a. To determine the values of the variable that make the denominators zero, we need to examine the denominators in the given equation:
[tex]\[
\frac{1}{x+2} + 2 = \frac{7}{x+2}
\][/tex]
From this equation, we can see that the denominators are [tex]\(x + 2\)[/tex]. We need to find the values of [tex]\(x\)[/tex] that make the denominator zero:
[tex]\[
x + 2 = 0
\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[
x = -2
\][/tex]
Therefore, the value of the variable that makes the denominator zero is:
[tex]\[
x = -2
\][/tex]
This is our restriction.
b. Next, we solve the equation while keeping in mind the restriction [tex]\(x \neq -2\)[/tex]:
[tex]\[
\frac{1}{x+2} + 2 = \frac{7}{x+2}
\][/tex]
To clear the denominators, we can multiply every term by [tex]\(x + 2\)[/tex]:
[tex]\[
(x + 2) \left( \frac{1}{x+2} \right) + 2(x + 2) = (x + 2) \left( \frac{7}{x+2} \right)
\][/tex]
This simplifies to:
[tex]\[
1 + 2(x + 2) = 7
\][/tex]
Expanding and simplifying:
[tex]\[
1 + 2x + 4 = 7
\][/tex]
[tex]\[
2x + 5 = 7
\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[
2x = 2
\][/tex]
[tex]\[
x = 1
\][/tex]
Therefore, the solution to the equation, keeping the restriction [tex]\(x \neq -2\)[/tex] in mind, is:
[tex]\[
x = 1
\][/tex]
