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Formula
Solve the system of equations
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$7 x - 4 y = 5$
$9 x + 8 y = 13$
$x$Intercept
$\left ( \dfrac { 5 } { 7 } , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 5 } { 4 } \right )$
$x$Intercept
$\left ( \dfrac { 13 } { 9 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 13 } { 8 } \right )$
$\begin{cases} 7x-4y = 5 \\9x+8y = 13 \end{cases}$
$x = 1 , y = \dfrac { 1 } { 2 }$
Solve quadratic equations using the square root
$\begin{cases} 7 x - 4 y = 5 \\ 9 x + 8 y = 13 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 7 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 7 } } \\ 9 x + 8 y = 13 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 7 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 7 } } \\ \color{#FF6800}{ 9 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ y } = \color{#FF6800}{ 13 } \end{cases}$
 Substitute the given $x$ value into the equation $9 x + 8 y = 13$
$\color{#FF6800}{ 9 } \left ( \color{#FF6800}{ \dfrac { 4 } { 7 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 7 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ y } = \color{#FF6800}{ 13 }$
$\color{#FF6800}{ 9 } \left ( \color{#FF6800}{ \dfrac { 4 } { 7 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 7 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ y } = \color{#FF6800}{ 13 }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$\color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 2 } }$
 Substitute the given $y$ value into the equation $x = \dfrac { 4 } { 7 } y + \dfrac { 5 } { 7 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 7 } } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 7 } }$
$x = \color{#FF6800}{ \dfrac { 4 } { 7 } } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } } + \dfrac { 5 } { 7 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ \dfrac { 2 } { 7 } } + \dfrac { 5 } { 7 }$
$x = \color{#FF6800}{ \dfrac { 2 } { 7 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 7 } }$
 Find the sum of the fractions 
$x = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 2 } }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 9 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \color{#FF6800}{ 13 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 9 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \color{#FF6800}{ 13 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 13 } = \color{#FF6800}{ 13 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 13 } = \color{#FF6800}{ 13 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 2 } }$
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