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Solve the system of equations

Answer

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$x - y = 3$

$x + y = 5$

$x$-intercept

$\left ( 3 , 0 \right )$

$y$-intercept

$\left ( 0 , - 3 \right )$

$x$-intercept

$\left ( 5 , 0 \right )$

$y$-intercept

$\left ( 0 , 5 \right )$

$\begin{cases} x-y = 3 \\x+y = 5 \end{cases}$

$x = 4 , y = 1$

Solve quadratic equations using the square root

$\begin{cases} x - y = 3 \\ x + y = 5 \end{cases}$

$ $ Solve a solution to $ x$

$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \\ x + y = 5 \end{cases}$

$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 5 } \end{cases}$

$ $ Substitute the given $ x $ value into the equation $ x + y = 5$

$\left ( \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 5 }$

$ $ Solve a solution to $ y$

$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$

$ $ Substitute the given $ y $ value into the equation $ x = y + 3$

$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$

$x = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$

$ $ Add $ 1 $ and $ 3$

$x = \color{#FF6800}{ 4 }$

$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$

$ $ The possible solutions are as follows $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$

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

$\begin{cases} \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 5 } \end{cases}$

$ $ Simplify the equality $ $

$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \end{cases}$

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

$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$

$\begin{cases} x = 4 \\ y = 1 \end{cases}$

Solve quadratic equations using the square root

$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 5 } \end{cases}$

$ $ Solve the system of linear equations for $ x , y $

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

$ $ Solve the system of linear equations for $ x , y $

$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 8 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$

$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 8 } \\ 2 y = 2 \end{cases}$

$ $ Solve a solution to $ x$

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

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

$ $ Solve a solution to $ y$

$\begin{cases} x = 4 \\ \color{#FF6800}{ y } = \color{#FF6800}{ 1 } \end{cases}$

$ $ 그래프 보기 $ $

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