### Video Transcript

In this video, we will learn how to
solve a system of two linear equations by considering their graphs and identifying
their point of intersection.

We recall first of all that a
linear equation is one in which the highest power of each variable that appears is
one and there are no terms in which the variables are multiplied together. For example, the equation two π₯
plus π¦ equals six is a linear equation.

A system of two linear equations is
simply a pair of two such equations. For example, if we also had the π₯
plus π¦ equals two equation, we now have a system of linear equations, sometimes
known as a pair of simultaneous equations. There are many different methods
that can be used to solve such systems of equations, but in this video, weβre
focusing on the graphical method. As a result, the two letters we use
to represent our variables will often be π₯ and π¦, but this doesnβt have to be the
case.

The solution to a system of two
linear equations can be found by plotting a graph of the two straight lines
represented by these equations and then identifying the coordinates of their point
of intersection. This is because this point lies on
both lines and therefore satisfies both equations simultaneously.

In our first example, weβll review
how to find an equation of a straight line from its graph. This will in turn enable us to
identify the system of linear equations that can be solved using a given graph.

Which of the following sets of
simultaneous equations could be solved using the given graph? (A) π¦ equals two π₯ minus four, π¦
equals π₯ plus five. (B) π¦ equals negative four π₯ plus
two, π¦ equals five π₯ minus one. (C) π¦ equals two π₯ minus four, π¦
equals negative π₯ plus five. (D) π¦ equals two π₯ plus four, π¦
equals negative π₯ plus five. Or (E) π¦ equals negative four π₯
plus two, π¦ equals five π₯ plus one.

Weβve been given a graph of two
straight lines. Letβs name them π one and π
two. We are asked to determine which
pair of simultaneous equations we could solve using this graph. This means that we need to
determine the equations of the two straight lines.

In order to do this, weβll recall
the general form of a straight line in its slopeβintercept form, π¦ equals ππ₯ plus
π. And we recall that the coefficient
of π₯, thatβs π, gives the slope of the line. And the constant term, thatβs π,
gives the π¦-intercept of the graph. Thatβs the π¦-value at which the
line intercepts the π¦-axis.

We can determine both of these
values from the diagram. First, line π one has a
π¦-intercept of five, and line π two has a π¦-intercept of negative four. Next, we determine the slope of
each line by using the fact that a line passing through two coordinate points β π₯
one, π¦ one and π₯ two, π¦ two β has the slope π equals π¦ two minus π¦ one over π₯
two minus π₯ one. π one passes through the points
zero, five and one, four. We could have selected any two
points from this line. But it makes our work easier if we
choose a second point near the π¦-intercept.

The first line has a slope of π
equals four minus five over one minus zero. So, the slope of π one is negative
one. The second line π two passes
through the points zero, negative four and one, negative two. Its slope is calculated by π
equals negative four minus negative two over zero minus one. So, the slope of π two is positive
two. Hence, the line π one has π equal
to five and π equal to negative one. Its slopeβintercept equation is π¦
equals negative π₯ plus five, which is one of the equations given in option (C) and
in option (D).

Line π two has π equal to
negative four and π equal to two. So, its equation is π¦ equals two
π₯ minus four. This gives us the system of
equations π¦ equals two π₯ minus four, π¦ equals negative π₯ plus five. The solutions to this system of
linear equations given by the coordinates of the point of intersection are π₯ equals
three and π¦ equals two. The given graph shows the system of
simultaneous linear equations in option (C): π¦ equals two π₯ minus four, π¦ equals
negative π₯ plus five.

Letβs now consider a second
example.

Use the shown graph to solve the
given simultaneous equations: π¦ equals four π₯ minus two, π¦ equals negative π₯
plus three.

We recall that the solution to a
system of equations is given by the coordinates of the point of intersection between
the graphs of all of the equations. This means the coordinates of the
point of intersection between the two lines tells us the solution to the
simultaneous equations. We see that the π₯-coordinate of
this point is one and the π¦-coordinate is two. This tells us π₯ equals one and π¦
equals two is a solution to the simultaneous equations.

We can verify this by substituting
these values into the equations. If we substitute π₯ equals one into
the first equation, we get π¦ equals four times one minus two equals two, which
agrees with our solution. Similarly, if we substitute π₯
equals one into the second equation, we get π¦ equals negative one plus three equals
two, which also agrees with our solution. Since both equations hold true,
this verifies the solution. Since this is the only point of
intersection, it is the only solution to the simultaneous equations. Therefore, the only solution is π₯
equals one and π¦ equals two.

In our next example, weβll need to
plot the graphs of the two equations that we wish to solve ourselves. So, weβll remind ourselves of some
of the key methods for doing this.

By plotting the graphs of π¦ equals
π₯ minus one and π¦ equals five π₯ plus seven, find the point that satisfies both
equations simultaneously.

We recall that if the pair of
coordinates π₯ and π¦ satisfy both equations simultaneously, then the point must lie
on the graphs of both equations. Hence, it is their point of
intersection, and so it is a solution to the system of equations. Therefore, we can find the
solutions to this system by finding the coordinates of the points of
intersection. We do this by sketching both graphs
on the same coordinate plane. We note that both lines are of the
form π¦ equals ππ₯ plus π. We recall that this line will have
a π¦-intercept of π and a slope of π, provided π does not equal zero.

