Hi, and welcome to this video on using substitution and elimination to solve linear systems!
Solving a system of equations means to determine the exact (x,y) coordinate that
satisfies both of the equations in the system.
The process of solving a system depends on the structure of the equations.
Some systems can easily be solved by graphing both equations and determining the exact point
of intersection, while other systems are more suitable to be solved algebraically.
This video will focus on two of the algebraic approaches, namely substitution and elimination.
To solve a system of two equations with two variables, we are trying to find the exact
point that satisfies both equations.
What do I mean by satisfy?
That means that if you substitute that point into both equations, you will have a true statement.
We also already mentioned that the structure of the system will determine which of the
solving methods to use.
The substitution method is typically used when one of the equations in the system is
already solved for one of the variables.
When this is the case, you can substitute the expression that the variable is equal to into the other equation.
For example:
y equals x plus 2,
and x plus 2y equals negative 5
Because we know that the variable y equals x plus 2, we can substitute (x plus 2) for the variable y in
the second equation.
Note that after the substitution, the y-variable is no longer in the equation.
At this point, we solve for x by distributing the coefficient of 2 into the parentheses.
Combine like terms, x plus 2x, and gather the constants by subtracting 4 from both sides of the equation:
Then we’re going to divide both sides by 3 and this results in x equals negative 3.
At this point, we are half-way to our solution.
In order to determine the y-variable, simply substitute x equals negative 3 into either of the original equations.
Substituting into the first equation gives the quickest solution:
The solution for this system is the ordered pair negative 3, negative 1.
Let’s practice the substitution method by looking at one more problem together.
y equals negative 2x plus 4
and 3x plus 2y equals 1
Because the “y” is solved in terms of “x” in the first equation, substitute the expression
negative 2x plus 4, for y in the second equation.
Distribute the 2 into the parentheses,
combine like terms, and solve for x.
Solve for y by substituting x equals 7 into the first equation of the system:
So the ordered pair that is the solution to this system is 7, negative 10.
To check your work, simply substitute this ordered pair into both equations to verify
that you get a true statement.
Remember, these types of systems were well suited for the substitution method because
one of the variables was already solved.
For systems of equations that have two equations in “standard form”, the elimination method
may be preferred.
Recall that standard-form equations have both variable terms on one side of the equation
and constant terms on the other.
The overall concept of this method is to “eliminate” one of the variables from the system by adding
like terms of the equation together.
A straightforward example of the elimination method is shown:
2x plus y equals 1,
and x minus y equals 8
Both of these equations are in standard form.
If the equations are added by combining like terms, the y-variable is eliminated because
the leading coefficients of the y-variables are of opposite signs.
Solve for x by dividing both sides by 3, which results in x equals 3.
Now, as before, use this value to determine the value of y by substituting x equals 3 into either
of the original equations.
Let’s use the second equation of the system because it is fairly simple:
To solve for y, subtract 3 from both sides
and divide by negative 1 for a solution of y equals negative 5.
So the final solution to this system is the ordered pair 3, negative 5.
As mentioned, that example was fairly straightforward because the coefficients of the y-variables
canceled out when added.
When coefficients do not cancel, you can use the Multiplication Property of Equality to
multiply the equation by whatever value you need to eliminate.
To wrap up, let’s see if you remember which method of solving would be most appropriate
for the following systems:
For this one you would use substitution, since y is solved in the second equation.
For the second one, we’d use elimination, since both equations are in the standard form
and the y-term is eliminated by adding.
Thanks for watching, and happy studying!