Explanation
We have the simultaneous equation
[tex]\begin{gathered} 6x-2y=30---equation\text{ 1} \\ -3x+3y=15---equation\text{ 2} \end{gathered}[/tex]To check if (10,15) is a solution, we will have to substitute
x=10 and y=15
into both equations and check if it is true
So for equation 1
[tex]\begin{gathered} 6(10)-2(15)=30 \\ True \end{gathered}[/tex]Next, for the second equation
[tex]\begin{gathered} -3(10)+3(15)=15 \\ True \end{gathered}[/tex]Thus,
We can conclude that (10,15) is a solution