Solve Equations By Substitution: A Step-by-Step Guide

Aug 4, 2025 by Sebastian Müller 54 views

Solving Systems of Equations with Substitution: A Step-by-Step Guide

Hey everyone! Today, we're diving into the world of solving systems of equations using the substitution method. It might sound intimidating, but trust me, it's a super useful tool in your math arsenal. We'll break down the process step-by-step, and by the end of this guide, you'll be a substitution pro!

What are Systems of Equations?

First, let's quickly recap what a system of equations actually is. Simply put, it's a set of two or more equations that contain the same variables. The goal is to find the values for those variables that make all the equations in the system true simultaneously. Think of it like finding the sweet spot that satisfies all the conditions.

For instance, consider the following system:

y = 5x + 2
4y - 16x = -4

Here, we have two equations, and both of them contain the variables 'x' and 'y'. Our mission, should we choose to accept it, is to find the values of 'x' and 'y' that work for both equations.

The Substitution Method: A Detailed Walkthrough

The substitution method is a clever technique that allows us to solve for one variable in terms of the other, and then substitute that expression into the other equation. This reduces the problem to a single equation with a single variable, which is much easier to solve. Let's break it down into clear steps.

Step 1: Solve for One Variable in Terms of the Other

The first crucial step is to pick one of the equations and solve it for one of the variables. Ideally, you want to choose an equation where one of the variables already has a coefficient of 1 (or -1). This makes the isolation process much cleaner. In our example system:

y = 5x + 2
4y - 16x = -4

The first equation, y = 5x + 2, is already solved for 'y'. This is fantastic! It means we can directly move on to the next step. However, just for illustration, let's say we didn't have an equation already solved. We could pick the second equation, 4y - 16x = -4, and solve it for 'y'. Here's how that would look:

Add 16x to both sides: 4y = 16x - 4
Divide both sides by 4: y = 4x - 1

So, we could express 'y' in terms of 'x' as y = 4x - 1. See? It's all about isolating one variable.

Step 2: Substitute the Expression into the Other Equation

This is where the magic of substitution really happens. We take the expression we found in Step 1 and substitute it into the other equation (the one we didn't use in Step 1). This is crucial because it eliminates one of the variables, leaving us with a single equation in a single variable. Using our original system and the already solved equation y = 5x + 2, we'll substitute this expression for 'y' into the second equation, 4y - 16x = -4:

4(5x + 2) - 16x = -4

Notice how we replaced 'y' with the entire expression (5x + 2). This is super important! Make sure you use parentheses to ensure the distribution is done correctly.

Step 3: Solve the Resulting Equation

Now we have a single equation with just one variable ('x' in this case). It's time to put your algebra skills to work and solve for that variable. Let's continue with our example:

4(5x + 2) - 16x = -4

Distribute the 4: 20x + 8 - 16x = -4
Combine like terms: 4x + 8 = -4
Subtract 8 from both sides: 4x = -12
Divide both sides by 4: x = -3

Great! We've found the value of x: x = -3.

Step 4: Substitute the Value Back to Find the Other Variable

We're halfway there! We now know the value of one variable ('x'). To find the value of the other variable ('y'), we simply substitute the value of 'x' back into either of the original equations (or the equation we solved for 'y' in Step 1). It usually makes sense to pick the simpler equation to make the calculation easier. In our case, the equation y = 5x + 2 looks pretty straightforward. Let's substitute x = -3 into this equation:

y = 5(-3) + 2
y = -15 + 2
y = -13

Awesome! We've found the value of y: y = -13.

Step 5: Check Your Solution

This is a crucial step that many people skip, but it's essential to ensure you have the correct answer. To check your solution, substitute the values you found for 'x' and 'y' into both of the original equations. If both equations are true, then your solution is correct! Let's check our solution x = -3 and y = -13:

Equation 1: y = 5x + 2
- -13 = 5(-3) + 2
- -13 = -15 + 2
- -13 = -13 (True!)
Equation 2: 4y - 16x = -4
- 4(-13) - 16(-3) = -4
- -52 + 48 = -4
- -4 = -4 (True!)

Since our solution satisfies both equations, we can confidently say that it's correct. Our solution is x = -3 and y = -13.

Putting It All Together: Solving the System

Okay, let's apply these steps to the given system of equations:

y = 5x + 2
4y - 16x = -4

Step 1: Solve for one variable. The first equation, y = 5x + 2, is already solved for 'y'.
Step 2: Substitute. Substitute 5x + 2 for 'y' in the second equation:
```
4(5x + 2) - 16x = -4
```

Step 3: Solve for x.

20x + 8 - 16x = -4
4x + 8 = -4
4x = -12
x = -3

Step 4: Solve for y. Substitute x = -3 into y = 5x + 2:
```
y = 5(-3) + 2
y = -15 + 2
y = -13
```
Step 5: Check. We already checked this solution in the previous section!

Therefore, the solution to the system is x = -3 and y = -13, which corresponds to the ordered pair (-3, -13).

Special Cases: No Solution and Infinite Solutions

Sometimes, when solving systems of equations, you might encounter some special cases:

No Solution

If, during the substitution process, you arrive at a contradiction (a statement that is always false, like 0 = 5), it means the system has no solution. This happens when the lines represented by the equations are parallel and never intersect.

Infinite Solutions

On the other hand, if you arrive at an identity (a statement that is always true, like 0 = 0), it means the system has infinitely many solutions. This happens when the two equations represent the same line.

Let's Look at an Example with Infinite Solutions

Consider the following system of equations:

x + y = 2
2x + 2y = 4

Let's use the substitution method to solve this. Solve the first equation for y:

y = 2 - x

Substitute this into the second equation:

2x + 2(2 - x) = 4
2x + 4 - 2x = 4
4 = 4

Notice that the variables canceled out, and we ended up with 4 = 4, which is a true statement. This indicates that the system has infinitely many solutions. Both equations represent the same line.

What About an Example with No Solution?

Consider this system:

y = 2x + 1
y = 2x + 3

Substitute the first equation into the second:

2x + 1 = 2x + 3
1 = 3

Here, we arrived at a false statement, 1 = 3. This means there is no solution. The lines are parallel and do not intersect.

Practice Makes Perfect

The best way to master the substitution method is to practice, practice, practice! Work through various examples, and you'll become more comfortable with the steps and the special cases. Don't be afraid to make mistakes – they're a valuable part of the learning process. Each time you solve a new problem, you strengthen your understanding and build your confidence.

Tips and Tricks for Substitution Success

Choose wisely: When deciding which variable to solve for in Step 1, look for the equation where a variable has a coefficient of 1 or -1. This will simplify your calculations.
Use parentheses: When substituting an expression, always use parentheses to avoid errors with distribution.
Check your work: Never skip the check step! Substituting your solution back into the original equations is the best way to catch mistakes.
Stay organized: Keep your work neat and organized. This will help you avoid errors and make it easier to follow your steps.

Conclusion

The substitution method is a powerful tool for solving systems of equations. By mastering this technique, you'll be well-equipped to tackle a wide range of mathematical problems. Remember the steps: solve for one variable, substitute, solve the resulting equation, substitute back to find the other variable, and always check your solution. With practice and a bit of patience, you'll be solving systems of equations like a pro in no time! So go forth, conquer those equations, and keep learning!

The correct answer to the initial problem is A. (-3, -13).