Step-by-step explanation
We have the simultaneous equation
![\begin{gathered} 6x-2y=30---equation\text{ 1} \\ -3x+3y=15---equation\text{ 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cyukdw48os9w13ixl960m2gomczt5vj0kf.png)
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
![\begin{gathered} 6(10)-2(15)=30 \\ True \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tk888fo9l6f6nm3vfehnpz92c3nabc890k.png)
Next, for the second equation
![\begin{gathered} -3(10)+3(15)=15 \\ True \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/uy2bsxcyup904zjnki6pm2qi8cofdupu9w.png)
Thus,
We can conclude that (10,15) is a solution