We want to solve the system of equations. Here we have a system have two linear
equations
with two unknowns. There are two main ways to solve a system of equations
algebraically.
One method is the elimination method, the other method is the substitution method.
We'll look at both methods for this example. Let's begin by looking at the
elimination method.
The idea behind the elimination method is, we want to add the equations together
and eliminate either the x-terms, or the y-terms. But if we're going to add the equations
together
and eliminate the extra y-terms because the x-terms, or y-terms must be opposites.
So looking at the two equations, notice how right now
the two y-terms are the same. So if we multiplied one equation by negative one,
then the y-terms would be opposite. One would be positive three-y,
the other would have a negative three-y. So let's go ahead and multiply the second equation by
negative one and we'll leave the first equation
the same. So the first equation is going to stay
four-x plus three-y equals fourteen.
And now we're going to multiply both sides of the second equation by negative one.
So negative one times two-x is negative two-x.
Negative one times three-y is negative three-y, or minus three-y.
And negative one times four is equal to negative four.
And now we're going to add the equations together.
So four-x plus negative two-x is two-x.
Next we have three-y plus negative three-y, which is zero,
so the y-terms are eliminated. And we have the equation two-x
equals fourteen plus negative four is ten.
Notice now we have an equation with one variable, which we can now solve for x.
So if we divide both sides by two we have x
equals ten divided by two, which equals five. We need to be careful though, we're not done.
The solution consists of one x and one y value,
or an ordered pair. So as an ordered pair, the ordered pair would be x, comma, y. So now we know
x is five. So we have five, comma. We still need to find the y value,
or we could just say x equals five. And now we need to find the value of y.
So to find the value of y, we need to perform substitution for x
into one of the original equations. So we'll substitute five for x and solve for y,
in either the first, or second equation. Let's go ahead and use the second equation here
and substitute five for x. So we'd have two times five,
plus three-y equals four. Two times five is
ten.
So solving for y, we would subtract ten on both sides
it would give us three-y
equals, four minus ten is negative six,
dividing both sides by three. We get y equals negative two.
So our solution to the system is x equals five
and y equals negative two. Or as an ordered pair, five, comma negative two.
Now let's take a look at the same system
and use the method of substitution. Using the method of substitution,
the first step is to solve one of the equations
for one of the variables. It doesn't matter which equation we use,
or which variable we solve for. Let's use the second equation,
two-x plus three-y equals four and we'll solve for x.
So we'll subtract three-y on both sides, which would give us two-x
equals four minus three-y. Then we'll divide everything by two.
Simplifying we get x
equals. Well four divided by two is two, so x equals two
minus three-halves y. Now that we've solved the second equation
for x, we're now going to go to the first equation
and substitute two minus three-halves y for x.
This will give us one equation that only contains y.
So performing this substitution we would have four
times x, which is two, minus three-halves y.
Then we sill have plus three-y
equals fourteen. So now we'll solve for y, we'll include
the parenthesis by distributing here. So four times two is eight
and then we'll have minus. Well four times three-halves that would be
twelve-halves, or six. So we have minus six-y
plus three-y equals fourteen.
Combining like terms we have eight, negative six-y
plus three-y is negative three-y, so minus three-y
equals fourteen.
Subtracting eight on both sides
we have negative three-y equals, fourteen minus eight is six.
Dividing both sides by negative three, we have
y equals negative two. So we had the same value last time.
As an ordered pair we know the y-coordinate would be
negative two and we don't know x but we know y is equal
to negative two. So now to find x,
we need to perform substitution and we'll substitute negative two for y.
And let's use this equation here already solved for x.
So if we substutute negative two for y, we would have
x equals two minus three-halves, times y, which is negative two,
which we'll write as negative two over one. So we have x
equals, notice here we can simplify.
The twos simplify to ones. But this simplifies to negative one
we we need to be careful about the sign. So we have two,
and then minus three times negative one,
that'd be minus negative three, or plus three, which equals five.
So now we know x equals five by using the substitution method.
So you can probably see for this example, using the substitution was actually
more work,
but the result is the same. A third method we could use would be to graph the two
linear equation on the coordinate plane
and verify this would be the point of intersection. I've already done this to save some time.
Here's a graph of the two lines and notice how the point of intersection,
this point here, does have an x-coordinate of five
and a y-coordinate of negative two.
Verifying our solution is x equals five
and y equals negative two.
I hope you found this helpful.