Using substitution to solve linear equations is an important skill in Year 8 math. It helps students understand variables and algebra. Let’s go through this step by step in a simple way.

Imagine we have two equations, which is what we usually look at when using substitution. Here’s an example:

1. ( y = 2x + 3 )
2. ( 3x + 2y = 12 )

The first equation tells us what (y) is in terms of (x). This is a great start because we can replace (y) in the second equation with what we found in the first. That’s what substitution means — we swap one variable with its value from another equation.

Let’s solve these equations using substitution. Follow these steps:

Step 1: Write Down the Equations

First, let’s make sure we have our equations:

• ( y = 2x + 3 ) (Equation 1)
• ( 3x + 2y = 12 ) (Equation 2)

Step 2: Substitute the Expression

Now, we substitute what we found in Equation 1 into Equation 2. This means we will change (y) in Equation 2 to (2x + 3):

$$3x + 2(2x + 3) = 12$$

Step 3: Simplify the Equation

Next, let’s simplify this equation. Start by multiplying out the (2):

$$3x + 4x + 6 = 12$$

Now, put together the like terms (the (3x) and (4x)):

$$7x + 6 = 12$$

Step 4: Solve for (x)

Continue solving for (x) by taking away (6) from both sides:

$$7x = 12 - 6$$ $$7x = 6$$

Now, divide both sides by (7):

$$x = \frac{6}{7}$$

Step 5: Find the Value of (y)

Now that we have (x), we need to find (y). Plug (x = \frac{6}{7}) back into Equation 1:

$$y = 2\left(\frac{6}{7}\right) + 3$$ $$y = \frac{12}{7} + 3$$

We can change (3) to ( \frac{21}{7} ) so that we can add them easily:

$$y = \frac{12}{7} + \frac{21}{7} = \frac{33}{7}$$

Step 6: Final Answer

So, the answers to the system of equations are:

$$x = \frac{6}{7} \quad \text{and} \quad y = \frac{33}{7}$$

Quick Review of the Process

To sum up, here’s how substitution works:

1. Rearrange to express one variable in terms of another.
2. Substitute this value into another equation.
3. Simplify and solve for the variable you isolated.
4. Substitute back to find the other variable.

Practice Example

To practice this, try solving the following equations using substitution:

1. ( y = 4x - 1 )
2. ( x + y = 10 )

Solution Steps:

1. Substitute (y) from the first equation into the second: $$x + (4x - 1) = 10$$
2. Combine the similar terms: $$5x - 1 = 10$$
3. Add (1) to both sides: $$5x = 11$$
4. Divide by (5): $$x = \frac{11}{5}$$
5. Put this back into the first equation to find (y): $$y = 4\left(\frac{11}{5}\right) - 1 = \frac{44}{5} - \frac{5}{5} = \frac{39}{5}$$

Conclusion

Using substitution to solve linear equations makes the process easier and clearer. For Year 8 students, getting this technique right is super important. It helps you get ready for more complicated math later. Plus, working with variables and expressions helps you practice problem-solving. Keep at it, and you'll get better with practice!