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Formula
Solve the system of equations
Graph
$8 x - y = 29$
$2 x + y = 11$
$x$Intercept
$\left ( \dfrac { 29 } { 8 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 29 \right )$
$x$Intercept
$\left ( \dfrac { 11 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 11 \right )$
$\begin{cases} 8x-y = 29 \\2x+y = 11 \end{cases}$
$x = 4 , y = 3$
Solve quadratic equations using the square root
$\begin{cases} 8 x - y = 29 \\ 2 x + y = 11 \end{cases}$
 Solve a solution to $y$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 29 } \\ 2 x + y = 11 \end{cases}$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 29 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 11 } \end{cases}$
 Substitute the given $y$ value into the equation $2 x + y = 11$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 29 } \right ) = \color{#FF6800}{ 11 }$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 29 } \right ) = \color{#FF6800}{ 11 }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
 Substitute the given $x$ value into the equation $y = 8 x - 29$
$\color{#FF6800}{ y } = \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 29 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 29 }$
 Organize the expression 
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 29 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 11 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 29 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 11 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 29 } = \color{#FF6800}{ 29 } \\ \color{#FF6800}{ 11 } = \color{#FF6800}{ 11 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 29 } = \color{#FF6800}{ 29 } \\ \color{#FF6800}{ 11 } = \color{#FF6800}{ 11 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
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