We can interpret the important
information about each line by referring to this formula. First, the line π¦ equals π₯ minus
one will have a π¦-intercept of negative one. We know that the coefficient of π₯
is one, so the slope of the first line must be one. We recall that the slope of a line
is the change in π¦ with respect to the change in π₯. Therefore, a slope of one
represents an increase of one in π¦ and an increase of one in π₯. Following the change of
coordinates, up one and right one, we get another coordinate point on the line: one,
zero. Connecting these points allows us
to sketch the line with a slope of one and a π¦-intercept of negative one.

Moving on to the second linear
equation, we have a π¦-intercept of seven and a slope of five. This means the change in the
π¦-coordinate is five and the change in the π₯-coordinate is one. So, we start at the π¦-intercept
then move up five and right one. However, because of the position of
the π¦-intercept near the edge of our coordinate plane, we can reverse these
directions as needed to find a point on the line to the left of zero, seven. So, we will move down five and left
one instead. We found a point on the line with
coordinates π₯ equals negative one and π¦ equals two.

Finally, we connect these points to
sketch the line π¦ equals five π₯ plus seven. We can see that both lines contain
the point negative two, negative three. This is the point of intersection,
meaning π₯ equals negative two and π¦ equals negative three satisfy both
equations.

We can verify that these
coordinates satisfy both equations by substituting π₯ equals negative two. We note that both equations give π¦
equals negative three. Since both equations hold true,
this verifies the solution we found by graphing. Hence, π₯ equals negative two and
π¦ equals negative three satisfies both equations. And we can say that the point that
satisfies both equations is negative two, negative three.

Plot the graphs of the simultaneous
equations π¦ equals two π₯ plus seven, π¦ equals two π₯ minus four, and then solve
the system.

We recall that the points of
intersection of the graphs of both equations will tell us the solutions to the
system of equations. This means we can solve the system
by sketching both equations on one coordinate plane. Since these are linear equations
given in slopeβintercept form, we will plot their graphs using their slope and
π¦-intercepts.

Weβll begin with the first
equation, π¦ equals two π₯ plus seven. We see that two is the coefficient
of π₯ and seven is the constant. That means that two is the slope
and seven is the π¦-intercept. So, we plot the π¦-intercept of the
first line at seven. The second line has the same slope
as the first line but a different π¦-intercept, negative four.

Now, we recall that slope is the
change in π¦ with respect to the change in π₯. Therefore, a slope of two can be
interpreted as adding two to the π¦-coordinate and adding one to the
π₯-coordinate. To find another point on the second
line, we simply move up two and right one. Then, to plot the graph of the line
π¦ equals two π₯ minus four, we connect the π¦-intercept to the new point with a
straight line. This is the line given by the
equation π¦ equals two π₯ minus four.

Since the first line has the same
slope as the second line, we can plot its graph by sketching a parallel line through
the π¦-intercept of seven. This is the line given by the
equation π¦ equals two π₯ plus seven. Seeing that these lines run
parallel, we know that there are no points of intersection, so this system has no
solutions.

In fact, we could have saved
ourselves the effort of drawing the graphs by just carefully examining the equations
of the two lines. We can see that both lines have the
same slope, but distinct π¦-intercepts. This tells us that these lines are
parallel; they have the same slope. And they are distinct; they pass
through different π¦-intercepts. Hence, the lines do not intersect
and the system has no solutions.

In our examples so far, weβve seen
two possibilities. Firstly, the lines could intersect
at a single point, in which case there is one solution to the system of linear
equations. We call this an independent system
of linear equations. Secondly, the lines could be
parallel, if they have the same slope and distinct π¦-intercepts. In this case, there is no solution
to the system of simultaneous equations, as the two lines will never intersect. We call this an inconsistent system
of equations. Systems with at least one solution
are considered consistent, such as in the first case.

There is in fact a third
option. Suppose we were asked to solve the
system of equations π¦ equals two π₯ minus four and four π₯ minus two π¦ equals
eight. If we were to plot these graphs, we
can see that these two equations describe the exact same line. This is because if we write the
second equation in slopeβintercept form, we get π¦ equals two π₯ minus four. This means the second equation is
an equivalent way of writing the equation of the first line. In this case, the lines are
described as being coincident. One line lies exactly on top of the
other. And every single point on this
infinitely long line will therefore satisfy the system of linear equations. We therefore say that there are
infinitely many solutions. We call this a dependent system of
linear equations, which is also considered consistent because this type of system
has solutions.

So, these are the three options
when solving a pair of linear, simultaneous equations graphically.

Letβs review the key points weβve
seen in this video. Firstly, we saw that systems of
linear equations can be solved by plotting their graphs and identifying the
coordinate points of their intersection. However, we also saw that not all
straight lines intersect. The three possibilities are that
the lines intersect at one point, meaning there is one solution consisting of an
π₯-coordinate and π¦-coordinate. The two lines are parallel. They never intersect, so there are
no solutions. Or the two lines are coincident, in
which case there are infinitely many solutions to the system of linear
equations.