Find the two numbers

There is a trick when the ratio changes is also $(5:6) \rightarrow (5-4:6-4)$

$\begin{aligned} &\hspace{5mm} a+4&: &b+4 &=5&:6 \\ +&\hspace{5mm} a-4&: &b-4 &=1&:2 \\ \hline &\hspace{5mm} 2a&: &2b&=6&:8 \\ &\hspace{5mm} a&: &b&=6&:8 \end{aligned}$

$(a,b) = (6,8)$

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Proof:

If $a,b,x,y > c$ are positive integers

$\left\{ \begin{array}{l}a+\textcolor{#3D99F6}{c}:b+\textcolor{#3D99F6}{c} =x:y \\ a-\textcolor{#3D99F6}{c}:b-\textcolor{#3D99F6}{c} =x-\textcolor{#3D99F6}{c}:y-\textcolor{#3D99F6}{c}\end{array} \right. \\ \left\{ \begin{array}{l}A+1:B+1 =x:y \\ A-1:B-1 =x-c:y-c\end{array} \right. \hspace{10mm} \textcolor{#3D99F6}{A=a/c, B=b/c} \\ \left\{ \begin{array}{l}Ay+y =Bx+x \\ Ay-Ac-y+\cancel{c} = Bx-Bc-x+\cancel{c}\end{array} \right. \\$

Sum up the two equations

$A(2y-c) = B(2x-c)\\ \implies a(2y-c) = b(2x-c)\\ \implies a:b = (2x-c) : (2y-c)\\$

Put $b=\cfrac{(2y-c)a}{(2x-c)}$ in $\cfrac{a+c}{b+c} = \cfrac{x}{y} \implies a = 2x-c = x + (x-c)$ and hence $b = y+(y-c)$

Let the two numbers be x and y. We are given:

(x - 4) / (y - 4) = 1/2 (equation 1) and

(x+4) / (y + 4) = 5/6 (equation 2)

Rearranging equation 1 yields 2 x - 8 = y - 4, so that y = 2 x - 4. Substitution into equation 2 gives:

(x + 4) / (2*x) = 5/6, or

6 x + 24 = 10 x. Hence, x = 6 and y = 2*6 - 4 = 8.