Can somebody help with this problem?

Two numbers are in the ratio 5 to 8. When 2 is added to each, the ratio of the resulting numbers is 2 to 3. Find the numbers.

Two numbers are in the ratio 5 to 8.

When 2 is added to each, the ratio of the resulting numbers is 2 to 3.

Find the numbers.

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{x}{y} &=& \dfrac{5}{8} \quad & | \quad \cdot y \\\\ & \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{5}{8}\cdot y } \\\\ \hline \\ (2) & \dfrac{x+2}{y+2} &=& \dfrac{2}{3} \quad & | \quad \cdot (y+2) \\\\ & x+2 &=& \dfrac{2}{3} \cdot (y+2) \quad & | \quad \cdot \dfrac{3}{2} \\\\ & \dfrac{3}{2}\cdot(x+2) &=& y+2 \quad & | \quad x =\dfrac{5}{8}\cdot y \\\\ & \dfrac{3}{2}\cdot\left(\dfrac{5}{8}\cdot y+2 \right) &=& y+2 \\\\ & \dfrac{3}{2}\cdot\dfrac{5}{8}\cdot y+2\cdot \dfrac{3}{2}&=& y+2 \\\\ & \dfrac{15}{16}\cdot y+3&=& y+2 \\\\ & y+2 &=& \dfrac{15}{16}\cdot y+3 \quad & | \quad -\dfrac{15}{16}\cdot y \\\\ & y-\dfrac{15}{16}\cdot y+2 &=& 3 \quad & | \quad - 2 \\\\ & y-\dfrac{15}{16}\cdot y &=& 3 - 2 \\\\ & \dfrac{1}{16}\cdot y &=& 1 \quad & | \quad \cdot 16 \\\\ & \mathbf{ y } &\mathbf{=}& \mathbf{16} \\\\ \hline \\ & x &=& \dfrac{5}{8}\cdot y \quad & | \quad y=16 \\\\ & x &=& \dfrac{5}{8}\cdot 16 \\\\ & \mathbf{ x } &\mathbf{=}& \mathbf{10} \\ \hline \end{array}\)

The numbers are 10 and 16

Let x be one number and y the second

We have

x /  y  = 5 / 8      multiply through by  y

x  = (5/8)y      (1)

Also

(x + 2) / ( y + 2)    =  2/3    cross-multiply

3(x + 2)  = 2(y + 2)     simplify

3x + 6   = 2y + 4         sub (1) in for x

3(5/8 * y) + 6  =  2y + 4   simplify

(15/8 ) y  + 6   = 2y  + 4       subtract 4, (15/8)y  from both sides

2  = (1/8)  y     multiply through by 8

16  = y      and x  = (5/8((16)   =  10

5x / 8x

(5x+2) / (8x+2) = 2/3      cross multiply

15x+6 = 16x+4

2= x   

5x= 10   8x = 16