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Formula
Solve the system of equations
Graph
$x = 2 y + 1$
$2 x + 2 y = 3$
$x$Intercept
$\left ( 1 , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 1 } { 2 } \right )$
$x$Intercept
$\left ( \dfrac { 3 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 3 } { 2 } \right )$
$\begin{cases} x = 2y+1 \\2x+2y = 3 \end{cases}$
$x = \dfrac { 4 } { 3 } , y = \dfrac { 1 } { 6 }$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \end{cases}$
 Substitute the given $x$ value into the equation $2 x + 2 y = 3$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 6 } }$
$\color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 6 } }$
 Substitute the given $y$ value into the equation $x = 2 y + 1$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$x = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 6 } } + 1$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ \dfrac { 1 } { 3 } } + 1$
$x = \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
 Add two numbers $\dfrac { 1 } { 3 }$ and $1$
$x = \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 6 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 6 } }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ \dfrac { 4 } { 3 } } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 6 } } = \color{#FF6800}{ 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 4 } { 3 } } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 6 } } = \color{#FF6800}{ 3 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ \dfrac { 4 } { 3 } } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 4 } { 3 } } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 6 } }$
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