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Solve the system of equations
$\begin{cases} 3a-2b = 4 \\a+b = -1 \end{cases}$
$a = \dfrac { 2 } { 5 } , b = - \dfrac { 7 } { 5 }$
Solve the system of equations
$\begin{cases} 3 a - 2 b = 4 \\ a + b = - 1 \end{cases}$
 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 
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } } , \color{#FF6800}{ b } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } }$
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