Today, we will be discussing systems of equations and how we can solve them using the two methods: Elimination and substitution.
So first off, what is a system of equations?
Currently, this is not a system. For us to have a system, we need to have more than one equation.
So if we write another equation underneath, this is now what we would call a system of equations.
For systems of equations, there usually exists a point (x, y) that we can plug in for x and y in either of these equations and they’ll both evaluate to the correct thing.
In this case, that point is (11, 6). We call a point like this a solution. Let’s plug this point into our first equation.
2(11) − 3(6) = 4
22 − 18 = 4
4 = 4
If we were to plug this point into our other equation, it would also evaluate to the correct solution.
So how do we find these solutions?
We can use two methods called elimination and substitution.
Let’s recall the system of equations from earlier.
2x − 3y = 4
−2x + 4y = 2
Let’s find the solution of this system using elimination. When we eliminate, we want to combine our two equations in some way so that we will get rid of one of our variables.
In this case, our x variables have coefficients that will cancel each other out if we add them together.
Let’s add the two equations together. 2x plus negative 2x is equal to zero x. Negative three y plus four y is equal to y. Four plus two is equal to six. This tells us that y is equal to six.
y = 6
We can plug this value of y into either of our equations and solve for x.
2x − 3(6) = 4
2x − 18 = 4
2x = 22
x = 11
In this case, we get that x is equal to eleven.
Let’s take a look at another situation.
x − 2y = 4
x + 4y = 4
In this case, if we were to add our equations together, we would not cancel out either of our variables. We could, however, subtract one equation from the other to cancel out a variable.
Let’s subtract the bottom equation from the top equation.
x minus x is equal to zero x, negative two y minus four y is equal to negative 6y and 4 minus 4 is zero. This means that negative 6 y is equal to zero so y must be equal to zero.
−6y = 0
y = 0
We can plug the value of y back into either equation to find x.
x − 2(0) = 4
x = 4
Sometimes, we might face a situation where we will be forced to multiply one or both of the equations in order to get the coefficients to cancel out.
4x + 6y = 8
−2x − 4y = −4
In this case, we can cancel out our x variables if we multiply the second equation by 2, and then add the two equations together.
4x + 6y = 8
−4x − 8y = −8
0x − 2y = 0
y = 0
If we plug the value of y back into either equation we can solve for x, and we will find that x = 2.
In some cases, it makes more sense to use the method of substitution rather than elimination. With substitution, we replace one of our variables with an expression that is equal to the variable.
Let’s take a look at this example:
x = 3y − 4
2x + 2y = 8
In this case, we have an expression which is equal to the variable x. To use substitution, rewrite the second equation except instead of writing the variable x we will write 3y − 4.
2(3y − 4) + 2y = 8
Simplify the expression to solve for y.
6y − 8 + 2y = 8
8y = 16
y = 2
We can plug this value back into either equation to solve for x.
If we are asked to use substitution to solve a problem that is not already set up to be used right away, we may have to spend some time isolating one of our variables.
With systems of equations, we sometimes have the special cases where we have no solutions and where we have infinitely many solutions.
3x − y = 2
3x − y = 1
In this case, we will always be forced to cancel both of our variables out.
If we subtract the two equations we will get:
0 = 3
Since this is impossible, we say that this system has no solutions.
In the case where both of our equations are actually the same, we will have infinitely many solutions.
Let’s take a look at a system which has infinitely many solutions.
2x + 4y = 4
x + 2y = 2
In this case, if we divide the first equation by 2, we will get exactly the same equation as the second one. If we try to subtract the same equation from itself, we will be left with 0 = 0.
In the case where both of our equations are the same, we have infinitely many solutions.
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