# Evaluate: {\text{begin}array l 3x+y=0 } x-3y+2=0\text{end}array .

## Expression: $\left\{\begin{array} { l } 3x+y=0 \\ x-3y+2=0\end{array} \right.$

Solve the equation for $y$

$\left\{\begin{array} { l } y=-3x \\ x-3y+2=0\end{array} \right.$

Substitute the given value of $y$ into the equation $x-3y+2=0$

$x-3 \times \left( -3x \right)+2=0$

Solve the equation for $x$

$x=-\frac{ 1 }{ 5 }$

Substitute the given value of $x$ into the equation $y=-3x$

$y=-3 \times \left( -\frac{ 1 }{ 5 } \right)$

Solve the equation for $y$

$y=\frac{ 3 }{ 5 }$

The possible solution of the system is the ordered pair $\left( x, y\right)$

$\left( x, y\right)=\left( -\frac{ 1 }{ 5 }, \frac{ 3 }{ 5 }\right)$

Check if the given ordered pair is the solution of the system of equations

$\left\{\begin{array} { l } 3 \times \left( -\frac{ 1 }{ 5 } \right)+\frac{ 3 }{ 5 }=0 \\ -\frac{ 1 }{ 5 }-3 \times \frac{ 3 }{ 5 }+2=0\end{array} \right.$

Simplify the equalities

$\left\{\begin{array} { l } 0=0 \\ 0=0\end{array} \right.$

Since all of the equalities are true, the ordered pair is the solution of the system

$\left( x, y\right)=\left( -\frac{ 1 }{ 5 }, \frac{ 3 }{ 5 }\right)$

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