Solve for the remaining variables

Oct 13, 2025 by recoveryshake.com 69 views

Introduction

Simultaneous equations are a set of equations that involve multiple variables and are solved simultaneously to find the values of the variables. In this article, we will discuss how to solve simultaneous equations using the elimination method. The elimination method is a popular method for solving simultaneous equations, and it involves eliminating one or more variables from the equations to solve for the remaining variables.

What are Simultaneous Equations?

Simultaneous equations are a set of equations that involve multiple variables and are solved simultaneously to find the values of the variables. For example, consider the following set of equations:

{ \begin{align*} x + y + z &= 4 \\ 2x - 3y + 4z &= 33 \\ 3x - 2y - 2z &= 2 \end{align*} \}

In this example, we have three equations and three variables (x, y, and z). We need to find the values of x, y, and z that satisfy all three equations simultaneously.

The Elimination Method

The elimination method is a popular method for solving simultaneous equations. It involves eliminating one or more variables from the equations to solve for the remaining variables. The basic steps involved in the elimination method are:

Write down the equations: Write down the set of equations that need to be solved.
Identify the variables: Identify the variables that need to be solved.
Eliminate one variable: Eliminate one variable from two or more equations.
Solve for the remaining variables: Solve for the remaining variables using the eliminated equations.

Step 1: Write Down the Equations

The first step in the elimination method is to write down the set of equations that need to be solved. In this case, we have the following set of equations:

{ \begin{align*} x + y + z &= 4 \\ 2x - 3y + 4z &= 33 \\ 3x - 2y - 2z &= 2 \end{align*} \}

Step 2: Identify the Variables

The next step is to identify the variables that need to be solved. In this case, we have three variables (x, y, and z) that need to be solved.

Step 3: Eliminate One Variable

To eliminate one variable, we need to multiply the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same in both equations. Let's eliminate the variable x from the first and second equations.

Multiply the first equation by 2 and the second equation by -1:

{ \begin{align*} 2(x + y + z) &= 2(4) \\ -1(2x - 3y + 4z) &= -1(33) \end{align*} \}

Simplifying the equations, we get:

{ \begin{align*} 2x + 2y + 2z &= 8 \\ -2x + 3y - 4z &= -33 \end{align*} \}

Now, add both equations to eliminate the variable x:

{ \begin{align*} (2x - 2x) + (2y + 3y) + (2z - 4z) &= 8 - 33 \\ 5y - 2z &= -25 \end{align*} \}

Step 4: Solve for the Remaining Variables

Now that we have eliminated the variable x, we can solve for the remaining variables (y and z) using the eliminated equations. Let's solve for y and z using the eliminated equation and the third equation.

We have the following equations:

{ \begin{align*} 5y - 2z &= -25 \\ 3x - 2y - 2z &= 2 \end{align*} \}

We can solve for y and z using these equations. Let's solve for y and z using substitution or elimination method.

Solving for y and z

We can solve for y and z using the substitution method. Let's solve for y and z using substitution method.

Rearrange the first equation to isolate y:

{ \begin{align*} 5y &= -25 + 2z \\ y &= \frac{-25 + 2z}{5} \end{align*} \}

Substitute the value of y into the third equation:

{ \begin{align*} 3x - 2\left(\frac{-25 + 2z}{5}\right) - 2z &= 2 \end{align*} \}

Simplifying the equation, we get:

{ \begin{align*} 3x + \frac{50 - 4z}{5} - 2z &= 2 \end{align*} \}

Multiply both sides by 5 to eliminate the fraction:

{ \begin{align*} 15x + 50 - 4z - 10z &= 10 \end{align*} \}

Combine like terms:

{ \begin{align*} 15x - 14z &= -40 \end{align*} \}

Now, we have two equations with two variables (x and z). We can solve for x and z using these equations.

Solving for x and z

We can solve for x and z using the substitution method. Let's solve for x and z using substitution method.

Rearrange the first equation to isolate x:

{ \begin{align*} 15x &= -40 + 14z \\ x &= \frac{-40 + 14z}{15} \end{align*} \}

Substitute the value of x into the second equation:

{ \begin{align*} 2\left(\frac{-40 + 14z}{15}\right) - 3y + 4z &= 33 \end{align*} \}

Simplifying the equation, we get:

{ \begin{align*} \frac{-80 + 28z}{15} - 3y + 4z &= 33 \end{align*} \}

Multiply both sides by 15 to eliminate the fraction:

{ \begin{align*} -80 + 28z - 45y + 60z &= 495 \end{align*} \}

Combine like terms:

{ \begin{align*} 88z - 45y &= 575 \end{align*} \}

Now, we have two equations with two variables (y and z). We can solve for y and z using these equations.

