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${2y=-2x+8x=y−1 $

To solve the system of equations, three steps must be followed.
1

Write the Equations in Slope-Intercept Form

Start by writing the equations in slope-intercept form by isolating the $y-$variables. For the first linear equation, divide both sides by $2.$ For the second equation, add $1$ to both sides.

${2y=-2x+8x=y−1 (I)(II) $

$(I):$ Solve for $y$

DivEqn

$(I):$ $LHS/2=RHS/2$

${y=2-2x+8 x=y−1 $

WriteSumFrac

$(I):$ Write as a sum of fractions

${y=2-2x +28 x=y−1 $

MovePartNumRight

$(I):$ $ca⋅b =ca ⋅b$

${y=2-2 x+28 x=y−1 $

MoveNegNumToFrac

$(I):$ Put minus sign in front of fraction

${y=-22 x+28 x=y−1 $

CalcQuot

$(I):$ Calculate quotient

${y=-1x+4x=y−1 $

IdPropMult

$(I):$ Identity Property of Multiplication

${y=-x+4x=y−1 $

$(II):$ Solve for $y$

${y=-x+4y=x+1 $

2

Graph the Lines

Now that the equations are both written in slope-intercept form, they can be graphed on the same coordinate plane.

3

Identify the Point of Intersection

The point where the lines intersect is the solution to the system.

The lines appear to intersect at $(1.5,2.5).$ Therefore, this is the solution to the system — the value of $x$ is $1.5$ and the value of $y$ is $2.5.$