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Solve the system of equations
Answer
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$x = 2 y + 1$
$2 x + 2 y = 3$
$x$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 $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 6 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-SIBLE POINTS: 4 
On a piece of paper, solve each system using substitution. Then match the system to its solution. 
$ \begin{cases} y=x-2 \\ 2x+2y=4 \end{cases} $ 
$ \begin{cases} y=x+2 \\ 2x+y=2 \end{cases} $ 
$ \begin{cases} x=y+8 \\ x+2y=2 \end{cases} $ 
$ \begin{cases} x=y-2 \\ 3x+y=10 \end{cases} $ 
$=\left(0,2\right)$ $=\left(2,4\right)$ $∴\left(2,0\right)$ $\left(6,-2\right)$
7th-9th grade
Algebra
search-thumbnail-$d\right) \begin{cases} 3 \\ \dfrac {1} {5}x--y-1 \\ 2 \\ 2x=15+10 \end{cases} $ $ \begin{cases} x-2y+1 \\ 2x+y=3 \end{cases} $
10th-13th grade
Algebra
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