Welcome to a lesson on using a table of values
to graph a linear equation in two variables.
We'll be following the steps outlined here for graphing
linear equations by plotting points.
As we go through each step, we'll look at the work example
below, and then afterwards, we'll take a look
at our own examples.
So, step one, we choose two or more values for the input
variable, normally x, and list them in a table of values.
Notice how it says choose two or more.
We only need two points to graph a line, but the third point
will serve as a check, so we'll always select three values
for the input variable x.
A typical table, shown below, has a column for x, the input,
y, the output, and a third column for the ordered pairs.
Notice in this case, we've selected negative two, zero,
and two, for the input values.
And, in general, it's always good to select zero,
as well as a negative value, and a positive value.
Step two, we substitute each input value, or x value,
into the equation, and compute the corresponding
output values, or corresponding y values.
Then, we list these values in the table.
So, our equation y=x+1, so those below we've substituted
negative two into the equation, zero, and positive two.
So, here are the corresponding y values,
or corresponding outputs.
We record these in the table, which we see here.
Step three, write each input-output pair on the table
as an ordered pair, the input-output pair is the xy pair.
So, looking at the table, notice how here are the ordered
pairs, negative two comma negative one, zero comma one,
and two comma three.
Step four, we plot the ordered pairs, connect them,
then extend the line beyond the points
to show that the pattern continues.
So, looking at the coordinate plane, notice how the points
have been plotted in red.
And notice how they do form a straight line, graphed here
in blue, this blue line is the graph
of our linear equation, y=x+1.
Now, let's look at our own example.
We want to graph the linear equation y=3x-2, the first step
is to select the input values, or the x values.
Let's go and select negative one, zero, and two.
Once again, notice how we have a negative value, zero,
and a positive value.
We don't have to use these values, but these should
work well.
The only thing to be careful about is when we go to find
the corresponding y value, if the y value would not
be on the vertical axis, we have to come back and change
the input value, or x value, until the corresponding y
value would occur on the vertical axis.
Otherwise, we'd go to plot the point, it would not
be on the coordinate plane.
So, step two, we substitute these values into the linear
equation, so for x=-1, we'd have y=3[-1]-2, that'd
be negative three minus two, or negative five.
When y=0, we have y=3[0]-2, this would be zero minus two,
or negative two.
And finally, when we have x=2, we'd have 3[2]-2,
which would be six minus two, or four.
We now record these in the table.
So, we have negative five, negative two, and positive four.
Next step, let's write the ordered pairs.
So for this first row we have the ordered pair,
negative one comma negative five, second row we have zero
comma negative two, and the third row
we have two comma four.
So, because we know these ordered pairs, satisfy the linear
equation each ordered pair gives us the coordinates
of a point that would be on the line.
So, for the next step we graph these points
on the coordinate plane.
So, for negative one comma negative five, from the origin
we would go left one unit, and down five units to here.
For zero comma negative two, this would be the vertical
intercept, from the origin we would just go down two
units to here.
And then, finally, we have two comma four,
so from the origin we'd go right two units,
and up four units to here.
Now, we graph a line through the three points.
This is why we always find three points.
If we only found two points we wouldn't catch a mistake,
but if we find three points, if we can't sketch a line
through all three points, we know we've made a mistake,
and we can go back and check our work.
So, here's the graph for our linear equation.
Let's look at another example.
Here we have y=2/3x-1, and notice here we have a fraction.
Whenever we have a fraction times an input variable,
we do want to try select values of x that would eliminate
the fraction in determining the value of y.
To make sure the fractions is eliminated, we're going to pay
close attention to the denominator.
Notice, here, we have the denominator of three, which means
if we select x values, or input values, that are multiples
of three, we could avoid fractions we determining the output
value, or y value.
So, the most common multiples of three would be negative
three, zero, and positive three.
Again, we don't have to select these values for the input
variable x, but these would be common values to use.
And again, we're selecting multiples of three for the input
variable x, because we want to avoid fractions
when determining the corresponding output value for y.
If our denominator was two, we'd select multiples of two
for x, if the denominator was five, we'd select multiples
of five for x.
Now, we perform the substitution for x.
So, when x is negative three, we'd have y=2/3[-3],
which I'll write as negative three over one, and then, minus
one, notice we can simplify before multiply, there's
a common factor of three here.
It's one, three, and three, and one, three, and three.
So, here we have two times negative one, that's negative
two, minus one, so the corresponding y value
is negative three.
Next, when x is zero, we have y=2/3[0]-1, which would
be zero minus one, or negative one.
And then, finally, we have y equals, when x is three, we'd
have 2/3 times three, or three over one, minus one.
Again, we can simplify, it's one, three, and three, and one,
three, and three.
So, we have two times one, or two, minus one, which equals
positive one.
So, again, notice by selecting multiples of three
for the input variable x, we were able to avoid fractions
for the y values, or corresponding outputs.
Let's record the y values, we have negative three, negative
one, and one.
And therefore the ordered pairs would be negative three
comma negative three, zero comma negative one,
and three comma one.
Next step, plot the points.
So, for the ordered pair negative three comma negative
three, left three units, and down three units.
So, this is negative three comma negative three.
Next, we have a vertical intercept of zero comma negative
one, which is here.
And finally, we have the ordered pair three comma one,
so right three up one to here, and notice how these three
points do form a line, and therefore this would be the graph
of our linear equation.
I think we'll stop here for this lesson, we'll look at one
more example in part two.