# Calculator search results

Formula
Solve the system of equations
Graph
$3 x - 4 y = 4$
$- 2 x + 3 y = - 2$
$x$-intercept
$\left ( \dfrac { 4 } { 3 } , 0 \right )$
$y$-intercept
$\left ( 0 , - 1 \right )$
$x$-intercept
$\left ( 1 , 0 \right )$
$y$-intercept
$\left ( 0 , - \dfrac { 2 } { 3 } \right )$
$\begin{cases} 3x-4y = 4 \\-2x+3y = -2 \end{cases}$
$x = 4 , y = 2$
Solve quadratic equations using the square root
$\begin{cases} 3 x - 4 y = 4 \\ - 2 x + 3 y = - 2 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ - 2 x + 3 y = - 2 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
 Substitute the given $x$ value into the equation $- 2 x + 3 y = - 2$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
 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 $x = \dfrac { 4 } { 3 } y + \dfrac { 4 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$x = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } + \dfrac { 4 } { 3 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ \dfrac { 8 } { 3 } } + \dfrac { 4 } { 3 }$
$x = \color{#FF6800}{ \dfrac { 8 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
 Find the sum of the fractions 
$x = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$\begin{cases} x = 4 \\ y = 2 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ 12 } \\ \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ 12 } \\ y = 2 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \\ y = 2 \end{cases}$
 그래프 보기 
Graph
Have you found the solution you wanted?
Try again
Try more features at QANDA!
Search by problem image
Ask 1:1 question to TOP class teachers
AI recommend problems and video lecture