Given the following system of equations:
[tex]\begin{gathered} 6x-5y=3\rightarrow(1) \\ 3x-8y=-15\operatorname{\rightarrow}(2) \end{gathered}[/tex]
We will use the elimination method to solve the system of equations
Multiply the second equation by (-2) to eliminate (x) then solve for (y)
[tex]\begin{gathered} \begin{cases}6x-5y=3 \\ 3x-8y=-15\rightarrow(\times-2){}\end{cases} \\ =============== \\ \begin{cases}6x-5y=3{} \\ -6x+16y={30}\end{cases} \\ =============== \\ 0+11y=33 \\ y=\frac{33}{11}=3 \end{gathered}[/tex]
Substitute with (y) into equation (1) to find (x):
[tex]\begin{gathered} 6x-5(3)=3 \\ 6x-15=3 \\ 6x=3+15 \\ 6x=18 \\ x=\frac{18}{6} \\ x=3 \end{gathered}[/tex]
So, the answer will be (x, y) = (3, 3)
