Answer:
The solution to the system of equations is;
[tex]\begin{gathered} x=0 \\ y=2 \end{gathered}[/tex]
Explanation:
Given the system of equations;
[tex]\begin{gathered} 2x+4y=8\text{ ----------1} \\ x=3y-6 \\ x-3y=-6\text{ ----------2} \end{gathered}[/tex]
Multiplying equation 2 by 2 and subtracting from equation 1;
[tex]\begin{gathered} x-3y=-6 \\ \times \\ 2 \\ = \\ 2x-6y=-12\text{ -----------2.1} \end{gathered}[/tex][tex]\begin{gathered} 2x+4y=8\text{ ----------1} \\ - \\ 2x-6y=-12\text{ -----------2.1} \\ = \\ 0+10y=20 \\ 10y=20 \\ y=\frac{20}{10} \\ y=2 \end{gathered}[/tex]
We can then substitute the value of y into equation 1, to get x;
[tex]\begin{gathered} 2x+4y=8 \\ 2x+4(2)=8 \\ 2x=8-8 \\ 2x=0 \\ x=0 \end{gathered}[/tex]
Therefore, the solution to the system of equations is;
[tex]\begin{gathered} x=0 \\ y=2 \end{gathered}[/tex]
