We have the system of equations:
[tex]\begin{gathered} \frac{x}{6}+\frac{y}{4}=6 \\ \frac{5x}{6}-\frac{y}{3}=11 \end{gathered}[/tex]
We can solve it by elimination: we will multiply the first equation by 5 and substract from the second equation. Then, we will eliminate x and we can solve for y.
We can write this as:
[tex]\begin{gathered} (\frac{5}{6}x-\frac{1}{3}y)-5(\frac{1}{6}x+\frac{1}{4}y)=11-5(6) \\ \frac{5}{6}x-\frac{1}{3}y-\frac{5}{6}x-\frac{5}{4}y=11-30 \\ 0x-(\frac{1}{3}+\frac{5}{4})y=-19 \\ -(\frac{4+15}{12})y=-19 \\ \frac{19}{12}y=19 \\ y=19\cdot\frac{12}{19} \\ y=12 \end{gathered}[/tex]
Now, with the value of y, we can solve for x as:
[tex]\begin{gathered} \frac{1}{6}x+\frac{1}{4}y=6 \\ \frac{1}{6}x+\frac{1}{4}\cdot12=6 \\ \frac{1}{6}x+3=6 \\ \frac{1}{6}x=6-3 \\ \frac{1}{6}x=3 \\ x=3\cdot6 \\ x=18 \end{gathered}[/tex]
Answer: x=18 and y=12
