Welcome to a lesson
on solving systems of linear equations by graphing.
Two linear equations that relate the same two variables
are called a system of equations.
Here's an example of a system of linear equations.
We have two linear equations that relate
the same two variables.
A solution to a system of linear equations
is an ordered pair that satisfies both equations.
So for this system, a solution would be in the form of
x comma y, where this ordered pair would satisfy
both the first equation and the second equation.
To solve a system of linear equations by graphing,
we graph both linear equations on the same coordinate plane.
And there are three types of solutions
to a system of linear equations.
First, we can have one unique solution,
which means when we graph the two lines,
the lines intersect at exactly one point,
which would look like this.
Notice how the two lines intersect at one point,
this point here, and therefore there is one solution.
The solution would be the ordered pair of this point,
and because we have one solution,
we say the system is consistent and independent.
We can also have no solution to our system.
This occurs when the two lines are parallel
and will never intersect.
So if our graph looks like this
because the lines are parallel,
there are no point of intersection,
and therefore, there's no selection to our system.
We can say the system is inconsistent.
The third possibility is that we have
infinitely many solutions.
This occurs when both lines graph as the same line.
So if we graph both lines, and they coincide
and are on top of each other,
meaning we have the same line twice,
they have an infinite number of points in common,
and therefore, our system has infinitely many solutions.
And we can say the system is consistent and dependent.
For example one, we're asked to verify
that the ordered pair five comma four
is a solution to the system of equations.
To do this, you want to substitute five for x and four for y
in both equations to make sure this ordered pair
satisfies both equations.
So for our first equation,
we would have four equals two times five minus six.
Simplifying, notice how we get four equals four,
which is true, which verifies five comma four
satisfies the first equation,
but we still must check five comma four
in the second equation.
So y equals x minus one would be four,
equals five minus one, so four equals four is true,
so we have just verified
that the ordered pair five comma four
is the solution to our system.
Example two, we're asked to solve
the system of equations by graphing
and then check our answer.
Notice how both equations are in slope-intercept form,
where we have the form y equals mx plus b
or y equals b plus mx.
So for the first equation, notice how b
is equal to positive six, which means
the vertical intercept is the ordered pair zero comma six,
and the slope m is equal to negative 2/3,
which we'll write as negative two over positive three.
Let's go ahead and graph this first line,
so zero comma six is here.
Because the slope is negative two over positive three,
we would go down to and right three.
Let's do that again, down two, right three.
When graphing by hand, we want to graph the line
as accurately as possible,
so we'll find multiple points on the line.
Let's go ahead and graph this first line.
And now, for the second equation,
we have y equals x plus one, or y equals one x plus one.
So notice how the slope m is equal to one,
which we'll write as one over one,
and b, the vertical intercept is also equal to one,
which would be the ordered pair zero comma one.
So we'll plot the ordered pair zero comma one.
Because the slope is positive one over positive one,
we would go up one, right one multiple times,
up one, right one, up one, right one, and so on.
Now if we wanted to find some points on the left,
we could write the slope as negative one over negative one.
This still simplifies to positive one,
but in this form, we can go down one unit and left one unit,
so down one, left one, down one, left one,
down one, left one.
Let's go ahead and graph the second line.
Now, analyzing the graph, notice how
we have one point of intersection,
this point here, and therefore, we have one solution.
The solution would be the ordered pair at this point,
which is three comma four.
So this is our solution,
the ordered pair three comma four.
To verify these, we have to substitute
three for x and four for y in both of our linear equations.
So let's go ahead and do that.
so for our first equation, we'd have four
equals six minus 2/3 times three,
I'll write three as three over one.
Simplifying, here was have a common factor of three,
so we have four equals six minus, here we have two,
and we have four equals four, which is true.
Now you must also check the ordered pair
in the second equation, y equals x plus one.
So for y equals x plus one,
we'd have four equals three plus one,
and notice how this is also true,
verifying our solution is correct.
