# Thread: linear equations

1. ## linear equations

Solve the following system of equations:
5x + 3y = 5(4 - x) AND 2x - 5y = -(y + 11)/2

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

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

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

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

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

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

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