### 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 𝑥
plus 𝑦 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 have the
equation 𝑥 plus 𝑦 equals two, 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, we’re focusing
on the graphical method. As a result, the two letters we use
to represent our variables will often be 𝑥 and 𝑦, but this doesn’t 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, we’ll review
how to find the equation of a straight line from its graph. This will, in turn, enable us to
identify the systems 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) 𝑦 equals two 𝑥 minus four
and 𝑦 equals 𝑥 plus five. (b) 𝑦 equals negative four 𝑥
plus two and 𝑦 equals five 𝑥 minus one. (c) 𝑦 equals two 𝑥 minus four
and 𝑦 equals negative 𝑥 plus five. (d) 𝑦 equals two 𝑥 plus four
and 𝑦 equals negative 𝑥 plus five. Or (e) 𝑦 equals negative four
𝑥 plus two and 𝑦 equals five 𝑥 plus one.

So, we’ve been given a graph of
two straight lines and 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, we’ll
recall the general form of a straight line in its slope–intercept form 𝑦 equals
𝑚𝑥 plus 𝑏. And we recall that the
coefficient of 𝑥, that’s 𝑚, gives the slope of the line. And the constant term, that’s
𝑏, gives the 𝑦-intercept of the graph. That’s the 𝑦-value at which
the line intercepts the 𝑦-axis.

Let’s consider the blue line
first of all. We can see that this line
intercepts the 𝑦-axis at five, which means the equation of this line will be in
the form 𝑦 equals some number of 𝑥 plus five. To find the slope of this line,
the value of 𝑚, we can draw in a little right-angled triangle anywhere below
this line. And in doing so, we see that
for every one unit the line moves across, that’s to the right, it also moves one
unit down. As the slope of a line can be
found using change in 𝑦 over change in 𝑥, we have negative one over one, which
is negative one. The equation of this line is,
therefore, 𝑦 equals negative 𝑥 plus five.

So, we found the equation of
our first line. Let’s now consider the red
line. And this time we see that this
line intercepts the 𝑦-axis at a value of negative four. The equation of this line is,
therefore, in the form 𝑦 equals some number of 𝑥 minus four. To find the slope, we again
sketch in a right-angled triangle anywhere below this line. And this time, we see that for
every one unit the line moves to the right, it moves two units up. The slope of the line, change
in 𝑦 over change in 𝑥, is therefore two over one, which is equal to two. So, the equation of this line
is 𝑦 equals two 𝑥 minus four.

Looking at the five possible
options we were given, we can see that this combination of equations of straight
lines is option (c) 𝑦 equals two 𝑥 minus four and 𝑦 equals negative 𝑥 plus
five. Although we aren’t actually
asked to solve the pair of simultaneous equations in this question, we could do
so by looking at the coordinates of the point of intersection of the two
lines. And we see that the coordinates
of this point are three, two. So, the solution to this pair
of simultaneous equations would be 𝑥 equals three and 𝑦 equals two.

Let’s now consider a second
example.

Use the shown graph to solve
the given simultaneous equations. 𝑦 equals four 𝑥 minus two and
𝑦 equals negative 𝑥 plus three.

Firstly, we notice that the
graph we have been given is the graph of the two straight lines represented by
these equations. Considering the blue line,
first of all, we see that it has a 𝑦-intercept of three and a slope of negative
one. So, substituting these values
into the general equation of a straight line in its slope–intercept form, we can
see that the equation of the blue line is 𝑦 equals negative 𝑥 plus three. That’s the second of the two
equations we were given.

In the case of the green line,
we see that it has a 𝑦-intercept of negative two and a slope of four. So, substituting these values
into the general equation of a straight line, we find that the equation of this
line is 𝑦 equals four 𝑥 minus two. That’s the first equation that
we were given.

So, this figure represents
these two equations graphically, and we can therefore use it to solve them
simultaneously. The solution to this pair of
linear simultaneous equations will be the coordinates of the point that lies on
both lines. That is the point where the two
lines intersect. We can see from the graph that
the lines intersect at this single point here. And it’s a point with integer
coordinates. The 𝑥-coordinate is one, and
the 𝑦-coordinate is two. The solution to this pair of
simultaneous equations, then, is 𝑥 equals one and 𝑦 equals two.

