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Solving Systems of Linear Equations Graphically

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Algebra 1
Linear Systems
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Solving Systems of Linear Equations Graphically 1.10 - Solution

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a
By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope and yy-intercept.

Write in Slope-Intercept Form

Let's rewrite the equation in slope-intercept form by isolating y.y.
2x+4y=102x+4y=10
Write in slope-intercept form
SubEqnLHS−2x=RHS−2x\text{LHS}-2x=\text{RHS}-2x
4y=10−2x4y=10-2x
DivEqnLHS/4=RHS/4\left.\text{LHS}\middle/4\right.=\left.\text{RHS}\middle/4\right.
y=10−2x4y=\dfrac{10-2x}{4}
WriteDiffFracWrite as a difference of fractions
y=104−2x4y=\dfrac{10}{4}-\dfrac{2x}{4}
CommutativePropAddCommutative Property of Addition
y=-2x4+104y=\text{-}\dfrac{2x}{4}+\dfrac{10}{4}
MovePartNumRighta⋅bc=ac⋅b\dfrac{a\cdot b}{c}=\dfrac{a}{c}\cdot b
y=-24x+104y=\text{-}\dfrac{2}{4}x+\dfrac{10}{4}
ReduceFracab=a/2b/2\dfrac{a}{b}=\dfrac{\left.a\middle/2\right.}{\left.b\middle/2\right.}
y=-12x+104y=\text{-}\dfrac{1}{2}x+\dfrac{10}{4}
Now, the second equation!
-x+2y=-1\text{-} x+2y=\text{-}1
Write in slope-intercept form
AddEqnLHS+x=RHS+x\text{LHS}+x=\text{RHS}+x
2y=-1+x2y=\text{-}1+x
DivEqnLHS/2=RHS/2\left.\text{LHS}\middle/2\right.=\left.\text{RHS}\middle/2\right.
y=-1+x2y=\dfrac{\text{-}1+x}{2}
WriteDiffFracWrite as a difference of fractions
y=-12+x2y=\dfrac{\text{-}1}{2}+\dfrac{x}{2}
MoveNegNumToFracPut minus sign in front of fraction
y=-12+x2y=\text{-}\dfrac{1}{2}+\dfrac{x}{2}
CommutativePropAddCommutative Property of Addition
y=x2−12y=\dfrac{x}{2}-\dfrac{1}{2}

Graphing the System

To graph the system we'll start by plotting the yy-intercepts, 104=2.5\frac{10}{4}=2.5 and 12=0.5.\frac{1}{2}=0.5.

Use the slopes, -12\text{-}\frac{1}{2} and 12,\frac{1}{2}, to determine another point that satisfies each equation, and connect the points with a line.

We can see that the lines intersect at exactly one point, the solution to the system.

The solution to the system is (2,1.5).(2,1.5).

b
The first thing we need to do is to rewrite the equations in slope-intercept form.
{y−x=5y−52=3x\begin{cases}y-x=5 \\ y-\dfrac{5}{2}=3x \end{cases}
Write in slope-intercept form
AddEqn(I): {\color{#8C8C8C}{\text{(I): }}}LHS+x=RHS+x\text{LHS}+x=\text{RHS}+x
{y=5+xy−52=3x\begin{cases}y=5+x \\ y-\dfrac{5}{2}=3x \end{cases}
CommutativePropAdd(I): {\color{#8C8C8C}{\text{(I): }}}Commutative Property of Addition
{y=x+5y−52=3x\begin{cases}y=x+5 \\ y-\dfrac{5}{2}=3x \end{cases}
AddEqn(II): {\color{#8C8C8C}{\text{(II): }}}LHS+52=RHS+52\text{LHS}+\dfrac{5}{2}=\text{RHS}+\dfrac{5}{2}
{y=x+5y=3x+52\begin{cases}y=x+5 \\ y=3x+\dfrac{5}{2} \end{cases}
We can now graph the lines using their yy-intercept and slope.

The solution to the system is where the graphs intersect.

We can estimate the solution to (1.25,6.25).(1.25,6.25). To be sure, we'll substitute the point into the system of equations to see if it satisfies the equations.
{y−x=5 (I)y−52=3x(II)\begin{cases}y-x=5 & \, \text {(I)}\\ y-\dfrac{5}{2}=3x & \text {(II)}\end{cases}
SubstituteIIx=1.25x={\color{#0000FF}{1.25}}, y=6.25y={\color{#009600}{6.25}}
{6.25−1.25=?56.25−52=?3⋅1.25\begin{cases}{\color{#009600}{6.25}}-{\color{#0000FF}{1.25}}\stackrel{?}{=}5 \\ {\color{#009600}{6.25}}-\dfrac{5}{2}\stackrel{?}{=}3\cdot{\color{#0000FF}{1.25}} \end{cases}
WriteDec(I): {\color{#8C8C8C}{\text{(I): }}}Write as a decimal
{6.25−1.25=?56.25−2.5=?3⋅1.25\begin{cases}6.25-1.25\stackrel{?}{=}5 \\ 6.25-2.5\stackrel{?}{=}3\cdot1.25 \end{cases}
SubTermsSubtract terms
{5=53.75=?3⋅1.25\begin{cases}5=5 \\ 3.75\stackrel{?}{=}3\cdot1.25 \end{cases}
Multiply(II): {\color{#8C8C8C}{\text{(II): }}}Multiply
{5=53.75=3.75\begin{cases}5=5 \\ 3.75=3.75 \end{cases}
The point satisfies the system! Thus, it is the solution.