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Solve the system of equations
Answer
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$8 x + 8 y = 1$
$4 x + 10 y = 1$
$x$Intercept
$\left ( \dfrac { 1 } { 8 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 1 } { 8 } \right )$
$x$Intercept
$\left ( \dfrac { 1 } { 4 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 1 } { 10 } \right )$
$\begin{cases} 8x+8y = 1 \\4x+10y = 1 \end{cases}$
$x = \dfrac { 1 } { 24 } , y = \dfrac { 1 } { 12 }$
Solve the system of equations
$\begin{cases} 8 x + 8 y = 1 \\ 4 x + 10 y = 1 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 8 } } \\ 4 x + 10 y = 1 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 8 } } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ y } = \color{#FF6800}{ 1 } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ 4 x + 10 y = 1$
$\color{#FF6800}{ 4 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 8 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ 4 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 8 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 12 } }$
$\color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 12 } }$
$ $ Substitute the given $ y $ value into the equation $ x = - y + \dfrac { 1 } { 8 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 8 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 8 } }$
$ $ Find the sum or difference of the fractions $ $
$x = \color{#FF6800}{ \dfrac { 1 } { 24 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 24 } }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 24 } } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 12 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 24 } } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 12 } }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 24 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 12 } } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 24 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 12 } } = \color{#FF6800}{ 1 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 24 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 12 } } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 24 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 12 } } = \color{#FF6800}{ 1 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 24 } } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 12 } }$
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