# Calculator search results

Formula
Solve the system of equations
Graph
$0.3 x + 0.4 y = 1.7$
$\dfrac { 2 } { 3 } x + \dfrac { 1 } { 2 } y = 3$
$x$-intercept
$\left ( \dfrac { 17 } { 3 } , 0 \right )$
$y$-intercept
$\left ( 0 , \dfrac { 17 } { 4 } \right )$
$x$-intercept
$\left ( \dfrac { 9 } { 2 } , 0 \right )$
$y$-intercept
$\left ( 0 , 6 \right )$
$\begin{cases} 0.3x+0.4y = 1.7 \\ \dfrac{ 2 }{ 3 } x+ \dfrac{ 1 }{ 2 } y = 3 \end{cases}$
$x = 3 , y = 2$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ x } + 0.4 y = 1.7 \\ \dfrac { 2 } { 3 } x + \dfrac { 1 } { 2 } y = 3 \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \color{#FF6800}{ \dfrac { 3 x } { 10 } } + 0.4 y = 1.7 \\ \dfrac { 2 } { 3 } x + \dfrac { 1 } { 2 } y = 3 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \color{#FF6800}{ 0.4 } \color{#FF6800}{ y } = 1.7 \\ \dfrac { 2 } { 3 } x + \dfrac { 1 } { 2 } y = 3 \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \dfrac { 3 x } { 10 } + \color{#FF6800}{ \dfrac { 2 y } { 5 } } = 1.7 \\ \dfrac { 2 } { 3 } x + \dfrac { 1 } { 2 } y = 3 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { 2 y } { 5 } = \color{#FF6800}{ 1.7 } \\ \dfrac { 2 } { 3 } x + \dfrac { 1 } { 2 } y = 3 \end{cases}$
 Convert decimals to fractions 
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { 2 y } { 5 } = \color{#FF6800}{ \dfrac { 17 } { 10 } } \\ \dfrac { 2 } { 3 } x + \dfrac { 1 } { 2 } y = 3 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { 2 y } { 5 } = \dfrac { 17 } { 10 } \\ \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ x } + \dfrac { 1 } { 2 } y = 3 \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { 2 y } { 5 } = \dfrac { 17 } { 10 } \\ \color{#FF6800}{ \dfrac { 2 x } { 3 } } + \dfrac { 1 } { 2 } y = 3 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { 2 y } { 5 } = \dfrac { 17 } { 10 } \\ \dfrac { 2 x } { 3 } + \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ y } = 3 \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { 2 y } { 5 } = \dfrac { 17 } { 10 } \\ \dfrac { 2 x } { 3 } + \color{#FF6800}{ \dfrac { y } { 2 } } = 3 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { 2 y } { 5 } = \dfrac { 17 } { 10 } \\ \dfrac { 2 x } { 3 } + \dfrac { y } { 2 } = 3 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 17 } { 3 } } \\ \dfrac { 2 x } { 3 } + \dfrac { y } { 2 } = 3 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 17 } { 3 } } \\ \color{#FF6800}{ \dfrac { 2 x } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 2 } } = \color{#FF6800}{ 3 } \end{cases}$
 Substitute the given $x$ value into the equation $\dfrac { 2 x } { 3 } + \dfrac { y } { 2 } = 3$
$\color{#FF6800}{ \dfrac { 2 \left ( - \dfrac { 4 } { 3 } y + \dfrac { 17 } { 3 } \right ) } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 2 } } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ \dfrac { 2 \left ( - \dfrac { 4 } { 3 } y + \dfrac { 17 } { 3 } \right ) } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 2 } } = \color{#FF6800}{ 3 }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
 Substitute the given $y$ value into the equation $x = - \dfrac { 4 } { 3 } y + \dfrac { 17 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 17 } { 3 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } + \dfrac { 17 } { 3 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 8 } { 3 } } + \dfrac { 17 } { 3 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 8 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 17 } { 3 } }$
 Find the sum or difference of the fractions 
$x = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 0.4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 1.7 } \\ \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 0.4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 1.7 } \\ \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ \dfrac { 17 } { 10 } } = \color{#FF6800}{ \dfrac { 17 } { 10 } } \\ \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 17 } { 10 } } = \color{#FF6800}{ \dfrac { 17 } { 10 } } \\ \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$\begin{cases} x = 3 \\ y = 2 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 0.4 } \color{#FF6800}{ y } = \color{#FF6800}{ 1.7 } \\ \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 0.4 } \color{#FF6800}{ y } = \color{#FF6800}{ 1.7 } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 60 } } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 30 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 0.4 } \color{#FF6800}{ y } = \color{#FF6800}{ 1.7 } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 60 } } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 30 } } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 200 } } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 21 } { 200 } } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 60 } } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 30 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 200 } } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 21 } { 200 } } \\ - \dfrac { 7 } { 60 } y = - \dfrac { 7 } { 30 } \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \\ - \dfrac { 7 } { 60 } y = - \dfrac { 7 } { 30 } \end{cases}$
$\begin{cases} x = 3 \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 60 } } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 30 } } \end{cases}$
 Solve a solution to $y$
$\begin{cases} x = 3 \\ \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$
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