Let's solve in $ℝ$: $\frac{12}{A}+\frac{24}{A}=A$

ⓘ Like terms are grouped by adding their coefficients: $\frac{12}{A}+\frac{24}{A}$ = $\frac{36}{A}$

$\frac{36}{A}=A$

We multiply both sides by $A$, nonzero for the fractions to be defined

$\frac{36}{A}·A=A·A$

ⓘ Multiplying a fraction by a term multiplies its numerator: $\frac{36}{A}·A$ = $\frac{36A}{A}$■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor $A$: $\frac{36A}{A}$ = $36$■Multiplying powers with the same base adds the exponents: $A·A$ = $A^2$

$36=A^2$

$A^2=36$

The two numbers whose square is $36$ are $-6$ and $6$

Candidates, pending the domain check: $A=\{-6,6\}$

For the expression $\frac{12}{A}$ to be defined, we must have: $A\ne 0$

Domain constraints: $A\ne 0$

▶Solution:   $A=\{-6,6\}$