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Math Help - linear equations

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    Exclamation linear equations

    Solve the following system of equations:
    5x + 3y = 5(4 - x) AND 2x - 5y = -(y + 11)/2
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    Quote Originally Posted by jsimms View Post
    Solve the following system of equations:
    5x + 3y = 5(4 - x) AND 2x - 5y = -(y + 11)/2
    \left\{<br />
\begin{array}{rcl}<br />
5x + 3y & = & 5(4 - x)\\<br />
2x - 5y & = & -\frac12(y + 11)<br />
\end{array}<br />
\right.

    \Rightarrow\left\{<br />
\begin{array}{rcll}<br />
5x + 3y & = & 20 - 5x & \text{\color{red}\quad(distribution)}\\<br />
-4x + 10y & = & y + 11 & \text{\color{red}\quad(multiplication by -2)}<br />
\end{array}<br />
\right.

    \Rightarrow\left\{<br />
\begin{array}{rcl}<br />
10x + 3y - 20 & = & 0\\<br />
-4x + 9y - 11 & = & 0<br />
\end{array}<br />
\right.\text{\color{red}\quad(moving terms over to left-hand side)}

    \Rightarrow\left\{<br />
\begin{array}{rcll}<br />
30x + 9y - 60 & = & 0 & \text{\color{red}\quad(multiplication by 3)}\\<br />
-4x + 9y - 11 & = & 0 &<br />
\end{array}<br />
\right.

    \Rightarrow\left\{<br />
\begin{array}{rcl}<br />
34x + 0y - 49 & = & 0\\<br />
-4x + 9y - 11 & = & 0<br />
\end{array}<br />
\right.\text{\color{red}\quad(subtracting equation 2 from equation 1)}

    \Rightarrow 34x -49 = 0\Rightarrow x = \frac{49}{34}

    Now, substitute back into equation 2 and solve for y.
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