We could, of course, check this
by substituting this pair of 𝑥-, 𝑦-values into each equation and confirming
that they do indeed satisfy each equation. Although we can see quite
clearly from our graph that this point does lie on both lines.

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

By plotting the graphs of 𝑦
equals negative two 𝑥 plus one and 𝑦 equals 𝑥 plus four, find the point that
satisfies both equations simultaneously.

We’re told in this problem that
we need to approach it by plotting the graphs of these two equations. So, let’s consider two
different methods for plotting straight-line graphs. In our first method, we’ll plot
the graph of 𝑦 equals negative two 𝑥 plus one by comparing its equation to the
general equation of a straight line in its slope–intercept form, 𝑦 equals 𝑚𝑥
plus 𝑏. We recall that, in this form,
the value of 𝑚, that’s the coefficient of 𝑥, represents the slope of the
line. So, the slope of the line we’re
looking to plot is negative two. This means that for every one
unit the line moves across to the right, it will move two units downwards.

We also recall that, in this
general form, the value of 𝑏, the constant term, represents the 𝑦-intercept of
the line. So in our equation, the
𝑦-intercept is positive one. This is the value at which the
line intersects the 𝑦-axis. Let’s plot this line then. We know that it intercepts the
𝑦-axis at a value of one. As the slope is negative two,
we know that if we move one unit across, we then need to move two units
downwards. So, we can plot our next point
at the coordinates one, negative one. We then move one across and two
down again and plot our next point at two, negative three.

We can also go back the other
way from our 𝑦-intercept. If we move one unit to the
left, we need to move two units up. So, we can also plot a point
with coordinates negative one, three. Joining all of these points
together with a straight line, and we have our first line 𝑦 equals negative two
𝑥 plus one.

We’ll use a different method to
plot the second line, which has equation 𝑦 equals 𝑥 plus four. This time, we’ll use a table of
values. We’ll choose a range of
different 𝑥-values. I’ve chosen the integer values
from negative two to two. And we’ll then use the equation
of the line to work out the corresponding 𝑦-values. For example, when 𝑥 is equal
to zero, 𝑦 will be equal to zero plus four, which is four. When 𝑥 is equal to negative
one, 𝑦 will be equal to negative one plus four or four minus one, which is
three.

In the same way, we can then
complete the rest of our table. We can then plot these points
or at least those which will fit on the axes I’ve got here and then join them
together with a straight line to give our second line 𝑦 equals 𝑥 plus
four. Notice that the 𝑦-intercept of
this line is four, and our line does indeed cross the 𝑦-axis at this point. And the slope is one, and our
line does indeed have a slope of one. So, we’ve plotted the two
lines. And now, we need to find the
point that satisfies both equations simultaneously. This is the point that lies on
both lines. It’s the coordinates of their
point of intersection.

From our graph, we can see that
the lines intersect at the point with coordinates negative one, three. As we we’re asked to find the
point that satisfies both equations simultaneously, we’ll give our answer as a
coordinate. So, our solution to the problem
is the coordinate negative one, three.

Plot the graphs of the
simultaneous equations 𝑦 equals two 𝑥 plus seven and 𝑦 equals two 𝑥 minus
four and then solve the system.

We’ll plot each of these graphs
by comparing their equations with the general equation of a straight line in its
slope–intercept form, 𝑦 equals 𝑚𝑥 plus 𝑏. The first line 𝑦 equals two 𝑥
plus seven has a slope of two and a 𝑦-intercept of seven. We can plot this line by first
plotting the 𝑦-intercept. And then for every one unit we
go to the right, we go two units up. In this case though, we’ll go
the other way. For every one unit we go to the
left, we go two units down. So that gives a point at
negative one, five; and then a point at negative two, three; and a point at
negative three, one. We can then join all of these
points up to give our first plotted line.

In the same way, we see that
the line 𝑦 equals two 𝑥 minus four has a slope of two and a 𝑦-intercept of
negative four. We can plot the
𝑦-intercept. And then, we go one unit to the
right and two units up, one unit right and two units up again, and again. And then, we join these points
up to give our second line. So, we’ve plotted the graphs of
these two simultaneous equations. But now, we’re asked to solve
the system, which means we’re looking for the point of intersection of these two
lines.

