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Solve the system of equations
Answer
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$\begin{cases} 6+5y = 11 \\6-2y = a \end{cases}$
$a = 4 , y = 1$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ 11 } \\ 6 - 2 y = a \end{cases}$
$ $ Solve a solution to $ y$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 1 } \\ 6 - 2 y = a \end{cases}$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ a } \end{cases}$
$ $ Substitute the given $ y $ value into the equation $ 6 - 2 y = a$
$\color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ a }$
$\color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ a }$
$ $ Solve a solution to $ a$
$\color{#FF6800}{ a } = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ a } = \color{#FF6800}{ 4 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ a } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ a } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ 11 } \\ \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ 4 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ 11 } \\ \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ 4 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 11 } = \color{#FF6800}{ 11 } \\ \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 11 } = \color{#FF6800}{ 11 } \\ \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ a } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
Solution search results
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