# Solving systems of linear equations by substitution

##### Intros
###### Lessons
• What is a system of equations?
• What does it mean by "solving system of equations by substitution"?
##### Examples
###### Lessons
1. Solve each linear system algebraically by substitution
1. 6x - 1y = 7
-9x + 2y = 7
2. 3(x+2) - (y+7) = 4
5(x+1) + 4(y+3) = 31
3. ${{x+1} \over 2} + {{y-2} \over 14} = 3$
${{x-2} \over 24} - {{y+1} \over2} = -{11 \over 4}$
4. x - y = - 1

3x + 5y = 21
###### Topic Notes
Another way to solve systems of linear equations is by substitution. We can substitute the y/x in one of the equations with the y/x in the other equation.

## Introduction to Solving Systems of Linear Equations by Substitution

The substitution method is a powerful technique for solving systems of linear equations. Our introduction video provides a comprehensive overview of this essential mathematical concept. By watching this video, students will gain a solid foundation in understanding how to approach and solve these types of problems. The substitution method involves replacing a variable in one equation with an expression from another equation, allowing us to simplify and solve the system step-by-step. This approach is crucial for developing problem-solving skills and logical thinking in mathematics. Mastering the substitution method is not only important for solving systems of linear equations but also serves as a building block for more advanced mathematical concepts. As students progress in their math studies, they will find this technique invaluable in various applications, from algebra to calculus and beyond. The introduction video serves as an excellent starting point for grasping this fundamental method and its significance in mathematical problem-solving.

## Understanding the Substitution Method

The substitution method is a fundamental technique used in solving systems of linear equations. As its name suggests, this method involves substituting one variable with an expression from another equation, hence the term 'substitution.' This approach is widely used due to its versatility and effectiveness, especially when dealing with more solving complex equations beyond linear systems.

The substitution method differs from other techniques, such as elimination or graphing, in its approach to solving equations. While elimination focuses on canceling out variables and graphing relies on visual representation, substitution uses algebraic manipulation to express one variable in terms of another. This unique characteristic makes it particularly useful in various mathematical scenarios.

To illustrate the substitution method, let's consider a simple example. Suppose we have two equations:

1. y = 2x + 3

2. y = -x + 7

The substitution method would proceed as follows:

1. Take the expression for y from the first equation: y = 2x + 3
2. Substitute this expression into the second equation: (2x + 3) = -x + 7
3. Solve the resulting equation for x: 2x + 3 = -x + 7, 3x = 4, x = 4/3
4. Substitute the value of x back into either original equation to find y: y = 2(4/3) + 3 = 5 2/3

This example demonstrates the core principle of the substitution method: replacing a variable with an equivalent expression to create a solvable equation. The method's power lies in its ability to reduce the number of variables we're working with at any given time, simplifying the problem-solving process.

It's important to note that the substitution method is often the most frequently used approach, especially when moving beyond linear equations. Its flexibility allows it to be applied to a wide range of mathematical problems, including quadratic equations, polynomial systems, and even some non-linear equations. This versatility makes it an essential tool in a mathematician's toolkit.

The substitution method's effectiveness stems from its logical and step-by-step nature. It breaks down complex systems into manageable parts, allowing for a systematic approach to problem-solving. This methodical process not only aids in finding solutions but also helps in developing a deeper understanding of the relationships between variables in a system of equations.

As students progress in their mathematical studies, they'll find that the substitution method becomes increasingly valuable. It forms the foundation for more advanced techniques in algebra, calculus, and beyond. Mastering this method early on provides a solid basis for tackling more complex mathematical challenges in the future.

In conclusion, the substitution method is a powerful and versatile technique for solving systems of equations. Its name reflects its core principle of substituting expressions, setting it apart from other solving methods. Through practice and application, students can harness the full potential of this method, preparing them for more advanced mathematical concepts and problem-solving strategies.