But I also want to verify the solution graphically
on the graphing calculator.
To check this solution graphically on the calculator,
we press Y=, enter in both of your equations,
so we have y equals six minus,
I'm going to put 2/3 in parentheses, x, Enter.
Second equation is y equals x plus one.
Now, to make sure we have the standard window,
let's just press Zoom six to graph the line, or both lines.
There's the first line, there's the second line.
To verify our solution, we want to determine
the point of intersection here.
To do this, we press 2nd, Trace for the calculation menu
option five for intersection,
and because we have only two lines on the coordinate plane,
we can simply press Enter three times.
So Enter, Enter, Enter.
And notice how our solution is correct.
The intersection point is the ordered pair three comma four.
Example three's the same type of problem,
but now our equations are in general, or standard, form.
So to graph these two lines, we'll solve for y
and write the equation in slope-intercept form.
So our first equation is
four x minus three y equals negative 18.
So we'll isolate the y term
by subtracting four x on both sides.
Notice here, I wrote negative four x minus 18,
but you can also write negative 18 minus four x.
And now we divide both sides by negative three.
Simplifying, we have y equals, this'd be positive 4/3 x.
And here, we're subtracting a negative.
That's the same as adding a positive,
so we have plus 18 divide by three is six.
Notice here, we have a slope of 4/3,
and the vertical intercept b is equal to six,
so the vertical intercept would be
the ordered pair zero comma six.
So to graph this line, we'll first plot
the vertical intercept of zero comma six.
Now using the slope because
it's positive four over positive three,
you'd go up four and right three,
which is off the coordinate plane.
So let's write this slope in a different form.
Let's write 4/3 as negative four over negative three.
This still simplifies to positive 4/3, but in this form,
it tells us to go down four and left three.
So down four, left three.
Let's do that again, down four, left three.
And now we can graph our first line.
Now for the second equation,
we have two x plus y equals negative four.
So to solve for y, we subtract two x on both sides,
so we'd have y equals either negative two x minus four
or negative four minus two x.
Let's write negative two x minus four.
So notice how the slope m is equal to negative two.
Let's write that as negative two over positive one.
And b, the vertical intercept is equal to negative four,
which is the ordered pair zero comma negative four.
So we'll plot the vertical intercept here,
and because the slope is negative two over positive one,
we'd go down two, right one.
If we go down two and right one again,
we're off the coordinate plane,
so let's also write the slope
as positive two over negative one.
In this form, we can go up two and left one,
so up two, left one, up two, left one,
up two, left one, and so on.
Again, when graphing by hand, we want to find
the multiple points on our lines
to graph them as accurately as possible.
So now we graph the second line.
Once again, notice how we have one point of intersection,
and therefore, we have one solution.
The solution would be the ordered pair
of this point of intersection,
which notice would be negative three comma two.
So our solution
is the ordered pair negative three comma two,
but we are asked to check our solution,
so we substitute negative three for x
and two for y in both of our linear equations.
So for the first equation, we'd have four
times x, which is negative three,
minus three times y, which is two, equals negative 18.
So we have negative 12 minus six
equals negative 18, which is true.
We have negative 18 equals negative 18,
and for the second equation, we have
two times negative three plus two equals negative four,
so negative six plus two equals negative four,
and negative six plus two is negative four,
so our solution checks.
Let's also check on the calculator.
So we'll press Y=, clear out the old equations.
Notice how to check on the calculator,
we do have to have the linear equation
in slope-intercept form.
So for our first equation,
we have y equals 4/3 x plus six.
We'll put 4/3 in parentheses,
Enter.
For our second equation,
we have y equals negative two x minus four.
Now we already know we have this in our window,
so we can just press graph this time.
There's the first equation,
there's the second,
so we want to find the point of intersection,
so we press 2nd, Trace for the calculation menu,
option five for intersection,
and press Enter three times, Enter, Enter, Enter,
verifying our solution is correct.
I think we'll stop here for this part.
We'll take a look at at two more examples in part two.