Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
menu_open
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
search Use offline Tools apps
Login account_circle menu_open

Solving Systems of Linear Equations Graphically

Solving Systems of Linear Equations Graphically 1.10 - Solution

arrow_back Return to Solving Systems of Linear Equations Graphically
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
4y=102x4y=10-2x
y=102x4y=\dfrac{10-2x}{4}
y=1042x4y=\dfrac{10}{4}-\dfrac{2x}{4}
y=-2x4+104y=\text{-}\dfrac{2x}{4}+\dfrac{10}{4}
y=-24x+104y=\text{-}\dfrac{2}{4}x+\dfrac{10}{4}
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
2y=-1+x2y=\text{-}1+x
y=-1+x2y=\dfrac{\text{-}1+x}{2}
y=-12+x2y=\dfrac{\text{-}1}{2}+\dfrac{x}{2}
y=-12+x2y=\text{-}\dfrac{1}{2}+\dfrac{x}{2}
y=x212y=\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.
{yx=5y52=3x\begin{cases}y-x=5 \\ y-\dfrac{5}{2}=3x \end{cases}
Write in slope-intercept form
{y=5+xy52=3x\begin{cases}y=5+x \\ y-\dfrac{5}{2}=3x \end{cases}
{y=x+5y52=3x\begin{cases}y=x+5 \\ y-\dfrac{5}{2}=3x \end{cases}
{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.
{yx=5(I)y52=3x(II)\begin{cases}y-x=5 & \, \text {(I)}\\ y-\dfrac{5}{2}=3x & \text {(II)}\end{cases}
{6.251.25=?56.2552=?31.25\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}
{6.251.25=?56.252.5=?31.25\begin{cases}6.25-1.25\stackrel{?}{=}5 \\ 6.25-2.5\stackrel{?}{=}3\cdot1.25 \end{cases}
{5=53.75=?31.25\begin{cases}5=5 \\ 3.75\stackrel{?}{=}3\cdot1.25 \end{cases}
{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.