To solve the given rational equation:
[tex]\[
\frac{4}{x+3}+\frac{3}{x-4}=\frac{21}{(x+3)(x-4)}
\][/tex]
we need to determine the restrictions on the variable [tex]\( x \)[/tex].
### Step-by-Step Solution:
1. Identify the Denominators:
- The denominators are [tex]\( x + 3 \)[/tex] and [tex]\( x - 4 \)[/tex].
2. Find When Denominators Are Zero:
- The values of [tex]\( x \)[/tex] that make [tex]\( x + 3 \)[/tex] and [tex]\( x - 4 \)[/tex] equal to zero will cause the denominators to be undefined.
3. Set Each Denominator Equal to Zero and Solve:
- For [tex]\( x + 3 = 0 \)[/tex]:
[tex]\[
x + 3 = 0 \implies x = -3
\][/tex]
- For [tex]\( x - 4 = 0 \)[/tex]:
[tex]\[
x - 4 = 0 \implies x = 4
\][/tex]
### Conclusion:
The variable [tex]\( x \)[/tex] cannot take the values that make the denominators zero. Hence, the restrictions on the variable [tex]\( x \)[/tex] are:
[tex]\[
x \neq -3 \quad \text{and} \quad x \neq 4
\][/tex]
Therefore, the values of [tex]\( x \)[/tex] that cause the denominators to be zero and must be excluded from the solution are [tex]\(-3\)[/tex] and [tex]\(4\)[/tex].
