# Calculate: x^2+2y=2x+2y=6

## Expression: ${x}^{2}+2y=2x+2y=6$

Write as a system of equations

$\left\{\begin{array} { l } {x}^{2}+2y=6 \\ 2x+2y=6\end{array} \right.$

Multiply both sides of the equation by $-1$

$\left\{\begin{array} { l } {x}^{2}+2y=6 \\ -2x-2y=-6\end{array} \right.$

Sum the equations vertically to eliminate at least one variable

${x}^{2}-2x=0$

Solve the equation for $x$

$\begin{array} { l }x=0,\\x=2\end{array}$

Substitute the given value of $x$ into the equation $2x+2y=6$

$\begin{array} { l }2 \times 0+2y=6,\\x=2\end{array}$

Substitute the given value of $x$ into the equation $2x+2y=6$

$\begin{array} { l }2 \times 0+2y=6,\\2 \times 2+2y=6\end{array}$

Solve the equation for $y$

$\begin{array} { l }y=3,\\2 \times 2+2y=6\end{array}$

Solve the equation for $y$

$\begin{array} { l }y=3,\\y=1\end{array}$

The possible solutions of the system are the ordered pairs $\left( x, y\right)$

$\begin{array} { l }\left( x_1, y_1\right)=\left( 0, 3\right),\\\left( x_2, y_2\right)=\left( 2, 1\right)\end{array}$

Check if the given ordered pairs are the solutions of the system of equations

$\begin{array} { l }{0}^{2}+2 \times 3=2 \times 0+2 \times 3=6,\\{2}^{2}+2 \times 1=2 \times 2+2 \times 1=6\end{array}$

Simplify the expression

$\begin{array} { l }6=6=6,\\{2}^{2}+2 \times 1=2 \times 2+2 \times 1=6\end{array}$

Simplify the expression

$\begin{array} { l }6=6=6,\\6=6=6\end{array}$

Since all of the equalities are true, the ordered pairs are the solution of the system

$\begin{array} { l }\left( x_1, y_1\right)=\left( 0, 3\right),\\\left( x_2, y_2\right)=\left( 2, 1\right)\end{array}$

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