Solving for y and z

We can solve for y and z using the substitution method. Let's solve for y and z using substitution method.

Rearrange the first equation to isolate y:

{ \begin{align*} 5y &= -25 + 2z \\ y &= \frac{-25 + 2z}{5} \end{align*} \}

Substitute the value of y into the second equation:

{ \begin{align*} 88z - 45\left(\frac{-25 + 2z}{5}\right) &= 575 \end{align*} \}

Simplifying the equation, we get:

{ \begin{align*} 88z + 225 - 18z &= 575 \end{align*} \}

Combine like terms:

{ \begin{align*} 70z &= 350 \end{align*} \}

Divide both sides by 70:

{ \begin{align*} z &= 5 \end{align*} \}

Now that we have found the value of z, we can substitute it into one of the equations to find the value of y.

Substitute the value of z into the first equation:

{ \begin{align*} 5y - 2(5) &= -25 \\ 5y - 10 &= -25 \end{align*} \}

Add 10 to both sides:

{ \begin{align*} 5y &= -15 \end{align*} \}

Divide both sides by 5:

{ \begin{align*} y &= -3 \end{align*} \}

Now that we have found the values of y and z, we can substitute them into one of the equations to find the value of x.

Substitute the values of y and z into the third equation:

{ \begin{align*} 3x - 2(-3) - 2(5) &= 2 \end{align*} \}

Simplifying the equation, we get:

{ \begin{align*} 3x + 6 - 10 &= 2 \end{align*} \}

Combine like terms:

{ \begin{align*} 3x - 4 &= 2 \end{align*} \}

Add 4 to both sides:

{ \begin{align*} 3x &= 6 \end{align*} \}

Divide both sides by 3:

{ \begin{align*} x &= 2 \end{align*} \}

Introduction

In our previous article, we discussed how to solve simultaneous equations using the elimination method. In this article, we will answer some frequently asked questions about solving simultaneous equations using the elimination method.

Q: What is the elimination method?

A: The elimination method is a popular method for solving simultaneous equations. It involves eliminating one or more variables from the equations to solve for the remaining variables.

Q: How do I choose which variable to eliminate first?

A: To choose which variable to eliminate first, look for the variables that have the same coefficient in two or more equations. Eliminate the variable with the same coefficient first.

Q: How do I eliminate a variable?

A: To eliminate a variable, multiply the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same in both equations. Then, add or subtract the equations to eliminate the variable.

Q: What if I have three or more variables?

A: If you have three or more variables, you can use the elimination method to eliminate one or more variables, and then use substitution or elimination method to solve for the remaining variables.

Q: Can I use the elimination method to solve non-linear equations?

A: No, the elimination method is used to solve linear equations. If you have non-linear equations, you may need to use other methods such as substitution, elimination, or graphing.

Q: What if I have a system of equations with no solution?

A: If you have a system of equations with no solution, it means that the equations are inconsistent. In this case, you will not be able to find a solution that satisfies all the equations.

Q: What if I have a system of equations with infinitely many solutions?

A: If you have a system of equations with infinitely many solutions, it means that the equations are dependent. In this case, you will be able to find a solution that satisfies all the equations, but there will be infinitely many solutions.

Q: Can I use the elimination method to solve systems of equations with fractions?

A: Yes, you can use the elimination method to solve systems of equations with fractions. To do this, multiply both sides of the equations by the least common multiple of the denominators to eliminate the fractions.

Q: What are some common mistakes to avoid when using the elimination method?

A: Some common mistakes to avoid when using the elimination method include:

Not checking if the equations are linear or non-linear before using the elimination method.
Not multiplying the equations by necessary multiples to eliminate the variable.
Not adding or subtracting the equations correctly to eliminate the variable.
Not checking if the equations are consistent or inconsistent before finding a solution.

Conclusion

Solving simultaneous equations using the elimination method can be a powerful tool for solving systems of equations. By following the steps outlined in this article and avoiding common mistakes, you can use the elimination method to solve systems of equations with confidence.

Additional Resources

If you are looking for additional resources to help you learn more about solving simultaneous equations using the elimination method, here are some suggestions:

Khan Academy: Solving Systems of Equations using the Elimination Method
Mathway: Solving Systems of Equations using the Elimination Method
Wolfram Alpha: Solving Systems of Equations using the Elimination Method

By following these resources and practicing the elimination method, you can become proficient in solving systems of equations and tackle more complex problems with confidence.