How to Solve Systems of Equations

An equation is a statement that two mathematical expressions are equal. Expressions represent quantities and can be strictly numerical or contain some combination of numerical terms and terms containing variables. For instance, \(6 + 3\) is a numerical expression, and \(2x – 7\) is an expression that contains a variable.

Equations are like sentences, and they can be evaluated as true or false. For example, \(2 × 5 = 10\) is a true statement because the quantity \(2\times 5\) is equal to the quantity 10. On the other hand, \(25 = 4\times 5\) is a false statement, because the quantity 25 is not equal to the quantity \(4\times 5\).

When equations contain variables, the job of solving is to discover the value(s) of the variables that make the equation true. For example, \(x = 10\) is only true when \(x\) has the value of 10, while \(2x = 10\) is only true when \(x\) has the value of 5.

Typically, equations are approached from the standpoint of “balancing,” “performing inverse operations,” and “isolating the variable.”

Solving Systems of Equations

Here’s an example of how the process of solving equations might be approached.

Equations as Scenarios

Expressions often represent quantities that are related in a certain way. Some common examples are\(l\), \(x\), and \(w\), which often represent the area of a rectangle, and \(\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \), which often represents the distance between two points on the coordinate plane. Sometimes, we need to create our own equations based on given information.

Example: Max is Ruby’s little brother. Ruby is half of three times Max’s age, and the sum of their ages is 25 years. How old is each sibling?

Let the expression \(a\) represent Max’s age. Then Ruby’s age is represented by the expression \(\frac{3a}{2}\) and the equation we need to solve is \(a+ \frac{3a}{2} = 25\). In other words, Max’s age plus Ruby’s age is 25.

Sometimes, there are multiple ways to visualize an equation. Sometimes, the order we use depends on how we see the terms. For example, let’s look at a couple of different ways we can visualize and solve \(\frac{3x}{4} = 30 \):

Visualizing the problem this way will lead you down two different paths to solving it:

Finally, let’s look at a couple of different ways we can visualize and solve \(\frac{3}{4x} = 30:\)

Hey guys! Welcome to this video over systems of equations.

A system of equations is a group of two or more equations, and each of the equations within the group have at least one unknown variable.

When given a system of equations, the goal is to find the value for each of the unknown variables.

In this video, we will discuss two tools to help you solve for the unknown variables: substitution, and elimination.

Substitution

Substitution is a way of solving the system by getting rid of all of the variables, except for one, then solving for that equation.

The best time to use substitution is when you have a variable that has a coefficient of 1 or -1. The reason why is, because if it has a coefficient of 1 or -1, then you don’t need to undo multiplication or division, you just need to undo addition or subtraction in order to isolate a variable.

There are three steps that you need to follow in order to be able to solve the system using substitution.

 
So let’s get started!

Example

 
In this example, we can see that our second equation has a variable with 1 as the coefficient. So, that lets us know to solve for \(x\). Now, you can actually solve for any of the variables. It will just always be easier to solve for one that has a coefficient of 1 or -1.

Now, to do this we just add \(9y\) to both sides.

 
So, we’ve done what our first step tells us to do and we solved for one of our variables. Step two now tells us to plug in the variable that we have solved for into our other equation, then solve. So we’re going to plug our \(x\) into \(4x+3y = 2\). Because we know that \(x=9y -19\), we’ll plug this into our \(x\)-value here and then solve.

 
Now that we have it down to one variable, we are able to solve for the value of that variable, so in this case, \(y\).

Let’s rewrite this by multiply our 4 by everything inside our parentheses.

 
Now we can add our 76 to both sides. You can do this multiple ways. If you wanted to add your \(y\)s together first, then do that. I’m just doing what is easier in my mind.

Once I add my 76 to both sides and add my \(y\)s together, I get \(39y= 78\).

Now, I divide both sides by 39.

 
Use the number that you get when you solve to solve for the other variable, or variables.

So now all we have to do is take our \(y\)-value and plug it into one of our original equations. It doesn’t matter which one! You can plug it into the first or the second, but I am going to plug it into the second equation, just because.

So that gives you:

 
And we’re done! We have found the value of both of our variables using substitution.

Now, let’s take a look at how to solve for a system using elimination.

Elimination

The reason this tool is called elimination is because you add together the two equations in order to eliminate one of your variables.

Example 1

 
We can tell just by looking at it, that our \(y\)s will cancel out. So once we add the two equations together. We have \(10x = 30\).

And we can divide both sides by 10: \(x=3\)

We can take our \(x\)-value and plug it into either of our original equations. I’ll plug it into our first one here. So, we have:

 
So, and then I’ll rewrite this:

 
Now, we need to move our 9 to the other side by subtracting 9.

 
Then we divide both sides by our -4 here.

 
So now we’ve solved and found both of our variables. That was a good example to learn how elimination works, but it may not always be that our terms cancel so easily.

Like in this example:

Example 2

 
None of our terms cancel right off the bat, so we’ll need to do a little manipulation in order to get them to do what we want them to do.

So, what could we do to get our terms to cancel? Well, there are a couple of things, but the easiest, as it appears to me, is to multiply our first equation by 2. That will allow us to cancel our \(y\)s, because we will be adding a \(-6\) and a \(+6\). So, let’s do that.

We have this equation (\(9x – 3y = -57\)) all being multiplied by 2, so let’s rewrite that and we get:

 
Now, we need to take this and add it to our other equation so we can get our \(y\)s to cancel out. So that would give us \(20x = -80\).

Now, we need to divide both of our sides by 20, which gives us \(x=-4\).

So, now that we know that \(x=-4\), we can plug in our \(-4\) where our \(x\)s are in either one of the original equations.

I’ll plug it into the second one. So we have:

 
And I’ll simplify this and rewrite it.

 
Now I’ll move and my 8 to both sides: \(6y=42\).

Now we divide both sides by 6 here: \(y=7\)

So our final answer to this problem is this:

 
I hope that this video on how to solve a system of equations using substitution and elimination has been helpful.

See you guys next time!

Frequently Asked Questions

System of Equations Practice Problems