Now, our two lines don’t
intersect on the graph I’ve drawn. Does this mean that I’ve just
chosen the wrong ranges for the 𝑥- and 𝑦-axes and they would intersect if I’d
chosen a larger range? Well, the answer to that is no,
these two lines will actually never intersect. And the reason for this is
because they are parallel lines. They both have the same slope
of two. We know that parallel lines
will never meet. And therefore, these two lines
will have no point of intersection. So, in fact, there are no
solutions to this system of simultaneous equations. And our reasoning for this is
that the two equations represent parallel lines.

In our examples, so far, we’ve 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. Secondly, the lines could be
parallel if they have the same slope. And, in this case, there is no
solution to a system of simultaneous equations as the two lines will never
intersect. There is, in fact, a third
option.

Suppose we were asked to solve the
system of equations 𝑦 equals two 𝑥 minus four and two 𝑦 equals four 𝑥 minus
eight. If we were to attempt to plot
these, we can see that the two equations actually describe the exact same line. This is because if we divide both
sides of the second equation by two, then it reduces to 𝑦 equals two 𝑥 minus
four. It’s just 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 to the system of linear equations.

So, these are the three options
when solving a pair of linear simultaneous equations graphically. The lines either intersect at a
single point, meaning there is one solution, a single pair of coordinates. The lines are parallel, which means
they never intersect. And there is no solution to the
system of linear equations. Or the lines are coincident,
meaning one lies directly on top of the other. And there are infinitely many
solutions to the system of simultaneous equations.

Sometimes, it may not be possible
to solve a set of simultaneous equations exactly using a graph if the solutions are
noninteger values. In this case, we can only give
approximate values for the solution, as we’ll see in our next example.

Using the graph, determine
which of the following is a sensible estimate for the solution to the
simultaneous equations two 𝑥 plus three 𝑦 equals 20, four 𝑥 minus four 𝑦
equals 11. (a) 𝑥 equals 5.4 and 𝑦 equals
3.1. (b) 𝑥 equals 5.4 and 𝑦 equals
2.9. (c) 𝑥 equals 5.6 and 𝑦 equals
2.4. (d) 𝑥 equals 5.7 and 𝑦 equals
2.9. Or (e) 𝑥 equals 5.9 and 𝑦
equals 2.7.

Now, It may not be immediately
obvious, but the two lines we’ve been given on the graph do represent the two
equations given in the question. The red line is two 𝑥 plus
three 𝑦 equals 20, and the blue line is four 𝑥 minus four 𝑦 equals 11. The solution to this pair of
simultaneous equations then will be the coordinates of the point where these two
lines intersect. But from looking at the figure,
we can see that they intersect in the middle of one of the smaller squares. So, we can’t find an exact
value for the solution. Instead, we’re going to be
looking for an estimate.

Let’s first make sure we’re
clear on the scale that has been used on each of our axes. In each case, it’s the
same. There are four small squares to
represent two units. Dividing by four, we see that
each small square on each axes represents 0.5 units. Looking at the horizontal
placement of this point first of all then, we can see that it is located between
the line three small squares to the right of four and then the 𝑥-value six. If each small square is 0.5,
then three small squares is 1.5, which means that the 𝑥-value to the left of
this point is 5.5. And so, our 𝑥-value is between
5.5 and six.

In the same way, looking at the
vertical placement of this point, we can see that it’s located between one small
square above two, that’s 2.5, and two small squares above two, that’s three. So, the 𝑦-value is between 2.5
and three.

Looking at the five options, we
can rule out options (a) and (b), as their 𝑥-values are out of range, and
option (c), as its 𝑦-value is out of range. To decide between the two
remaining options then, we know that the point is vertically very close to
three, and it seems to be closer to 5.5 than to six. So, option (d) is the most
sensible estimate. 𝑥 is approximately equal to
5.7, and 𝑦 is approximately equal to 2.9.

Let’s review the key points that
we’ve seen in this lesson. Firstly, we saw that systems of
linear equations can be solved by plotting the straight-line graphs represented by
each equation and identifying the coordinates of their point of intersection. However, we also saw that not all
pairs of straight lines intersect. The three possibilities are that
the lines intersect at one point, meaning there is one solution to the system of
linear equations. The two lines are parallel, they
never intersect, and so there is no solution. Or the two lines are coincident, in
which case there’re infinitely many solutions to the system of equations. Finally, we also saw that we can
use graphs to find approximate solutions to systems of linear equations when the
solutions are noninteger values.