Please help me I have been stuck on this question for ages

Answer:28 green and 35 red
Step-by-step explanation:
Given
If there are r red counter and g green counter then
Probability of drawing a green counter is [tex]P(g)=\frac{4}{9}[/tex]
and [tex]P(g)=\frac{\text{No of g counter}}{\text{Total no of counter}}[/tex]
Thus [tex]\frac{\text{No of g counter}}{\text{Total no of counter}}=\frac{4}{9}[/tex]
[tex]\frac{g}{g+r}=\frac{4}{9}[/tex]
[tex]\Rightarrow 9g=4g+4r[/tex]
[tex]\Rightarrow 5g=4r\quad \ldots(i)[/tex]
Also if 4 red and 2 green counter is added the probability of drawing a green counter is
[tex]P(g)=\frac{10}{23}=\frac{\text{No of g counter}}{\text{Total no of counter}}[/tex]
[tex]\Rightarrow \frac{10}{23}=\frac{g+2}{g+2+r+4}[/tex]
[tex]\Rightarrow \frac{10}{23}=\frac{g+2}{g+r+6}[/tex]
[tex]\Rightarrow 10g+10r+60=23g+46[/tex]
[tex]\Rightarrow 10r+14=13g\ quad \ldots(ii)[/tex]
Substitute the value of g in equation (ii)[/tex]
[tex]\Rightarrow 10\times \frac{5}{4}g+14=13g[/tex]
[tex]\Rightarrow \frac{25}{2}g+14=13g[/tex]
[tex]\Rightarrow g=28[/tex]
Therefore [tex]r=35[/tex]
Thus there 28 green counter and 35 red counter