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Solve the system of equations
Answer
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$\begin{cases} w-2z = 10 \\2w+3z = -8 \end{cases}$
$w = 2 , z = - 4$
Solve quadratic equations using the square root
$\begin{cases} w - 2 z = 10 \\ 2 w + 3 z = - 8 \end{cases}$
$ $ Solve a solution to $ w$
$\begin{cases} \color{#FF6800}{ w } = \color{#FF6800}{ 2 } \color{#FF6800}{ z } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \\ 2 w + 3 z = - 8 \end{cases}$
$\begin{cases} \color{#FF6800}{ w } = \color{#FF6800}{ 2 } \color{#FF6800}{ z } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ w } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ z } = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{cases}$
$ $ Substitute the given $ w $ value into the equation $ 2 w + 3 z = - 8$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ z } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ z } = \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ z } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ z } = \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$ $ Solve a solution to $ z$
$\color{#FF6800}{ z } = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ z } = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$ $ Substitute the given $ z $ value into the equation $ w = 2 z + 10$
$\color{#FF6800}{ w } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 10 }$
$\color{#FF6800}{ w } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 10 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ w } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ w } = \color{#FF6800}{ 2 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ w } = \color{#FF6800}{ 2 } , \color{#FF6800}{ z } = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ w } = \color{#FF6800}{ 2 } , \color{#FF6800}{ z } = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) = \color{#FF6800}{ 10 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) = \color{#FF6800}{ 10 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 10 } = \color{#FF6800}{ 10 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 8 } = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 10 } = \color{#FF6800}{ 10 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 8 } = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ w } = \color{#FF6800}{ 2 } , \color{#FF6800}{ z } = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
Solution search results
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$ \begin{cases} 2x+y=10 \\ 5x-y=18 \end{cases} $ 
$ \begin{cases} 2x+5y=31 \\ 6x-2y=-26 \end{cases} $
7th-9th grade
Algebra
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