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Formula
Solve the system of equations
Answer
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$\begin{cases} 3a-2b = 4 \\a+b = -1 \end{cases}$
$a = \dfrac { 2 } { 5 } , b = - \dfrac { 7 } { 5 }$
Solve quadratic equations using the square root
$\begin{cases} 3 a - 2 b = 4 \\ a + b = - 1 \end{cases}$
$ $ Solve a solution to $ a$
$\begin{cases} \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ b } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ a + b = - 1 \end{cases}$
$\begin{cases} \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ b } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{cases}$
$ $ Substitute the given $ a $ value into the equation $ a + b = - 1$
$\left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ b } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$\left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ b } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$ $ Solve a solution to $ b$
$\color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } }$
$\color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } }$
$ $ Substitute the given $ b $ value into the equation $ a = \dfrac { 2 } { 3 } b + \dfrac { 4 } { 3 }$
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$a = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \right ) + \dfrac { 4 } { 3 }$
$ $ Calculate the product of rational numbers $ $
$a = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 14 } { 15 } } + \dfrac { 4 } { 3 }$
$a = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 14 } { 15 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$ $ Find the sum or difference of the fractions $ $
$a = \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } } , \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } }$
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } } , \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \right ) = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \right ) = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } } , \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } }$
$\begin{cases} a = \dfrac { 2 } { 5 } \\ b = - \dfrac { 7 } { 5 } \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ a } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ b } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{cases}$
$ $ Solve the system of linear equations for $ a , b $
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ a } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ b } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ 7 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ a } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ b } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ 7 } \end{cases}$
$ $ Solve the system of linear equations for $ a , b $
$\begin{cases} \color{#FF6800}{ 15 } \color{#FF6800}{ a } = \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ 7 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 15 } \color{#FF6800}{ a } = \color{#FF6800}{ 6 } \\ 5 b = - 7 \end{cases}$
$ $ Solve a solution to $ a$
$\begin{cases} \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } } \\ 5 b = - 7 \end{cases}$
$\begin{cases} a = \dfrac { 2 } { 5 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ 7 } \end{cases}$
$ $ Solve a solution to $ b$
$\begin{cases} a = \dfrac { 2 } { 5 } \\ \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \end{cases}$
Solution search results
search-thumbnail-$∩$ $∩1$ 
$ \begin{cases} 3\left(x-1\right)-2\left(1+x\right)<1 \\ 3x-11>0 \end{cases} $ $ \begin{cases} 3x-3-2+2x47 \\ 3x-4>0 \end{cases} $ 
$ \begin{cases} x<7+5 \\ 222l+4 \end{cases} $ $ \begin{cases} x<6 \\ 3x>4 \end{cases} $
1st-6th grade
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