Answer
$\{1,7\}$.
Work Step by Step
Replace $x$ with $a$ in the given functiions:
$f(a)=\frac{5}{a-4},g(a)=\frac{3}{a-3}$ and $h(a)=\frac{a^2-20}{a^2-7a+12}$
The given expression is
$\Rightarrow (f+g)(a)=h(a)$
Or we can write using the formula for the sum of two functions:
$\Rightarrow f(a)+g(a)=h(a)$
Substitute all fucntions.
$\Rightarrow \frac{5}{a-4}+\frac{3}{a-3}=\frac{a^2-20}{a^2-7a+12}$
In order to clear fractions we determine the Least Common Denominator (LCD) of all denominators.
Factor the denominator on the right hand side.
Factor $a^2-7a+12$
Rewrite the middle term $-7a$ as $-4a-3a$.
$\Rightarrow a^2-4a-3a+12$
Group terms.
$\Rightarrow (a^2-4a)+(-3a+12)$
Factor each group.
$\Rightarrow a(a-4)-3(a-4)$
Factor out $(a-4)$.
$\Rightarrow (a-4)(a-3)$
Back substitute into the equation.
$\Rightarrow \frac{5}{a-4}+\frac{3}{a-3}=\frac{a^2-20}{(a-4)(a-3)}$....(1)
Multiply both sides by the LCD $(a-4)(a-3)$
$\Rightarrow (a-4)(a-3)\left (\frac{5}{a-4}+\frac{3}{a-3}\right )=(a-4)(a-3)\left(\frac{a^2-20}{(a-4)(a-3)}\right )$
Use the distributive property.
$\Rightarrow (a-4)(a-3)\left (\frac{5}{a-4}\right )+(a-4)(a-3)\left (\frac{3}{a-3}\right )=(a-4)(a-3)\left(\frac{a^2-20}{(a-4)(a-3)}\right )$
Cancel common terms.
$\Rightarrow 5(a-3)+3(a-4)=a^2-20$
Use distributive property.
$\Rightarrow 5a-15+3a-12=a^2-20$
Add like terms.
$\Rightarrow 8a-27=a^2-20$
Add $-8a+27$ to both sides.
$\Rightarrow 8a-27-8a+27=a^2-20-8x+27$
Simplify.
$\Rightarrow 0=a^2-8a+7$
Or we can write.
$\Rightarrow a^2-8a+7=0$
Rewrite the middle term $-8a$ as $-7a-1a$.
$\Rightarrow a^2-7a-1a+7=0$
Group terms.
$\Rightarrow (a^2-7a)+(-1a+7)=0$
Factor each group.
$\Rightarrow a(a-7)-1(a-7)=0$
Factor out $(a-7)$.
$\Rightarrow (a-7)(a-1)=0$
Set each factor equal zero.
$\Rightarrow a-7=0$ or $a-1=0$
Isolate $a$.
$\Rightarrow a=7$ or $a=1$
The solution set is $\{1,7\}$.
Note: We have to check if the values of $a$ that we got check the equation. From (1) we find that the equation is defined for all real values of $a$ except those for which denominators are zero, therefore for all real $a$ except $3$ and $4$. Since we got the values $1$ and $7$ for $a$, which are different from $3$ and $4$, it means they are correct.
