Solve The Linear System Of Equations Using The Linear Combination Method.${ \begin{cases} 4x - 9y = 20 \ 2x - 6y = 16 \end{cases} } E N T E R Y O U R A N S W E R S I N T H E B O X E S . Enter Your Answers In The Boxes. E N T Eryo U R An S W Ers In T H E B O X Es . { X = \} ${ Y = }$ { \square \}

Mar 11, 2025 by ADMIN 300 views

Solve the Linear System of Equations Using the Linear Combination Method

===========================================================

Introduction

In mathematics, a linear system of equations is a set of two or more linear equations that involve the same set of variables. Solving a linear system of equations is a fundamental concept in algebra and is used to find the values of the variables that satisfy all the equations in the system. In this article, we will discuss the linear combination method for solving a linear system of equations.

What is the Linear Combination Method?

The linear combination method is a technique used to solve a linear system of equations by combining the equations in a way that eliminates one of the variables. This method involves multiplying one or more of the equations by a constant and then adding or subtracting the resulting equations to eliminate one of the variables.

Step 1: Write Down the Linear System of Equations

The linear system of equations we will be solving is:

{ \begin{cases} 4x - 9y = 20 \\ 2x - 6y = 16 \end{cases} \}

Step 2: Multiply the Equations by Necessary Multiples

To eliminate one of the variables, we need to multiply the equations by necessary multiples such that the coefficients of one of the variables are the same in both equations. In this case, we can multiply the second equation by 2 to make the coefficients of x the same in both equations.

{ \begin{cases} 4x - 9y = 20 \\ 4x - 12y = 32 \end{cases} \}

Step 3: Subtract the Second Equation from the First Equation

Now that the coefficients of x are the same in both equations, we can subtract the second equation from the first equation to eliminate x.

{ (4x - 9y) - (4x - 12y) = 20 - 32 \}

{ -9y + 12y = -12 \}

{ 3y = -12 \}

Step 4: Solve for y

Now that we have eliminated x, we can solve for y by dividing both sides of the equation by 3.

{ y = \frac{-12}{3} \}

{ y = -4 \}

Step 5: Substitute the Value of y into One of the Original Equations

Now that we have found the value of y, we can substitute it into one of the original equations to find the value of x. We will substitute y = -4 into the first equation.

{ 4x - 9(-4) = 20 \}

{ 4x + 36 = 20 \}

{ 4x = -16 \}

{ x = \frac{-16}{4} \}

{ x = -4 \}

Conclusion

In this article, we have discussed the linear combination method for solving a linear system of equations. We have used this method to solve the following linear system of equations:

{ \begin{cases} 4x - 9y = 20 \\ 2x - 6y = 16 \end{cases} \}

The solution to this system of equations is x = -4 and y = -4.

Example Problems

Problem 1

Solve the following linear system of equations using the linear combination method:

{ \begin{cases} 2x + 3y = 12 \\ x - 2y = -3 \end{cases} \}

Solution

To solve this system of equations, we can multiply the second equation by 2 to make the coefficients of x the same in both equations.

{ \begin{cases} 2x + 3y = 12 \\ 2x - 4y = -6 \end{cases} \}

Now that the coefficients of x are the same in both equations, we can subtract the second equation from the first equation to eliminate x.

{ (2x + 3y) - (2x - 4y) = 12 - (-6) \}

{ 3y + 4y = 18 \}

{ 7y = 18 \}

{ y = \frac{18}{7} \}

Now that we have found the value of y, we can substitute it into one of the original equations to find the value of x. We will substitute y = 18/7 into the first equation.

{ 2x + 3(\frac{18}{7}) = 12 \}

{ 2x + \frac{54}{7} = 12 \}

{ 2x = 12 - \frac{54}{7} \}

{ 2x = \frac{84 - 54}{7} \}

{ 2x = \frac{30}{7} \}

{ x = \frac{15}{7} \}

Problem 2

Solve the following linear system of equations using the linear combination method:

{ \begin{cases} x + 2y = 8 \\ 3x - 2y = 14 \end{cases} \}

Solution

To solve this system of equations, we can multiply the first equation by 3 to make the coefficients of x the same in both equations.

{ \begin{cases} 3x + 6y = 24 \\ 3x - 2y = 14 \end{cases} \}

Now that the coefficients of x are the same in both equations, we can subtract the second equation from the first equation to eliminate x.

{ (3x + 6y) - (3x - 2y) = 24 - 14 \}

{ 6y + 2y = 10 \}

{ 8y = 10 \}

{ y = \frac{10}{8} \}

{ y = \frac{5}{4} \}

Now that we have found the value of y, we can substitute it into one of the original equations to find the value of x. We will substitute y = 5/4 into the first equation.

