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# Solving Systems of Linear Equations using Elimination

## Solving Systems of Linear Equations using Elimination 1.4 - Solution

a
Before we can solve the system, we have to simplify the first equation by using the Distributive Property. $\begin{cases}3(x+2y)=\text{-} 3 \\ 3x+4y=1 \end{cases} \quad \Leftrightarrow \quad \begin{cases}3x+6y=\text{-} 3 & \, \text {(I)}\\ 3x+4y=1 & \text {(II)}\end{cases}$ To solve a system of linear equations using the elimination method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. Here if we subtract Equation (II) from Equation (I) the $x$-terms in this system will eliminate each other. $\begin{array}{c r c l}& 3x+6y & = & \text{-} 3 \\ - & 3x+4y & = & 1 \\ \hline & 0+2y & = & \text{-}4\end{array}$ We can find $y$ by solving the resulting equation. $0+2y=\text{-} 4 \quad \Leftrightarrow \quad y=\text{-} 2$ To find $x$ we can substitute $\text{-} 2$ for $y$ in any of the original equations. Let's use Equation (II).
$3x+4y=1$
$3x+4({\color{#0000FF}{\text{-}2}})=1$
$3x-8=1$
$3x=9$
$x=3$
The solution to the system of equations is $(3,\text{-}2).$
b
Let's begin by simplifying the second equation. $\begin{cases}6x-4y=28 \\ 4(x+y)=12 \end{cases} \quad \Leftrightarrow \quad \begin{cases}6x-4y=28 & \, \text {(I)}\\ 4x+4y=12 & \text {(II)}\end{cases}$ When using the elimination method one of the variable terms needs to be eliminated by adding or subtracting the equations. We can see that the $y$-terms will eliminate each other if we add the equations to each other. $\begin{array}{c r c l}& 6x-4y & = & 28 \\ + & 4x+4y & = & 12 \\ \hline & 10x+\phantom{4}0 & = & 40\end{array}$ Let's solve the resulting equation. $10x+0=40 \quad \Leftrightarrow \quad x=4$ We can find $y$ by substituting $4$ for $x$ into any of the original equations, for example Equation (I), and solving the resulting equation.
$6x-4y=28$
$6\cdot {\color{#0000FF}{4}}-4y=28$
Solve for $y$
$24-4y=28$
$\text{-}4y=4$
$y=\text{-}1$
The solution to the system is $(4,\text{-}1).$