Answer:
Step-by-step explanation:
We have
We are going to analyze both sides of the equation separately.
Part a:
Applying common denominator:
Part b:
In the numerator we can apply common factor 4. And in the denominator we can apply difference of squares.
<em>Remember: </em><u><em>Difference of squares:</em></u><em> </em>
Then,
An extraneous solution is <em>not a valid solution for a problem. </em>We know that the denominator can't be zero.
The denominator of the first side of the equation is:
and we have to see for which values of the denominator is zero:
⇒ and
The denominator of the second side of the equation is:
And we have to see for which values of the expression is zero,
⇒ and
Then the extraneous solutions of the equation are:
Because those are the values of that make the denominator zero.