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Solve the system of equations
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$3 x - 2 y = 16$
$3 y - 2 x = 6$
$x$Intercept
$\left ( \dfrac { 16 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 8 \right )$
$x$Intercept
$\left ( - 3 , 0 \right )$
$y$Intercept
$\left ( 0 , 2 \right )$
$\begin{cases} 3x-2y = 16 \\3y-2x = 6 \end{cases}$
$x = 12 , y = 10$
Solve the system of equations
$\begin{cases} 3 x - 2 y = 16 \\ 3 y - 2 x = 6 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } } \\ 3 y - 2 x = 6 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \end{cases}$
 Substitute the given $x$ value into the equation $3 y - 2 x = 6$
$\color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } } \right ) = \color{#FF6800}{ 6 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } } \right ) = \color{#FF6800}{ 6 }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
 Substitute the given $y$ value into the equation $x = \dfrac { 2 } { 3 } y + \dfrac { 16 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } }$
$x = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } + \dfrac { 16 } { 3 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ \dfrac { 20 } { 3 } } + \dfrac { 16 } { 3 }$
$x = \color{#FF6800}{ \dfrac { 20 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } }$
 Find the sum of the fractions 
$x = \color{#FF6800}{ 12 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 12 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 12 } , \color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 12 } , \color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 12 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } = \color{#FF6800}{ 16 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 12 } = \color{#FF6800}{ 6 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 12 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } = \color{#FF6800}{ 16 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 12 } = \color{#FF6800}{ 6 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 16 } = \color{#FF6800}{ 16 } \\ \color{#FF6800}{ 6 } = \color{#FF6800}{ 6 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 16 } = \color{#FF6800}{ 16 } \\ \color{#FF6800}{ 6 } = \color{#FF6800}{ 6 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 12 } , \color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
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