Respuesta :

We are given the following system of equations:

[tex]\begin{gathered} 3x+3y=42,\text{ (1)} \\ x-3y=-22\text{, (2)} \end{gathered}[/tex]

Equation (1) can be rewritten by dividing by 3 on both sides as:

[tex]x+y=14,\text{ (1)}[/tex]

Solving for "x" in equation (1), we get:

[tex]x=14-y[/tex]

Replacing in equation (2)

[tex]\begin{gathered} x-3y=-22 \\ 14-y-3y=-22 \end{gathered}[/tex]

simplifying

[tex]14-4y=-22[/tex]

Now we solve for "y", by subtracting 14 on both sides:

[tex]\begin{gathered} 14-14-4y=-22-14 \\ -4y=-36 \end{gathered}[/tex]

Dividing by -4 on both sides:

[tex]y=-\frac{36}{-4}=9[/tex]

Replacing the value of "y" in equation (1)

[tex]\begin{gathered} x+y=14 \\ x+9=14 \end{gathered}[/tex]

Now we subtract 9 on both sides:

[tex]\begin{gathered} x+9-9=14-9 \\ x=5 \end{gathered}[/tex]

Therefore, the solution of the system is:

[tex](x,y)=(5,9)[/tex]

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