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Judge the identity
$\dfrac{ x }{ 2+i } + \dfrac{ y }{ 2-i } = 4-i$
$x = \dfrac { 15 } { 2 } , y = \dfrac { 5 } { 2 }$
Find the unknown when the unknown is a rational number
$\color{#FF6800}{ \dfrac { x } { 2 + i } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 2 - i } } = 4 - i$
 Organize the expression 
$\color{#FF6800}{ \dfrac { 2 x } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { i x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 y } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { i y } { 5 } } = 4 - i$
$\dfrac { 2 x } { 5 } - \dfrac { \color{#FF6800}{ i } \color{#FF6800}{ x } } { 5 } + \dfrac { 2 y } { 5 } + \dfrac { i y } { 5 } = 4 - i$
 Organize the coefficients 
$\dfrac { 2 x } { 5 } - \dfrac { \color{#FF6800}{ x } \color{#FF6800}{ i } } { 5 } + \dfrac { 2 y } { 5 } + \dfrac { i y } { 5 } = 4 - i$
$\dfrac { 2 x } { 5 } - \dfrac { x i } { 5 } + \dfrac { 2 y } { 5 } + \dfrac { \color{#FF6800}{ i } \color{#FF6800}{ y } } { 5 } = 4 - i$
 Organize the coefficients 
$\dfrac { 2 x } { 5 } - \dfrac { x i } { 5 } + \dfrac { 2 y } { 5 } + \dfrac { \color{#FF6800}{ y } \color{#FF6800}{ i } } { 5 } = 4 - i$
$\color{#FF6800}{ \dfrac { 2 x } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x i } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 y } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y i } { 5 } } = 4 - i$
 Calculate the similar terms 
$\color{#FF6800}{ \dfrac { 2 x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 y } { 5 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 5 } } \right ) \color{#FF6800}{ i } = 4 - i$
$\color{#FF6800}{ \dfrac { 2 x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 y } { 5 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 5 } } \right ) \color{#FF6800}{ i } = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ i }$
 If the real number part and the imaginary number part are the same, the two complex numbers are the same 
$\begin{cases} \color{#FF6800}{ \dfrac { 2 x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 y } { 5 } } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 5 } } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 2 x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 y } { 5 } } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 5 } } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{cases}$
 Solve the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 15 } { 2 } } , \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 5 } { 2 } }$
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