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Formula

Solve the system of equations

Answer

Graph

$x + 2 y = 2$

$2 x - 3 y = 4$

$x$Intercept

$\left ( 2 , 0 \right )$

$y$Intercept

$\left ( 0 , 1 \right )$

$x$Intercept

$\left ( 2 , 0 \right )$

$y$Intercept

$\left ( 0 , - \dfrac { 4 } { 3 } \right )$

$\begin{cases} x+2y = 2 \\2x-3y = 4 \end{cases}$

$x = 2 , y = 0$

Solve the system of equations

$\begin{cases} x + 2 y = 2 \\ 2 x - 3 y = 4 \end{cases}$

$ $ Solve a solution to $ x$

$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \\ 2 x - 3 y = 4 \end{cases}$

$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ 4 } \end{cases}$

$ $ Substitute the given $ x $ value into the equation $ 2 x - 3 y = 4$

$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ 4 }$

$ $ Solve a solution to $ y$

$\color{#FF6800}{ y } = \color{#FF6800}{ 0 }$

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

$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 0 } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$

$ $ Organize the expression $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$

$ $ The possible solutions are as follows $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 0 }$

$ $ Check if it is the solution to the system of equations $ $

$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 0 } = \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 0 } = \color{#FF6800}{ 4 } \end{cases}$

$ $ Simplify the equality $ $

$\begin{cases} \color{#FF6800}{ 2 } = \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \end{cases}$

$ $ Since it is true in both equations, it is the solution of the system of equations $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 0 }$

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