Welcome to a lesson on solving
systems of linear equations using the substitution method.
The substitution method is an algebraic technique
for solving a system of equations.
It's important that we have algebraic techniques
for solving a system, because sometimes graphing
isn't practical.
The substitution method gets its name because we
perform substitution in order to solve the system.
Remember, the solution to a system of equations
is an ordered pair that satisfies both equations.
To understand how this process is going to work
before we formalize our steps, let's consider the given
system of equations.
Notice how the first equation, we have y equals 2x.
Since y must equal 2x, let's consider what happens
when we substitute 2x for y in the second equation.
So if we perform the substitution,
the second equation would be x plus not y,
but x plus 2x equals three.
So by performing this substitution we get a single equation
with a single variable, which we can now solve.
X plus 2x equals 3x, so we have 3x equals three,
dividing both sides by three,
notice how we get x equals one.
So for our ordered pair solution,
the first value is going to be one.
Now that we know x equals one, and we also know y equals 2x,
because that's one of the given equations,
if we substitute one for x,
we'd have y equals two times one,
in there for y equals two.
So the solution to this system is the ordered pair
one comma two.
Let's take a moment and verify this.
So we'll substitute one for x, and two for y,
in both equations.
So for the first equation we would have
two equals two times one, which is true.
And for the second equation we would have
one plus two equals three, which is also true,
verifying our solution is correct.
But going back to the system just for a moment,
this worked out very nicely because notice how
one of the equations was solved for y.
But in many cases we won't have one of the equations
solve for one of the variables,
and therefore we'll follow the steps given below.
Let's look at an example and walk through each step.
For step one, we solve one of the equations of the system
for one of the variables.
So looking at our system here below,
notice how it would be easiest to solve the second equation
for y, because we can solve for y in one step.
If we solve for any of the other variables,
in either equation, it would take two steps,
we might end up with fractions,
which would complicate things, so for step one,
let's go ahead and solve this equation for y,
so we'll subtract 2x on both sides which will give us
y equals 20 minus 2x.
So this was step one.
Step two, we want to substitute the expression for the
variable obtained in step one into the other equation.
So because we solve this equation for y,
we know y equals 20 minus 2x,
we'll substitute 20 minus 2x for y in the first equation.
So when we do this for step two,
we would have 3x minus two times, not y, but 20 minus 2x,
and then equals 16.
Step three we want to solve this equation,
so we would clear the parentheses by distributing,
because of the subtraction we would distribute negative two.
So step three we'd have 3x minus 40 plus 4x equals 16.
Combine like terms, 3x plus 4x is 7x.
Adding 40 to both sides, we'd have 7x equals 56,
dividing both sides by seven, we have x equals eight,
since 56 divided by seven equals eight.
Now we've made some good progress,
we know that the x value must be eight,
but remember, the solution is an ordered pair,
so for our solution, for the ordered pair,
the first value is going to be eight.
Step four, substitute the result back into one of
the original equations to find the ordered pair solution.
So we want to substitute eight for x
into one of the equations here,
let's go ahead and use the second equation
and substitute eight for x.
So for step four, we would have two times eight,
plus y equals 20, 16 plus y equals 20,
subtracting 16 on both sides, y equals four.
So our ordered pair solution is eight comma four.
Step five says check the result into both
of the original equations, so we'll substitute eight for x,
and four for y in both of these equations here.
So for the first equation, in step five,
we have three times x which is eight,
minus two times y which is four, equals 16.
So 24 minus eight equals 16, 24 minus eight is 16,
so our solution works in the first equation.
And now for the second equation we would have
two times eight plus y is four,
equals 20, we have 16 plus four equals 20,
which also checks.
Let's look at another example.
Again, our first step is to determine which equation
to solve for which variable,
and notice how it's going to be easiest to solve
this second equation here for x by adding 2y to both sides.
So if we add 2y to both sides,
we have x equals negative three plus 2y,
so this is step one.
Step two, we'll substitute negative three plus 2y for x,
and the first equation, which would be here,
so for step two, we'll have five times the quantity
negative three plus 2y, minus 4y, equals nine.
Step three, we want to solve this equation,
so we'll distribute five here, and then solve for y.
So we'd have negative 15 plus 10y, minus 4y equals nine.
Combine like terms, here we have 10y minus 4y, that's 6y.
So negative 15 plus 6y equals nine.
To isolate 6y we would add 15 to both sides,
nine plus 15 is 24, so we have 6y equals 24,
divide both sides by six, 24 divided by six is four.
So we know y is equal to positive four, so for our solution,
which we know is going to be an ordered pair,
the second value will be four.
Step four, we want to substitute this value here for y into
one of the equations to determine our ordered pair solution.
Because we're trying to find the value of x,
let's use equation two, so for step four,
using equation two we'd have x minus two times four,
because we now know y is four, equals negative three,
so x minus eight equals negative three,
adding eight to both sides, x equals five.
So our ordered pair solution would be five comma four.
Step five, we want to check our solution.
So we'll substitute these values into both equations.
With the first equation we have five times five,
minus four times four, equals nine.
So 25 minus 16 equals nine, which is true.
So our solution works in the first equation.
And now for the second equation, we have x which is five,
minus two times y which is four, equals negative three.
So five minus eight equals negative three, which is true.
So we verify this ordered pair does satisfy both
of our equations in the system.
Example three, first step, solve for one of the variables.
Looking at example three, notice how for our first step,
it'll be easiest if we solve this first equation for y.
So if we solve this equation for y we would subtract 3x
on both sides, so y equals five minus 3x,
this would be our first step.
Step two, because we solved this equation for y,
we want to substitute five minus 3x for y
in the second equation, so step two,
we would have 6x plus two times, for y we're substituting
five minus 3x, this is equal to positive 11.
Step three, we solve this equation,
so we'll distribute here, so we'll distribute positive two,
so step three would be 6x plus 10, minus 6x equals 11.
Now here, when we combine like terms,
something special happens, notice 6x minus 6x would be zero,
so when we simplify this equation, we get 10 equals 11,
we know 10 does not equal 11, this is not true.
So whenever the variables drop out,
and we have a statement that's not true,
this indicates our system has no solution.
So let's look at our last example, example four.
Looking at our system of equations,
notice how the second equation here is already solved for y,
so step one has already been done for us for step one,
we already know y is equal to x plus one.
But because we're using the second equation here,
we'll now substitute x plus one for y in the first equation.
So step two we would have x minus the quantity x plus one,
equals negative one.
Step three is to solve this equation,
we need to be careful here, if it's helpful,
they can think of having a one here,
and therefore to clear the parentheses,
we'll distribute negative one because of the subtraction.
So step three, we'd have x,
minus x, minus one equals negative one.
Notice here the variables simplify out again,
x minus x is equal to zero, so this simplifies to
negative one equals negative one.
But notice here we have a true statement,
of course negative one equals negative one,
so whenever the variables drop out,
and we have a true statement,
this indicates we have infinitely many solutions.
So in this lesson, we saw all three types of
systems of linear equations where it's possible
to have one solution, like example one and example two.
It's possible to have no solution like example three.
And it's also possible to have infinitely many solutions
like we see here in example four.
I hope you found this helpful.