Multiplying the Equations by Suitable Constants

Oct 13, 2025 by wbrum.org 129 views

Introduction

When dealing with a system of linear equations, it's often necessary to eliminate one of the variables to solve for the other. In this case, we're tasked with eliminating the $$y$$ terms and solving for $$x$$ in the fewest steps possible. To achieve this, we need to multiply the equations by suitable constants before adding them together. In this article, we'll explore the process of eliminating $$y$$ terms and solving for $$x$$ using the given equations.

The Given Equations

We have two linear equations:

$4x - 3y = 34$
$3x + 2y = 7$

Our goal is to eliminate the $$y$$ terms and solve for $$x$$ in the fewest steps possible.

Multiplying the Equations by Suitable Constants

To eliminate the $$y$$ terms, we need to multiply the equations by constants that will make the coefficients of $$y$$ in both equations opposite in sign but equal in magnitude. This will allow us to add the equations together and eliminate the $$y$$ terms.

Let's analyze the coefficients of $$y$$ in both equations:

In the first equation, the coefficient of $$y$$ is $$-3$$.
In the second equation, the coefficient of $$y$$ is $$2$$.

To make the coefficients of $$y$$ opposite in sign but equal in magnitude, we need to multiply the first equation by $$\frac{2}{3}$$ and the second equation by $$\frac{3}{2}$$.

Multiplying the First Equation by $$\frac{2}{3}$$

Multiplying the first equation by $$\frac{2}{3}$$ gives us:

\frac{2}{3}(4x - 3y) = \frac{2}{3}(34)

Simplifying the equation, we get:

\frac{8}{3}x - 2y = \frac{68}{3}

Multiplying the Second Equation by $$\frac{3}{2}$$

Multiplying the second equation by $$\frac{3}{2}$$ gives us:

\frac{3}{2}(3x + 2y) = \frac{3}{2}(7)

Simplifying the equation, we get:

\frac{9}{2}x + 3y = \frac{21}{2}

Adding the Equations Together

Now that we have the equations multiplied by suitable constants, we can add them together to eliminate the $$y$$ terms.

Adding the two equations, we get:

\frac{8}{3}x - 2y + \frac{9}{2}x + 3y = \frac{68}{3} + \frac{21}{2}

Simplifying the equation, we get:

\frac{16}{6}x + \frac{9}{2}x - 2y + 3y = \frac{136}{6} + \frac{63}{6}

Combining like terms, we get:

\frac{16}{6}x + \frac{27}{6}x = \frac{199}{6}

Simplifying further, we get:

\frac{43}{6}x = \frac{199}{6}

Solving for $$x$$

Now that we have the equation $$\frac{43}{6}x = \frac{199}{6}$$, we can solve for $$x$$.

Dividing both sides of the equation by $$\frac{43}{6}$$, we get:

x = \frac{\frac{199}{6}}{\frac{43}{6}}

Simplifying the equation, we get:

x = \frac{199}{43}

Conclusion

In this article, we explored the process of eliminating $$y$$ terms and solving for $$x$$ using the given equations. By multiplying the equations by suitable constants, we were able to add the equations together and eliminate the $$y$$ terms. We then solved for $$x$$ by dividing both sides of the equation by the coefficient of $$x$$. The final answer is $$x = \frac{199}{43}$$.

Final Answer

The final answer is $$\boxed{\frac{199}{43}}$$.

Discussion

The process of eliminating $$y$$ terms and solving for $$x$$ is a common technique used in linear algebra. By multiplying the equations by suitable constants, we can add the equations together and eliminate the $$y$$ terms. This technique is useful when dealing with systems of linear equations with multiple variables.

Example Problems

Here are a few example problems that demonstrate the process of eliminating $$y$$ terms and solving for $$x$$:

Solve the system of linear equations:

2x + 3y = 7

x - 2y = -3

Solve the system of linear equations:

4x - 2y = 10

2x + 3y = 8

Solve the system of linear equations:

3x + 2y = 12

x - 4y = -5

Tips and Tricks

Here are a few tips and tricks that can help you eliminate $$y$$ terms and solve for $$x$$:

Make sure to multiply the equations by suitable constants that will make the coefficients of $$y$$ opposite in sign but equal in magnitude.
Simplify the equations by combining like terms.
Use the distributive property to expand the equations.
Check your work by plugging the solution back into the original equations.

By following these tips and tricks, you can eliminate $$y$$ terms and solve for $$x$$ with ease.

Q: What is the purpose of eliminating $$y$$ terms in a system of linear equations?

A: The purpose of eliminating $$y$$ terms is to solve for one of the variables, in this case, $$x$$. By eliminating the $$y$$ terms, we can isolate $$x$$ and find its value.

Q: How do I determine which constants to multiply the equations by?

A: To determine which constants to multiply the equations by, you need to make the coefficients of $$y$$ in both equations opposite in sign but equal in magnitude. This will allow you to add the equations together and eliminate the $$y$$ terms.

Q: What if the coefficients of $$y$$ in both equations are not opposite in sign but equal in magnitude?

A: If the coefficients of $$y$$ in both equations are not opposite in sign but equal in magnitude, you need to multiply one or both of the equations by a constant that will make the coefficients of $$y$$ opposite in sign but equal in magnitude.

Q: How do I add the equations together to eliminate the $$y$$ terms?

A: To add the equations together, you need to multiply the equations by the suitable constants and then add them together. Make sure to combine like terms and simplify the equation.

Q: What if I get a fraction as the solution for $$x$$?

A: If you get a fraction as the solution for $$x$$, you can simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor.

Q: Can I use this method to solve systems of linear equations with more than two variables?

A: Yes, you can use this method to solve systems of linear equations with more than two variables. However, you need to eliminate one variable at a time, and then use the resulting equation to eliminate another variable.

Q: What are some common mistakes to avoid when eliminating $$y$$ terms and solving for $$x$$?

A: Some common mistakes to avoid when eliminating $$y$$ terms and solving for $$x$$ include:

Not multiplying the equations by suitable constants
Not combining like terms
Not simplifying the equation
Not checking the solution by plugging it back into the original equations

Q: Can I use a calculator to solve systems of linear equations?

A: Yes, you can use a calculator to solve systems of linear equations. However, it's always a good idea to check the solution by plugging it back into the original equations to ensure that it's correct.

Q: What are some real-world applications of solving systems of linear equations?

A: Some real-world applications of solving systems of linear equations include:

Finding the intersection point of two lines
Determining the cost of producing a product
Calculating the interest rate on a loan
Solving problems in physics and engineering

Q: Can I use this method to solve systems of linear equations with complex numbers?

A: Yes, you can use this method to solve systems of linear equations with complex numbers. However, you need to be careful when multiplying and dividing complex numbers.

Q: What are some tips for solving systems of linear equations with multiple variables?

A: Some tips for solving systems of linear equations with multiple variables include:

Use the substitution method to eliminate one variable at a time
Use the elimination method to eliminate one variable at a time
Use a calculator to check the solution
Check the solution by plugging it back into the original equations

By following these tips and avoiding common mistakes, you can successfully eliminate $$y$$ terms and solve for $$x$$ in systems of linear equations.