Calculator search results
Formula
Solve the system of equations
Answer
Graph
$y = 2 x - 1$
$x + 2 y = 8$
$x$-intercept
$\left ( \dfrac { 1 } { 2 } , 0 \right )$
$y$-intercept
$\left ( 0 , - 1 \right )$
$x$-intercept
$\left ( 8 , 0 \right )$
$y$-intercept
$\left ( 0 , 4 \right )$
$\begin{cases} y = 2x-1 \\x+2y = 8 \end{cases}$
$x = 2 , y = 3$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 8 } \end{cases}$
$ $ Substitute the given $ y $ value into the equation $ x + 2 y = 8$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 8 }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 8 }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$ $ Substitute the given $ x $ value into the equation $ y = 2 x - 1$
$\color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$y = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } - 1$
$ $ Multiply $ 2 $ and $ 2$
$y = \color{#FF6800}{ 4 } - 1$
$y = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$ $ Subtract $ 1 $ from $ 4$
$y = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ 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 $ $
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 8 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 8 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 8 } = \color{#FF6800}{ 8 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 8 } = \color{#FF6800}{ 8 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\begin{cases} x = 2 \\ y = 3 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ x + 2 y = 8 \end{cases}$
$ $ Do transposition $ $
$\begin{cases} \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ x + 2 y = 8 \end{cases}$
$\begin{cases} \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 8 } \end{cases}$
$ $ Solve the system of linear equations for $ x , y $
$\begin{cases} \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \end{cases}$
$\begin{cases} \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \end{cases}$
$ $ Solve the system of linear equations for $ x , y $
$\begin{cases} \color{#FF6800}{ 10 } \color{#FF6800}{ x } = \color{#FF6800}{ 20 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 10 } \color{#FF6800}{ x } = \color{#FF6800}{ 20 } \\ - 5 y = - 15 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \\ - 5 y = - 15 \end{cases}$
$\begin{cases} x = 2 \\ \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \end{cases}$
$ $ Solve a solution to $ y$
$\begin{cases} x = 2 \\ \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \end{cases}$
$ $ 그래프 보기 $ $
Graph
Have you found the solution you wanted?
Try againTry more features at QANDA!
Search by problem image
Ask 1:1 question to TOP class teachers
AI recommend problems and video lecture