## Step-by-Step Guide to Solving by Substitution

Solving systems of linear equations using the substitution method is a fundamental skill in algebra. This step-by-step guide will walk you through the process, using a simple example to illustrate each step. By the end, you'll understand how to solve substitution problems efficiently and identify which variable to isolate first for easier solving.

### What is the Substitution Method?

The substitution method involves replacing a variable in one equation with an equivalent expression from another equation. This allows us to solve for one variable and then use that solution to find the values of the other variables.

### Step-by-Step Substitution Guide

Let's use the following system of equations as our example:

• Equation 1: y = 2x + 3
• Equation 2: 3x - y = 7

#### Step 1: Choose a Variable to Isolate

In this case, y is already isolated in Equation 1, so we'll use that. If neither equation has an isolated variable, choose the one that's easiest to isolate.

#### Step 2: Substitute the Isolated Variable

Replace y in Equation 2 with the expression from Equation 1:

3x - (2x + 3) = 7

#### Step 3: Solve the Resulting Equation

Simplify and solve for x:

• 3x - 2x - 3 = 7
• x - 3 = 7
• x = 10

#### Step 4: Substitute the Solution

Now that we know x = 10, substitute this value into either of the original equations. Let's use Equation 1:

y = 2(10) + 3

y = 20 + 3

y = 23

#### Step 5: Check Your Solution

Verify that x = 10 and y = 23 satisfy both original equations:

• Equation 1: 23 = 2(10) + 3
• Equation 2: 3(10) - 23 = 7

### Tips for Identifying Which Variable to Isolate First

1. Look for pre-isolated variables: If one equation already has a variable isolated (like y = 2x + 3 in our example), use that equation for substitution.
2. Simplest coefficient: If no variable is isolated, choose the variable with the simplest coefficient (preferably 1 or -1) to isolate first.
3. Avoid fractions: When possible, choose a variable that won't result in fractional coefficients when isolated.
4. Consider the other equation: Look at both equations and determine which variable, when substituted, will lead to the simplest calculation in the other equation.

### Common Mistakes to Avoid

• Forgetting to distribute negative signs when substituting
• Making arithmetic errors during simplification
• Skipping the final check step
• Not solving for both variables

### Practice Makes Perfect

The key to mastering the substitution method is practice. Start with simple systems like our example, then gradually move to more

## Dealing with Fractions and Coefficients

When solving systems of equations, encountering fractions or complex coefficients can make the process more challenging. However, with the right strategies, you can simplify your work and solve these equations efficiently. This guide will explore effective methods for handling equations with fractions and complex coefficients, focusing on how to choose the best equation to work with first and how to simplify fractions during the substitution process.

One of the primary strategies for dealing with fractions in substitution is to choose the equation that will minimize fraction work. When faced with multiple equations, look for the one with the fewest fractions or the simplest coefficients. This approach can significantly reduce the complexity of your calculations and minimize the risk of errors.

For example, consider the following system of equations:

1/2x + 1/3y = 4
2x - y = 7

In this case, it's advisable to start with the second equation (2x - y = 7) as it doesn't contain any fractions. You can then use this equation to substitute for one of the variables in the first equation, simplifying the overall process.

When working with equations that contain fractions, it's often helpful to multiply both sides of the equation by the least common multiple (LCM) of the denominators. This technique eliminates fractions and makes the equation easier to manipulate. For instance:

1/2x + 1/3y = 4
Multiplying both sides by 6 (LCM of 2 and 3):
3x + 2y = 24

This transformed equation is now much simpler to work with during the substitution process.

