Multiplying the First Equation

Oct 13, 2025 by pinkandtrash.com 112 views

Introduction

In algebra, equations are often used to represent relationships between variables. When solving systems of linear equations, it is common to use techniques such as substitution or elimination to find the values of the variables. One method for eliminating variables is to multiply each equation by a number that produces opposite coefficients for $x$ or $y$ . This technique is useful when the coefficients of the variables are not already opposites, and it can help to simplify the equations and make them easier to solve.

The Problem

The following system of linear equations is given:

\begin{align*} 1x - 1y &= 100 \\ \frac{3}{8}x + \frac{7}{8}y &= 2000 \end{align*}

Our goal is to multiply each equation by a number that produces opposite coefficients for $x$ or $y$ , and then use the resulting equations to solve for the values of $x$ and $y$ .

Multiplying the First Equation

To eliminate the variable $y$ , we need to multiply the first equation by a number that will produce opposite coefficients for $y$ . Since the coefficient of $y$ in the first equation is $-1$ , we can multiply the entire equation by $-1$ to produce a coefficient of $1$ for $y$ .

\begin{align*} -1(1x - 1y) &= -1(100) \\ -1x + 1y &= -100 \end{align*}

Multiplying the Second Equation

To eliminate the variable $y$ , we also need to multiply the second equation by a number that will produce opposite coefficients for $y$ . Since the coefficient of $y$ in the second equation is $\frac{7}{8}$ , we can multiply the entire equation by $-8$ to produce a coefficient of $-7$ for $y$ .

\begin{align*} -8\left(\frac{3}{8}x + \frac{7}{8}y\right) &= -8(2000) \\ -3x - 7y &= -16000 \end{align*}

Adding the Equations

Now that we have multiplied both equations to produce opposite coefficients for $y$ , we can add the two equations together to eliminate the variable $y$ .

\begin{align*} (-3x - 7y) + (1x - 1y) &= -16000 - 100 \\ -2x - 8y &= -16100 \end{align*}

Solving for $x$

We now have a single equation with only one variable, $x$ . We can solve for $x$ by isolating it on one side of the equation.

\begin{align*} -2x - 8y &= -16100 \\ -2x &= -16100 + 8y \\ x &= \frac{-16100 + 8y}{-2} \\ x &= 8050 - 4y \end{align*}

Solving for $y$

We can substitute the expression for $x$ into one of the original equations to solve for $y$ . Let's use the first equation.

\begin{align*} 1x - 1y &= 100 \\ 1(8050 - 4y) - 1y &= 100 \\ 8050 - 4y - y &= 100 \\ 8050 - 5y &= 100 \\ -5y &= -7950 \\ y &= \frac{-7950}{-5} \\ y &= 1590 \end{align*}

Solving for $x$

Now that we have found the value of $y$ , we can substitute it into the expression for $x$ to find the value of $x$ .

\begin{align*} x &= 8050 - 4y \\ x &= 8050 - 4(1590) \\ x &= 8050 - 6360 \\ x &= 1690 \end{align*}

Conclusion

In this article, we used the technique of multiplying equations to eliminate variables to solve a system of linear equations. We multiplied each equation by a number that produced opposite coefficients for $x$ or $y$ , and then used the resulting equations to solve for the values of $x$ and $y$ . This technique is useful when the coefficients of the variables are not already opposites, and it can help to simplify the equations and make them easier to solve.

Example Use Cases

This technique can be used in a variety of situations, such as:

Solving systems of linear equations in physics and engineering problems
Finding the intersection points of two lines in geometry
Solving systems of linear equations in economics and finance problems

Tips and Tricks

When using this technique, it is essential to:

Multiply each equation by a number that produces opposite coefficients for $x$ or $y$
Use the resulting equations to solve for the values of $x$ and $y$
Check the solutions by substituting them back into the original equations

Introduction

In our previous article, we discussed the technique of multiplying equations to eliminate variables. This technique is useful when the coefficients of the variables are not already opposites, and it can help to simplify the equations and make them easier to solve. In this article, we will answer some frequently asked questions about this technique.

Q: What is the purpose of multiplying equations to eliminate variables?

A: The purpose of multiplying equations to eliminate variables is to simplify the equations and make them easier to solve. By multiplying each equation by a number that produces opposite coefficients for $x$ or $y$ , we can eliminate one of the variables and solve for the other variable.

Q: How do I know which equation to multiply by which number?

A: To determine which equation to multiply by which number, you need to look at the coefficients of the variables in each equation. You want to multiply each equation by a number that will produce opposite coefficients for $x$ or $y$ . For example, if the coefficient of $y$ in the first equation is $-1$ , you can multiply the entire equation by $-1$ to produce a coefficient of $1$ for $y$ .

Q: What if the coefficients of the variables are already opposites?

A: If the coefficients of the variables are already opposites, you do not need to multiply the equations. You can simply add or subtract the equations to eliminate one of the variables.

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

A: Yes, you can use this technique to solve systems of linear equations with more than two variables. However, you will need to multiply each equation by a number that produces opposite coefficients for each variable, and then use the resulting equations to solve for the values of the variables.

Q: What if I make a mistake when multiplying the equations?

A: If you make a mistake when multiplying the equations, you may end up with incorrect solutions. To avoid this, make sure to double-check your work and use a calculator or computer program to check your solutions.

Q: Can I use this technique to solve systems of linear equations with fractions?

A: Yes, you can use this technique to solve systems of linear equations with fractions. However, you will need to multiply each equation by a number that will eliminate the fractions. For example, if the coefficient of $y$ in the first equation is $\frac{1}{2}$ , you can multiply the entire equation by $2$ to produce a coefficient of $1$ for $y$ .

Q: What are some common mistakes to avoid when using this technique?

A: Some common mistakes to avoid when using this technique include:

Multiplying the equations by the wrong numbers
Not checking the solutions
Not using a calculator or computer program to check the solutions
Not following the steps carefully