Veda Solves The Following System Of Linear Equations By Elimination. What Is The Value Of X X X In The Solution Of The Equations?${ \begin{aligned} 6 + 4x - 2y &= 0 \ -3 - 7y &= 10x \end{aligned} }$A. -20 B. -1 C. 1 D. 30

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Introduction

In mathematics, a system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables. One of the methods used to solve a system of linear equations is the elimination method. This method involves adding or subtracting the equations to eliminate one of the variables, making it easier to solve for the other variable. In this article, we will use the elimination method to solve a system of linear equations and find the value of xx.

The System of Linear Equations

The given system of linear equations is:

{ \begin{aligned} 6 + 4x - 2y &= 0 \\ -3 - 7y &= 10x \end{aligned} \}

Step 1: 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 the variable to be eliminated are the same. Let's multiply the first equation by 7 and the second equation by 2.

{ \begin{aligned} 42 + 28x - 14y &= 0 \\ -6 - 14y &= 20x \end{aligned} \}

Step 2: Add or Subtract the Equations

Now, we can add or subtract the equations to eliminate one of the variables. Let's add the two equations.

{ \begin{aligned} (42 + 28x - 14y) + (-6 - 14y) &= 0 + 20x \\ 36 + 28x - 28y &= 20x \end{aligned} \}

Step 3: Simplify the Equation

Now, let's simplify the equation by combining like terms.

{ \begin{aligned} 36 + 28x - 28y &= 20x \\ 36 + 8x - 28y &= 0 \end{aligned} \}

Step 4: Solve for xx

Now, we can solve for xx by isolating it on one side of the equation.

{ \begin{aligned} 36 + 8x - 28y &= 0 \\ 8x &= 28y - 36 \\ x &= \frac{28y - 36}{8} \\ x &= \frac{7y - 9}{2} \end{aligned} \}

Step 5: Substitute the Value of xx in One of the Original Equations

Now, we can substitute the value of xx in one of the original equations to find the value of yy.

Let's substitute the value of xx in the first original equation.

{ \begin{aligned} 6 + 4x - 2y &= 0 \\ 6 + 4\left(\frac{7y - 9}{2}\right) - 2y &= 0 \\ 6 + 2(7y - 9) - 2y &= 0 \\ 6 + 14y - 18 - 2y &= 0 \\ 12y - 12 &= 0 \\ 12y &= 12 \\ y &= 1 \end{aligned} \}

Step 6: Find the Value of xx

Now that we have the value of yy, we can find the value of xx by substituting the value of yy in the equation x=7y−92x = \frac{7y - 9}{2}.

{ \begin{aligned} x &= \frac{7y - 9}{2} \\ x &= \frac{7(1) - 9}{2} \\ x &= \frac{7 - 9}{2} \\ x &= \frac{-2}{2} \\ x &= -1 \end{aligned} \}

Conclusion

In this article, we used the elimination method to solve a system of linear equations and find the value of xx. We multiplied the equations by necessary multiples, added or subtracted the equations, simplified the equation, solved for xx, substituted the value of xx in one of the original equations, and found the value of xx. The value of xx is −1-1.

Answer

Q: What is the elimination method?

A: The elimination method is a technique used to solve a system of linear equations by adding or subtracting the equations to eliminate one of the variables, making it easier to solve for the other variable.

Q: How do I know which variable to eliminate?

A: To determine which variable to eliminate, you need to look at the coefficients of the variables in the equations. If the coefficients of one variable are the same, you can eliminate that variable by adding or subtracting the equations.

Q: What if the coefficients of the variables are not the same?

A: If the coefficients of the variables are not the same, you need to multiply the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same.

Q: How do I multiply the equations by necessary multiples?

A: To multiply the equations by necessary multiples, you need to multiply each term in the equation by the same number. For example, if you want to multiply the first equation by 2, you need to multiply each term in the equation by 2.

Q: What if I get a negative value for one of the variables?

A: If you get a negative value for one of the variables, it means that the variable is not a solution to the system of linear equations. In this case, you need to re-examine your work and try again.

Q: Can I use the elimination method to solve a system of linear equations with three or more variables?

A: Yes, you can use the elimination method to solve a system of linear equations with three or more variables. However, it may be more complicated and require more steps.

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 multiplying the equations by necessary multiples
  • Not adding or subtracting the equations correctly
  • Not simplifying the equation correctly
  • Not solving for the correct variable

Q: How do I know if I have solved the system of linear equations correctly?

A: To know if you have solved the system of linear equations correctly, you need to check your work by plugging the values of the variables back into the original equations. If the equations are true, then you have solved the system correctly.

Q: Can I use a calculator or computer to solve a system of linear equations?

A: Yes, you can use a calculator or computer to solve a system of linear equations. Many calculators and computers have built-in functions to solve systems of linear equations.

Conclusion

In this article, we answered some frequently asked questions about solving a system of linear equations by elimination. We covered topics such as the elimination method, multiplying the equations by necessary multiples, and common mistakes to avoid. We also discussed how to know if you have solved the system correctly and whether you can use a calculator or computer to solve a system of linear equations.

Additional Resources

If you want to learn more about solving systems of linear equations, here are some additional resources:

  • Khan Academy: Solving Systems of Linear Equations
  • Mathway: Solving Systems of Linear Equations
  • Wolfram Alpha: Solving Systems of Linear Equations

Practice Problems

Here are some practice problems to help you practice solving systems of linear equations by elimination:

  1. Solve the system of linear equations:

{ \begin{aligned} 2x + 3y &= 5 \\ x - 2y &= -3 \end{aligned} \}

  1. Solve the system of linear equations:

{ \begin{aligned} x + 2y &= 4 \\ 3x - 2y &= 5 \end{aligned} \}

  1. Solve the system of linear equations:

{ \begin{aligned} x - 3y &= 2 \\ 2x + 5y &= 11 \end{aligned} \}

Answer Key

  1. x = 2, y = 1
  2. x = 1, y = 1.5
  3. x = 5, y = 1