# How to Solve Simultaneous Equations

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• 0:01 Simultaneous Equations
• 0:54 Graphing Simultaneous…
• 1:30 Elimination Method
• 5:12 Substitution Method
• 8:35 No Solution
• 9:44 Summary

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Lesson Transcript
Instructor: Joseph Vigil
In this lesson, you'll learn what simultaneous equations are and discover methods for solving them, including graphing, the elimination method and the substitution method.

## What Are Simultaneous Equations?

Simultaneous equations are a set of independent equations that involve one or more common variables. We use the word simultaneous because there is at least one solution that satisfies all the equations at the same time.

A set of simultaneous equations might look like this:

7x - 2y = 45

5x + y = 37

For two or more equations to share a solution, they must use the same variables. In general, those variables can be raised to any power, but for now, we're just going to focus on linear equations where the exponents are no greater than 1.

Solving a single equation can already be challenging; so how do we solve more than one equation at a time, especially when they involve multiple variables?

In this lesson, we'll discuss three different methods for solving simultaneous equations.

## Graphing the Equations

The first method is to graph the set of equations to find where they intersect. The coordinates at the point of intersection are the solution that satisfies all of the equations in the set.

Let's use our previous example to see this method in action:

7x - 2y = 45

5x + y = 37

When we graph these equations, this is the result:

The graph shows that the equations intersect at the point (7, 2), meaning that the solution that satisfies both of these equations is x = 7, y = 2.

## The Elimination Method

Sometimes it's convenient to find the solution algebraically by eliminating one of the equation's variables. This is known as the elimination method. Let's take a look at this example:

I've color-coded the equations so we can keep track of their parts as we manipulate them.

Notice that the equations contain y and -y, which would cancel each other out if we added them together. So let's add together the equations and eliminate that y.

All I've done is taken the like terms from the two equations and combined them. Notice that I had to add like terms together on both sides of the equations. So everything on the left sides of the original equations ends up on the left side of the new equation, and everything on the right sides of the original equations ends up on the right side of the new equation.

In this new equation, we have y + -y. Adding a negative is the same as subtracting, so what we really have is y - y. The y has been eliminated! Now we can add the rest of the like terms and we'll only have to deal with the variable x.

When we do that, we have the equation:

3x = 3

Which we can solve for x:

3x / 3 = 3 / 3

x = 1

We've found the value for x. Since x = 1, we can go back to one of the original equations and plug in 1 for x:

x + y = 2

1 + y = 2

y = 2 - 1

y = 1

In this set of simultaneous equations, x and y both equal 1. The solution is x = 1, y = 1.

But what if the equations don't match so conveniently?

Let's solve the following set of simultaneous equations:

x + 5y = 17

20x - 10y = 10

None of our coefficients, or the numbers attached to the variables, match. But take a look at 5y and -10y. If we multiply 5y by 2, we'd have 10y, which we could then use to cancel out the -10y.

We can't just multiply one part of the first equation by 2, however. We need to multiply the whole equation by 2. So instead of:

x + 5y = 17

We'll have:

2(x + 5y) = 2(17)

2x + 10y = 34

All I've done is multiply each part of the equation by two. Now our set of equations looks like this:

I'll combine all the like terms from both equations again, just like I did last time:

Remember, 10y + -10y is the same as 10y - 10y, so they cancel each other out. Now, I'll simplify by adding the like terms and then solve for x:

22x = 44

22x / 22 = 44 / 22

x = 2

Now we can plug 2 in for x in either of the original equations:

x + 5y = 17

2 + 5y = 17

5y = 15

5y / 5 = 15 / 5

y = 3

The solution to this set of equations is x = 2, y = 3.

## The Substitution Method

There's another algebraic method - the substitution method - to solve simultaneous equations that works well if one of the variables has a coefficient of 1. Let's consider this set of equations:

x + 5y = 55

10x - 2y = 30

The x in the first equation has a coefficient of 1, so let's solve for x in the first equation by subtracting 5y from both sides:

x + 5y - 5y = 55 - 5y

x = 55 - 5y

The 5y cancels itself out on the equation's left side, but it's now on the right side. In essence, we 'moved' the 5y to the other side of the equation.

Don't worry about finding the value for either of the variables yet. Instead, since we now know that x is the same thing as 55 - 5y, we'll do a substitution in the second original equation:

Once I distribute the 10 into the expression in parentheses, I have:

550 - 50y - 2y = 30

Now, I can combine the like terms:

550 - 52y = 30

And then solve for y:

-52y = -520

-52y / -52 = -520 / -52

y = 10

We now have the value for y and can plug it in to either of the original equations to solve for x:

x + 5y = 55

x + 5(10) = 55

x + 50 = 55

x + 50 - 50 = 55 - 50

x = 5

The solution for this set of equations is {x = 5, y = 10}.

Let's solve one more set of equations using substitution:

7x - 2y = 45

5x + y = 37

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