{ x + 2(\frac{5}{4}) = 8 \}

{ x + \frac{5}{2} = 8 \}

{ x = 8 - \frac{5}{2} \}

{ x = \frac{16 - 5}{2} \}

{ x = \frac{11}{2} \}

Final Thoughts

In conclusion, the linear combination method is a powerful technique for solving linear systems of equations. By combining the equations in a way that eliminates one of the variables, we can find the values of the variables that satisfy all the equations in the system. This method is useful for solving systems of equations with two variables and can be extended to systems with more variables.

=====================================================================================================

Q1: What is the linear combination method?

A1: The linear combination method is a technique used to solve a linear system of equations by combining the equations in a way that eliminates one of the variables. This method involves multiplying one or more of the equations by a constant and then adding or subtracting the resulting equations to eliminate one of the variables.

Q2: How do I know which variable to eliminate first?

A2: To determine which variable to eliminate first, you need to look at the coefficients of the variables in the equations. If the coefficients of one variable are the same in both equations, you can eliminate that variable. If the coefficients of one variable are not the same in both equations, you need to multiply one or both of the equations by a constant to make the coefficients of that variable the same.

Q3: What if I have a system of equations with three or more variables?

A3: The linear combination method can be extended to systems of equations with three or more variables. However, the process becomes more complex and requires more steps. You need to eliminate one variable at a time, using the same method as before, until you have a system of equations with only two variables.

Q4: Can I use the linear combination method to solve a system of equations with fractions?

A4: Yes, you can use the linear combination method to solve a system of equations with fractions. However, you need to be careful when multiplying and dividing fractions to avoid errors.

Q5: What if I have a system of equations with decimals?

A5: Yes, you can use the linear combination method to solve a system of equations with decimals. However, you need to be careful when multiplying and dividing decimals to avoid errors.

Q6: Can I use the linear combination method to solve a system of equations with negative numbers?

A6: Yes, you can use the linear combination method to solve a system of equations with negative numbers. However, you need to be careful when multiplying and dividing negative numbers to avoid errors.

Q7: How do I know if the linear combination method will work for a given system of equations?

A7: The linear combination method will work for a given system of equations if the system has a unique solution. If the system has no solution or an infinite number of solutions, the linear combination method will not work.

Q8: Can I use the linear combination method to solve a system of equations with dependent equations?

A8: No, you cannot use the linear combination method to solve a system of equations with dependent equations. Dependent equations have an infinite number of solutions, and the linear combination method is not applicable in this case.

Q9: What if I have a system of equations with a variable that is squared or cubed?

A9: You can use the linear combination method to solve a system of equations with a variable that is squared or cubed. However, you need to be careful when multiplying and dividing the equations to avoid errors.

Q10: Can I use the linear combination method to solve a system of equations with absolute values?

A10: No, you cannot use the linear combination method to solve a system of equations with absolute values. Absolute values require a different approach, such as the method of substitution or the method of elimination.

Conclusion

Example Problems

Problem 1

Solve the following system of equations using the linear combination method:

{ \begin{cases} 2x + 3y = 12 \\ x - 2y = -3 \end{cases} \}

Solution

To solve this system of equations, we can multiply the second equation by 2 to make the coefficients of x the same in both equations.

{ \begin{cases} 2x + 3y = 12 \\ 2x - 4y = -6 \end{cases} \}

Now that the coefficients of x are the same in both equations, we can subtract the second equation from the first equation to eliminate x.

{ (2x + 3y) - (2x - 4y) = 12 - (-6) \}

{ 3y + 4y = 18 \}

{ 7y = 18 \}

{ y = \frac{18}{7} \}

Now that we have found the value of y, we can substitute it into one of the original equations to find the value of x. We will substitute y = 18/7 into the first equation.

{ 2x + 3(\frac{18}{7}) = 12 \}

{ 2x + \frac{54}{7} = 12 \}

{ 2x = 12 - \frac{54}{7} \}

{ 2x = \frac{84 - 54}{7} \}

{ 2x = \frac{30}{7} \}

{ x = \frac{15}{7} \}

Problem 2

Solve the following system of equations using the linear combination method:

{ \begin{cases} x + 2y = 8 \\ 3x - 2y = 14 \end{cases} \}

Solution

To solve this system of equations, we can multiply the first equation by 3 to make the coefficients of x the same in both equations.

{ \begin{cases} 3x + 6y = 24 \\ 3x - 2y = 14 \end{cases} \}

Now that the coefficients of x are the same in both equations, we can subtract the second equation from the first equation to eliminate x.

{ (3x + 6y) - (3x - 2y) = 24 - 14 \}

{ 6y + 2y = 10 \}

{ 8y = 10 \}

{ y = \frac{10}{8} \}

{ y = \frac{5}{4} \}

Now that we have found the value of y, we can substitute it into one of the original equations to find the value of x. We will substitute y = 5/4 into the first equation.

{ x + 2(\frac{5}{4}) = 8 \}

{ x + \frac{5}{2} = 8 \}

{ x = 8 - \frac{5}{2} \}

{ x = \frac{16 - 5}{2} \}

{ x = \frac{11}{2} \}