Another effective strategy is to use cross-multiplication when dealing with equations that have fractions on both sides. This method can help simplify the equation and eliminate denominators. For example:

x/3 = y/4
Cross-multiplying:
4x = 3y

When faced with complex coefficients, it's often beneficial to factor out common terms or simplify expressions before proceeding with substitution. This can make the equations more manageable and reduce the likelihood of calculation errors. For instance:

5x + 10y = 30
Factoring out 5:
5(x + 2y) = 30

In some cases, you may encounter equations where variables appear in denominators. These situations require special care to avoid division by zero. It's often helpful to multiply both sides of the equation by the variable in the denominator to eliminate it. For example:

1/(x+1) + 1/(y-2) = 3
Multiplying by (x+1)(y-2):
(y-2) + (x+1) = 3(x+1)(y-2)

When substituting values during the solving process, it's crucial to simplify fractions whenever possible. This not only makes calculations easier but also helps in identifying potential errors. For example, if you've determined that x = 2/3, and you need to substitute this into the equation 3x + 4 = 10, simplify as follows:

3(2/3) + 4 = 10
2 + 4 = 10
6 = 10

By simplifying at each step, you can quickly verify if your solution is correct or if further calculations are needed.

In conclusion, handling equations with fractions and complex coefficients requires a strategic approach. By choosing the right equation to start with, simplifying fractions early in the process, and employing techniques like cross-multiplication and factoring, you can efficiently solve even the most challenging systems of equations. Remember to simplify at each step and always be mindful of potential division by zero scenarios. With practice, these strategies will become second nature, allowing you to tackle complex equations with confidence and precision.

## Common Challenges and Troubleshooting

The substitution method is a powerful tool for solving systems of equations, but students often encounter challenges when applying this technique. Understanding these common pitfalls and learning how to troubleshoot can significantly improve your problem-solving skills. Let's explore some of the most frequent substitution challenges and strategies to overcome them.

One of the primary obstacles students face is sign errors. When isolating a variable or substituting an expression, it's easy to mishandle positive and negative signs. This can lead to incorrect solutions or equations that don't balance. To avoid this, always double-check your work, paying close attention to signs when moving terms between sides of an equation.

Arithmetic mistakes are another common issue. Even simple calculation errors can derail the entire problem-solving process. To minimize these errors, break down complex calculations into smaller steps and use a calculator when necessary. It's also helpful to estimate your answer beforehand, so you can quickly identify if your final result seems reasonable.

Choosing the wrong variable to isolate can make the substitution process unnecessarily complicated. Look for equations where one variable is easily isolated, typically those with a coefficient of 1. This simplifies the substitution and reduces the likelihood of errors in subsequent steps.

Students sometimes struggle with properly substituting the isolated expression into the other equation. It's crucial to replace every instance of the variable with the entire expression, including parentheses where needed. Forgetting to do this can lead to incorrect equations and solutions.

Another challenge is losing track of the original variables when working with substituted expressions. Always clearly label your work and keep the original variables in mind throughout the process. This helps ensure you can interpret your final answer correctly.

To troubleshoot your work effectively, develop a systematic approach. After completing a problem, go back and check each step. Verify that you've correctly isolated variables, substituted expressions, and performed calculations accurately. It's also beneficial to plug your solution back into both original equations to confirm it satisfies the system.

Utilizing technology can be a valuable troubleshooting tool. Graphing calculators or online graphing tools can help you visualize the system of equations and confirm your solution visually. This can be particularly useful for identifying errors in your algebraic work.

Practice is key to overcoming substitution challenges. Work through a variety of problems, gradually increasing in complexity. As you encounter difficulties, take note of where you struggle and focus on improving those specific areas. Remember, even experienced problem-solvers make mistakes the key is learning how to identify and correct them efficiently.

By being aware of these common challenges and employing these troubleshooting strategies, you can enhance your proficiency with the substitution method. With practice and careful attention to detail, you'll find yourself tackling complex systems of equations with greater confidence and accuracy.

## Practice Problems and Solutions

To reinforce your understanding of substitution in algebra, let's work through a series of practice problems with step-by-step solutions. These problems range from basic to more complex, incorporating fractions and varied coefficients to provide comprehensive practice.

