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# Solving systems of linear equations by substitution

- Intro Lesson: a17:16
- Lesson: 1a2:45
- Lesson: 1b4:56
- Lesson: 1c2:48

## How to Solve System of Equations:

Before we get into solving systems of linear equations via the substitution method, let's first consider and understand what it means to "solve" a system of equations. When we say "solve", with regards to linear, quadratic, exponential, or any other type of equation, what we really mean is that we are trying to find values of 'x' – the dependent variable – that satisfy 'y' – the independent variable.

Take for example the following, simple, equation: y = 2x = 2

In this example equation, we know that y is equal to 2x and is also equal to 2. With that knowledge, since y is equal to both 2x and 2, we can say that 2x = 2. Then, the next natural step is to solve this equation using algebra, giving us the "solution" that x = 1.

In the case of systems of equations, the process isn't that different. In solving systems of equations, what we are trying to do is trying to find values of x and y that makes two distinct equations equal to each other – effectively "solving" both equations. Further information on system of equations can be founded in another lesson. In a system of equations, there are several outcomes which can occur with regards to the number of solutions. We have the specific lessons on how to determine the number of solutions to linear equations and system of linear-quadratic equations. We also have graphing systems of equations and inequalities covered!

To do so, there are two main methods: solving systems by substitution, and solving systems by elimination. In this article, we will focus on substitution, which is arguably slightly more simple than the other method, elimination. For elimination, please check out the video and articles that focus on that method in particular. To make sure you're ready for elimination, it is important to master adding and subtracting polynomials and adding and subtracting rational expressions.

Now that we've covered the basics, let's solve systems using substitution!

## Solving Systems of Equations By Substitution:

Before we get into using the method of substitution, make sure you're comfortable with your algebra by reviewing the lesson on solving linear equations with variables on both sides.

The basic procedure behind solving systems via substitution is simple: Given two linear equations, all we need to do is to "substitute" one in the pair of equations into its other by rearranging for variables. This procedure is better outlined below with the general example:

Consider the following equations, with (x,y) being coordinates and everything else representing constants.

1) $ty = ax$

2) $zy = x + b$

**Step 1: Rearrange one of the equations to get 'y' by itself**

1) $y = \frac{ax}{t}$

2) $zy = x + b$

**Step 2: Substitute the rearranged equation into its partner**

**Step 3: Solve for x**

Since this is just a general case, we can't solve for x. But note all we have to do is get x by itself.

**Step 4: Substitute the solution for x into either of the initially given equations to find y**

Once we have the value for x, we can substitute it into any of the two equations to find our solution for y.

**Step 5: Write final answer out as a point**

Therefore, our solution is (x,y)

Once again, this is just a general case. Also note that in this example we chose to solve for x first. It doesn't matter which variable you solve first, just note that x is often the easier one to solve for first, as it often involves less modification in the initial give equations. The best way to learn and master how to solve by substitution is to do some practice problems.

**Example 1:**

Take the following simultaneous equations and solve.

**Step 1: Rearrange one of the equations to get 'y' by itself**

Let's use the first equation and rearrange it so we can have y by itself. We could certainly take the second equation, but that would involve more work.

**Step 2: Substitute the rearranged equation into its partner**

Now, we are going to substitute our newly rearranged equation 6x - 7 = y into -9x + 2y = 7.

**Step 3: Solve for x**

Now that we have successfully performed substitution, let's solve for x.

**Step 4: Substitute the solution for x into either of the initially given equations to find y**

Now that we have x, we can put x=7 into either of the equations to solve for y. Let's chose the first equation because it is more simple.

**Step 5: Write final answer out as a point**

The final answer: (7, 35)

The following image below summarizes the work we've just done:

**Example 2:**

Solve the following linear system.

In some instances, we are going to need to do some simplification of both equations before we can carry on with substitution and solving. In this case, we must first expand and simplify both equations:

**Step 1: Rearrange one of the equations to get 'y' by itself**

Just like in the first example, let's use the first equation and rearrange it so we can have y by itself. We could certainly take the second equation, but that would involve more work.

**Step 2: Substitute the rearranged equation into its partner and solve for x**

Now, we are going to substitute our newly rearranged equation 3x - 5 = y into 5x + 4y = 14 and solve for x.

**Step 3: Substitute the solution for x into either of the initially given equations to find y**

Now that we have x, we can put x = 2 into either of the equations to solve for y. Let's chose the first equation because it is more simple.

**Step 4: Write final answer out as a point**

The final answer: (2, 1)

And that's all there is to it! Now, make sure you do lots of practice problems to get more comfortable using this method. As well, check out this great link, which will allow you to easily check your work.

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### Solving systems of linear equations by substitution

#### Lessons

- Introductiona)
- What is a system of equations?
- What does it mean by "solving system of equations by substitution"?

- 1.Solve each linear system algebraically by substitutiona)6x - 1y = 7

-9x + 2y = 7b)3(x+2) - (y+7) = 4

5(x+1) + 4(y+3) = 31c)x - y = - 1

3x + 5y = 21