

The correct answer is:
C) his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Explanation:
To solve this rational equation, we cross multiply:
8*(x-4)=2*(x+2).

Using the distributive property, we have
8*x-8*4=2*x+2*2;
8x-32=2x+4.

Subtract 2x from each side:
8x-32-2x=2x+4-2x;
6x-32=4.

Add 32 to each side:
6x-32+32=4+32;
6x=36.

Divide both sides by 6:
$\frac{6x}{6}$=$\frac{36}{6}$;
x=6.

Extraneous solutions are solutions that come about in the problem but are not valid solutions to the problem; the only values of x that would give this sort of answer are -2 and 4, since these are the two values that would make one of the denominators 0 (we cannot divide by 0).

Answer 2

Answer: His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Step-by-step explanation: Edg 2020