### Problem 1 (Easy)

Solve the system of equations:
x + y = 10
2x - y = 4

#### Solution:

1. From the first equation, express y in terms of x: y = 10 - x
2. Substitute this into the second equation: 2x - (10 - x) = 4
3. Simplify: 2x - 10 + x = 4
4. Combine like terms: 3x - 10 = 4
5. Add 10 to both sides: 3x = 14
6. Divide by 3: x = 14/3
7. Substitute x back into y = 10 - x: y = 10 - 14/3 = 16/3
8. Solution: x = 14/3, y = 16/3

### Problem 2 (Medium)

Solve the system of equations:
3x + 2y = 13
5x - 3y = 1

#### Solution:

1. From the first equation, express y in terms of x: y = (13 - 3x) / 2
2. Substitute this into the second equation: 5x - 3((13 - 3x) / 2) = 1
3. Multiply both sides by 2 to eliminate fractions: 10x - 3(13 - 3x) = 2
4. Distribute: 10x - 39 + 9x = 2
5. Combine like terms: 19x - 39 = 2
6. Add 39 to both sides: 19x = 41
7. Divide by 19: x = 41/19
8. Substitute x back into y = (13 - 3x) / 2: y = (13 - 3(41/19)) / 2 = 25/19
9. Solution: x = 41/19, y = 25/19

### Problem 3 (Challenging)

Solve the system of equations:
2/3x - 1/4y = 5
1/2x + 3/5y = 7

#### Solution:

1. Multiply the first equation by 12 to eliminate fractions: 8x - 3y = 60
2. Multiply the second equation by 10 to eliminate fractions: 5x + 6y = 70
3. From the first equation, express y in terms of x: y = (8x - 60) / 3
4. Substitute this into the second equation: 5x + 6((8x - 60) / 3) = 70
5. Multiply both sides by 3: 15x + 2(8x - 60) = 210
6. Distribute: 15x + 16x - 120 = 210
7. Combine like terms: 31x - 120 = 210
8. Add 120 to both sides: 31x = 330

## Conclusion

The substitution method for solving systems of linear equations is a powerful tool in algebra. Key points include isolating a variable in one equation, substituting its expression into the other equation, and solving for the remaining variable. This method is particularly useful when one equation is already solved for a variable. Remember, practice and patience are crucial for mastering the substitution method. Don't get discouraged if it seems challenging at first; with consistent effort, you'll improve. If you're still unsure, don't hesitate to rewatch the introduction video for a refresher. To solidify your understanding, attempt more practice problems of varying difficulty levels. As you work through different examples, you'll develop a better intuition for when and how to apply the substitution method effectively. Keep in mind that mastering this technique will serve you well in more advanced mathematical concepts and real-world problem-solving scenarios.

### Example:

Solve each linear system algebraically by substitution:
6x - 1y = 7
-9x + 2y = 7

#### Step 1: Isolate One Variable

To solve the system of equations by substitution, we first need to isolate one of the variables in one of the equations. In this case, we notice that the coefficient of $y$ in the first equation is -1, which makes it easier to isolate $y$.
Starting with the first equation:
6x - y = 7
We isolate $y$ by subtracting 6x from both sides:
-y = -6x + 7
To get a positive $y$, we divide both sides by -1:
y = 6x - 7

#### Step 2: Substitute the Isolated Variable

Now that we have $y$ isolated in terms of $x$, we substitute this expression into the second equation where $y$ appears. The second equation is:
-9x + 2y = 7
Substituting $y = 6x - 7$ into the equation:
-9x + 2(6x - 7) = 7

#### Step 3: Simplify and Solve for $x$

Next, we simplify the equation by distributing and combining like terms:
-9x + 12x - 14 = 7
Combine the $x$ terms:
3x - 14 = 7
Add 14 to both sides to isolate the term with $x$:
3x = 21
Divide by 3 to solve for $x$:
x = 7

#### Step 4: Substitute Back to Find $y$

Now that we have $x = 7$, we substitute this value back into the expression we found for $y$:
y = 6x - 7
Substitute $x = 7$:
y = 6(7) - 7
y = 42 - 7
y = 35

#### Step 5: Verify the Solution

Finally, we verify the solution by substituting $x = 7$ and $y = 35$ back into the original equations to ensure they are satisfied:
For the first equation:
6(7) - 35 = 7
42 - 35 = 7
7 = 7 (True)
For the second equation:
-9(7) + 2(35) = 7
-63 + 70 = 7
7 = 7 (True)
Since both equations are satisfied, the solution to the system is $x = 7$ and $y = 35$.

