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We want to solve the system
${3x−y=58x−y=20 (I)(II) $
using the elimination method. We then need to eliminate one of the variable terms by adding or subtracting the equations. We can see that the $y$-terms in this system will eliminate each other if we subtract Equation (II) from Equation (I). Let's do that!
$− 3x−y8x−y-5x+0 === 520-15 $
We find $x$ by solving the resulting equation.
$-5x+0=-15⇔x=3$
We find $y$ by substituting the value of $x$ into either of the original equations and solve the resulting equation for $y.$ Let's use Equation (I).
The solution to the system of equations is $(3,4).$

$3x−y=5$

Substitute$x=3$

$3⋅3−y=5$

MultiplyMultiply

$9−y=5$

SubEqn$LHS−9=RHS−9$

$-y=-4$

MultEqn$LHS⋅(-1)=RHS⋅(-1)$

$y=4$