Question

Classify the solutions of 3 over x plus 5, plus one fifth, equals 2 over x plus 5 as extraneous or non-extraneous.

Answer 1

The correct answer is:

non-extraneous

Explanation:

An extraneous solution is one that we arrive at that will not work in the equation.  For rational equations such as we have, extraneous solutions are ones that will cause the denominator to be 0.  For ours, that would mean x=-5.

The equation we have is:

$\frac{3}{x+5}+\frac{1}{5}=\frac{2}{x+5}$

We will multiply everything by (x+5) in order to get that off the bottom of the fractions:

$\frac{3}{x+5}\times (x+5)+\frac{1}{5}\times (x+5)=\frac{2}{x+5}\times (x+5) \\ \\3+\frac{x+5}{5}=2$

Multiply all terms by 5 to eliminate the fraction:

$3\times 5+\frac{x+5}{5}\times 5=2\times 5 \\ \\15+x+5=10$

Combine like terms:

20+x = 10

Subtract 20 from each side:

20+x-20 = 10-20

x = -10

Since this is not -5, this is not an extraneous solution.

Answer 2

3/(x + 5) + 1/5 = 2/(x + 5) 
1/(x + 5) + 1/5 = 0 
1/(x + 5) = -1/5 
5 = -(x + 5) 
5 = -x - 5 
10 = -x 
x = -10 
This is not extraneous.
hope it helps