### FAQs

Here are some frequently asked questions about solving systems of linear equations by substitution:

#### 1. What is the Substitution Method?

The substitution method is a technique for solving systems of linear equations. It involves expressing one variable in terms of another from one equation and substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be solved.

#### 2. How do you solve by substitution step by step?

To solve by substitution: 1. Isolate one variable in one equation. 2. Substitute that expression into the other equation. 3. Solve the resulting equation for the remaining variable. 4. Substitute the found value back into either original equation to find the other variable. 5. Check your solution in both original equations.

#### 3. What is the formula of substitution method?

There isn't a single formula for the substitution method. Instead, it's a process of manipulating equations. However, if you have two equations like y = mx + b and ax + by = c, you would substitute the expression for y (mx + b) into the second equation: a(x) + b(mx + b) = c, and then solve for x.

#### 4. What is an example of substitution method?

Consider the system: y = 2x + 1 and 3x - y = 5. Substitute y from the first equation into the second: 3x - (2x + 1) = 5. Simplify: x - 1 = 5. Solve: x = 6. Then substitute back to find y = 2(6) + 1 = 13. The solution is (6, 13).

#### 5. How many steps are in the substitution method?

The substitution method typically involves 4-5 main steps: isolating a variable, substituting the expression, solving the resulting equation, back-substituting to find the other variable, and checking the solution. However, the complexity of each step can vary depending on the specific equations involved.

### Prerequisite Topics for Solving Systems of Linear Equations by Substitution

Understanding the foundations of algebra is crucial when tackling more advanced concepts like solving systems of linear equations by substitution. One of the key prerequisites is mastering applications of linear equations. This knowledge provides a solid base for recognizing real-world scenarios where systems of equations are applicable, making the learning process more relevant and engaging.

Another important aspect is being comfortable with distance and time related questions in linear equations. These types of problems often involve solving complex equations, which is a skill directly transferable to the substitution method in systems of equations. By practicing these questions, students develop the analytical thinking required for more intricate algebraic manipulations.

The ability to solve polynomials with unknown constant terms is also highly relevant. This skill involves expressing one variable in terms of another, which is the core principle behind the substitution method. Mastering this concept allows students to approach systems of equations with confidence and clarity.

Perhaps the most directly related prerequisite is solving linear equations with variables on both sides. This skill is essential because the substitution method often requires rearranging equations and isolating variables. Students who are proficient in this area will find the transition to systems of equations much smoother.

Lastly, a solid grasp of basic arithmetic, particularly dividing integers, is crucial. While this may seem elementary, it's vital for simplifying equations and combining like terms, which are frequent steps in the substitution method. Strong arithmetic skills ensure that students can focus on the problem-solving aspects rather than getting bogged down by calculations.

By mastering these prerequisite topics, students build a strong foundation for tackling systems of linear equations using the substitution method. Each concept contributes to the overall understanding and proficiency in this more advanced topic. The applications of linear equations provide context, while skills in solving complex equations and manipulating variables directly apply to the substitution process. Arithmetic proficiency ensures smooth execution of the method.

Remember, mathematics is a cumulative subject. Each new concept builds upon previous knowledge. Therefore, taking the time to thoroughly understand these prerequisites will not only make learning about systems of equations easier but will also enhance overall mathematical problem-solving skills. As students progress, they'll find that these foundational concepts continue to be relevant in more advanced mathematical studies.