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Solve the system of equations
Answer
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Graph
$2 x + y = 7$
$x + 2 y = 5$
$x$Intercept
$\left ( \dfrac { 7 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 7 \right )$
$x$Intercept
$\left ( 5 , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 5 } { 2 } \right )$
$\begin{cases} 2x+y = 7 \\x+2y = 5 \end{cases}$
$x = 3 , y = 1$
Solve quadratic equations using the square root
$\begin{cases} 2 x + y = 7 \\ x + 2 y = 5 \end{cases}$
$ $ Solve a solution to $ y$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \\ x + 2 y = 5 \end{cases}$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 5 } \end{cases}$
$ $ Substitute the given $ y $ value into the equation $ x + 2 y = 5$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \right ) = \color{#FF6800}{ 5 }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \right ) = \color{#FF6800}{ 5 }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 }$
$ $ Substitute the given $ x $ value into the equation $ y = - 2 x + 7$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 7 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 7 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 7 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ 5 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 7 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ 5 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 7 } = \color{#FF6800}{ 7 } \\ \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 7 } = \color{#FF6800}{ 7 } \\ \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
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