Question

A student solves the following equation and determines that the solution is −2. Is the student correct? Explain. 3 a + 2 − 6a a2 − 4 = 1 a − 2

Answer 1

For this case, we have the following equation (according to the comments):

$\frac{3}{a+2} - \frac{6a}{a^2-4} = \frac{1}{a-2}$

First, we factor the quadratic expression in the denominator:

$\frac{3}{a+2} - \frac{6a}{(a-2)(a+2)} = \frac{1}{a-2}$

Then, we multiply both sides of the equation by (a-2) (a + 2):

$\frac{3(a-2)(a+2)}{a+2} - \frac{6a(a-2)(a+2)}{(a-2)(a+2)} = \frac{(a-2)(a+2)}{a-2}$

Later, canceling similar terms we have:

$3(a-2) - 6a = a+2$

We do distributive property on the left side of the equation:

$3a-6-6a=a+2$

By grouping variables and constant terms we have:

$3a - 6a -a = 2+6$

Rewriting we have:

$-4a=8$

Finally, by clearing "a" we have:

$a=-\frac{8}{4} a=-2$

Note: the value of a is a extraneous solution because it makes the denominator of the original equation equal to zero.

Answer:

the student correct, but the value of a= -2 is an extraneous solution

Answer 2

No, a possible solution is a = –2, but it does not check in the original equation because it makes two of the denominators equal to zero.
No, a = –2 is an extraneous solution, and there is no solution to the original rational equation.