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Solving Systems of Linear Equations using Elimination

Solving Systems of Linear Equations using Elimination 1.2 - Solution

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We want to solve the system {3x+y=17x+3y=11\begin{cases}3x+y=17 \\ x+3y=11 \end{cases} using the elimination method. We then need to eliminate one of the variable terms when one equation is added to or subtracted from the other. For that to happen we first need to manipulate one or both equations. Let's multiply Equation (II) by -3,\text{-}3, as this will make the xx-terms get opposite coefficients.
x+3y=11x+3y=11
-3(x+3y)=-3(11)\text{-} 3(x+3y)=\text{-} 3(11)
-3x9y=-3(11)\text{-} 3x-9y=\text{-} 3(11)
-3x9y=-33\text{-} 3x-9y=\text{-} 33
This give us the following system of equations. {3x+y=17-3x9y=-33\begin{cases}3x+y=17 \\ \text{-} 3x-9y=\text{-} 33 \end{cases} If we add the equations the xx-terms will eliminate each other. 3x+9y=17+-3x9y=-3308y=-16\begin{array}{c r c l}& 3x+\phantom{9}y & = & 17 \\ + & \text{-} 3x-9y & = & \text{-} 33 \\ \hline & 0-8y & = & \text{-}16\end{array} We can find yy by solving the resulting equation. 08y=-16y=2 0-8y=\text{-}16 \quad \Leftrightarrow \quad y=2 To find xx we can substitute 22 for yy in any of the original equations. Let's use Equation (II).
x+3y=11x+3y=11
x+32=11x+3\cdot {\color{#0000FF}{2}}=11
x+6=11x+6=11
x=5x=5
We have found the solution to the system of equations and it is (5,2).(5,2).