Substitute the expression for y into the other equation

Oct 13, 2025 by sorusgentrification.com 23 views

Introduction

In mathematics, solving a system of equations is a fundamental concept that involves finding the values of variables that satisfy multiple equations simultaneously. There are several methods to solve a system of equations, including substitution, elimination, and graphing. In this article, we will focus on solving a system of equations using the substitution method.

What is the Substitution Method?

The substitution method is a technique used to solve a system of equations by substituting the expression of one variable from one equation into the other equation. This method is particularly useful when one of the equations is linear and the other is quadratic or has a more complex form.

Step-by-Step Guide to Solving the System of Equations by Substitution

To solve the system of equations using the substitution method, follow these steps:

Identify the equations: The given system of equations is:
${ \begin{array}{l} y = 4x - 9 \\ y = x - 3 \end{array} \}$
We have two linear equations in the form of $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.
Solve one equation for y: We can choose either equation to solve for $y$ . Let's solve the first equation for $y$ :

$y = 4x - 9$

This equation is already solved for $y$ .
Substitute the expression for y into the other equation: Substitute the expression for $y$ from the first equation into the second equation:

$4x - 9 = x - 3$

Now we have a single equation with one variable, $x$ .
Solve for x: To solve for $x$ , we need to isolate the variable on one side of the equation. We can do this by adding $9$ to both sides of the equation and then subtracting $x$ from both sides:

$4x - x = -3 + 9$

This simplifies to:

$3x = 6$

Now, divide both sides by $3$ to solve for $x$ :

$x = \frac{6}{3}$

$x = 2$
Find the value of y: Now that we have the value of $x$ , we can substitute it into one of the original equations to find the value of $y$ . Let's use the first equation:

$y = 4x - 9$

Substitute $x = 2$ into the equation:

$y = 4(2) - 9$

This simplifies to:

$y = 8 - 9$

$y = -1$
Check the solution: To verify that our solution is correct, we can plug the values of $x$ and $y$ back into both original equations to see if they are true.

Conclusion

In this article, we solved a system of equations using the substitution method. We identified the equations, solved one equation for $y$ , substituted the expression for $y$ into the other equation, solved for $x$ , found the value of $y$ , and checked the solution. The final answer is $\boxed{(2, -1)}$ .

Discussion

The substitution method is a powerful tool for solving systems of equations. It is particularly useful when one of the equations is linear and the other is quadratic or has a more complex form. However, it requires careful substitution and simplification to avoid errors.

Example Problems

Here are a few example problems to practice solving systems of equations using the substitution method:

${ \begin{array}{l} y = 2x + 1 \\ y = x - 2 \end{array} \}$
A. (1, 3) B. (2, 1) C. (-1, 2) D. (3, 1)
${ \begin{array}{l} y = 3x - 2 \\ y = 2x + 1 \end{array} \}$
A. (1, 2) B. (2, 1) C. (-1, 3) D. (3, 2)
${ \begin{array}{l} y = x + 2 \\ y = 2x - 1 \end{array} \}$
A. (1, 3) B. (2, 1) C. (-1, 2) D. (3, 1)

Practice Problems

Try solving the following systems of equations using the substitution method:

${ \begin{array}{l} y = 2x - 3 \\ y = x + 1 \end{array} \}$
A. (1, 2) B. (2, 1) C. (-1, 3) D. (3, 2)
${ \begin{array}{l} y = 3x + 2 \\ y = 2x - 1 \end{array} \}$
A. (1, 3) B. (2, 1) C. (-1, 2) D. (3, 1)
${ \begin{array}{l} y = x - 1 \\ y = 2x + 2 \end{array} \}$
A. (1, 2) B. (2, 1) C. (-1, 3) D. (3, 2)

Solutions

Here are the solutions to the practice problems:

${ \begin{array}{l} y = 2x - 3 \\ y = x + 1 \end{array} \}$
The solution is $\boxed{(2, 1)}$ .
${ \begin{array}{l} y = 3x + 2 \\ y = 2x - 1 \end{array} \}$
The solution is $\boxed{(1, 3)}$ .
${ \begin{array}{l} y = x - 1 \\ y = 2x + 2 \end{array} \}$
The solution is $\boxed{(2, 1)}$ .
Solving the System of Equations by Substitution: Q&A =====================================================

Introduction

In our previous article, we discussed how to solve a system of equations using the substitution method. In this article, we will answer some frequently asked questions about solving systems of equations by substitution.

Q: What is the substitution method?

A: The substitution method is a technique used to solve a system of equations by substituting the expression of one variable from one equation into the other equation. This method is particularly useful when one of the equations is linear and the other is quadratic or has a more complex form.

Q: How do I choose which equation to solve for y?

A: You can choose either equation to solve for y. The key is to make sure that you substitute the correct expression for y into the other equation.

Q: What if I get stuck during the substitution process?

A: If you get stuck during the substitution process, try to simplify the equation by combining like terms or using algebraic properties. If you are still having trouble, try to visualize the problem by graphing the equations on a coordinate plane.

Q: Can I use the substitution method to solve a system of equations with more than two variables?

A: Yes, you can use the substitution method to solve a system of equations with more than two variables. However, it may be more challenging to solve the system, and you may need to use additional techniques such as elimination or graphing.

Q: What if I get a false solution?

A: If you get a false solution, it means that the solution does not satisfy both equations. In this case, you need to re-examine your work and try to find the correct solution.

Q: Can I use the substitution method to solve a system of equations with fractions or decimals?

A: Yes, you can use the substitution method to solve a system of equations with fractions or decimals. However, you may need to use additional techniques such as multiplying or dividing both sides of the equation by a common factor.

Q: How do I know if the substitution method is the best method to use?

A: The substitution method is a good choice when one of the equations is linear and the other is quadratic or has a more complex form. However, if the equations are both linear, the elimination method may be a better choice.

Q: Can I use the substitution method to solve a system of equations with absolute value or inequalities?

A: Yes, you can use the substitution method to solve a system of equations with absolute value or inequalities. However, you may need to use additional techniques such as solving for the absolute value or using a graphing calculator.

Conclusion

Solving systems of equations by substitution is a powerful technique that can be used to solve a wide range of problems. By following the steps outlined in this article and answering the frequently asked questions, you can become proficient in using the substitution method to solve systems of equations.

Practice Problems

Try solving the following systems of equations using the substitution method:

${ \begin{array}{l} y = 2x - 3 \\ y = x + 1 \end{array} \}$
A. (1, 2) B. (2, 1) C. (-1, 3) D. (3, 2)
${ \begin{array}{l} y = 3x + 2 \\ y = 2x - 1 \end{array} \}$
A. (1, 3) B. (2, 1) C. (-1, 2) D. (3, 1)
${ \begin{array}{l} y = x - 1 \\ y = 2x + 2 \end{array} \}$
A. (1, 2) B. (2, 1) C. (-1, 3) D. (3, 2)

Solutions

Here are the solutions to the practice problems:

${ \begin{array}{l} y = 2x - 3 \\ y = x + 1 \end{array} \}$
The solution is $\boxed{(2, 1)}$ .
${ \begin{array}{l} y = 3x + 2 \\ y = 2x - 1 \end{array} \}$
The solution is $\boxed{(1, 3)}$ .
${ \begin{array}{l} y = x - 1 \\ y = 2x + 2 \end{array} \}$
The solution is $\boxed{(2, 1)}$ .

Additional Resources

For more information on solving systems of equations by substitution, check out the following resources:

Khan Academy: Solving Systems of Equations by Substitution
Mathway: Solving Systems of Equations by Substitution
IXL: Solving Systems of Equations by Substitution