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Solve the system of equations
Answer
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Graph
$x = y - 1$
$2 x + y = 7$
$x$Intercept
$\left ( - 1 , 0 \right )$
$y$Intercept
$\left ( 0 , 1 \right )$
$x$Intercept
$\left ( \dfrac { 7 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 7 \right )$
$x = 2 , y = 3$
Solve the system of equations
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 7 } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ 2 x + y = 7$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 7 }$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 7 }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ Substitute the given $ y $ value into the equation $ x = y - 1$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$x = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$ $ Subtract $ 1 $ from $ 3$
$x = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 7 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 7 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 2 } = \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 7 } = \color{#FF6800}{ 7 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 2 } = \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 7 } = \color{#FF6800}{ 7 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
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