Solve The System Of Equations:$\[ \begin{array}{l} 3x + 8y = 15 \\ 2x - 8y = 10 \\ \end{array} \\]Find:$\[ X = \square \\]$\[ Y = \square \\]

by ADMIN 142 views

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. In this article, we will focus on solving a system of two linear equations with two variables. We will use the method of substitution and elimination to find the values of the variables.

The System of Equations

The system of equations we will be solving is:

3x+8y=152x−8y=10\begin{array}{l} 3x + 8y = 15 \\ 2x - 8y = 10 \\ \end{array}

Step 1: Write Down the System of Equations

The first step in solving a system of linear equations is to write down the system of equations. In this case, we have two linear equations with two variables, x and y.

Step 2: Multiply the Equations by Necessary Multiples

To eliminate one of the variables, we need to multiply the equations by necessary multiples. We can multiply the first equation by 2 and the second equation by 3 to make the coefficients of x in both equations equal.

6x+16y=306x−24y=30\begin{array}{l} 6x + 16y = 30 \\ 6x - 24y = 30 \\ \end{array}

Step 3: Subtract the Second Equation from the First Equation

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

16y+24y=30−3040y=0\begin{array}{l} 16y + 24y = 30 - 30 \\ 40y = 0 \\ \end{array}

Step 4: Solve for y

Now that we have eliminated the variable x, we can solve for y.

40y=0y=0\begin{array}{l} 40y = 0 \\ y = 0 \\ \end{array}

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

Now that we have the value of y, we can substitute it into one of the original equations to solve for x.

3x+8(0)=153x=15x=5\begin{array}{l} 3x + 8(0) = 15 \\ 3x = 15 \\ x = 5 \\ \end{array}

Conclusion

In this article, we solved a system of two linear equations with two variables using the method of substitution and elimination. We first multiplied the equations by necessary multiples to make the coefficients of x in both equations equal. Then, we subtracted the second equation from the first equation to eliminate the variable x. Finally, we solved for y and substituted the value of y into one of the original equations to solve for x.

Final Answer

The final answer is:

  • x=5x = 5
  • y=0y = 0

Discussion

This system of linear equations can be solved using other methods such as the method of substitution or the method of elimination. However, the method of substitution and elimination is a more efficient method for solving systems of linear equations.

Example Problems

Here are some example problems that can be solved using the method of substitution and elimination:

  • Solve the system of equations: 2x+3y=74x−2y=12\begin{array}{l} 2x + 3y = 7 \\ 4x - 2y = 12 \end{array}
  • Solve the system of equations: x+2y=53x−4y=11\begin{array}{l} x + 2y = 5 \\ 3x - 4y = 11 \end{array}

Tips and Tricks

Here are some tips and tricks for solving systems of linear equations:

  • Make sure to multiply the equations by necessary multiples to make the coefficients of x in both equations equal.
  • Subtract the second equation from the first equation to eliminate the variable x.
  • Solve for y and substitute the value of y into one of the original equations to solve for x.
  • Use the method of substitution and elimination to solve systems of linear equations.

Conclusion

Introduction

In our previous article, we solved a system of two linear equations with two variables using the method of substitution and elimination. In this article, we will answer some frequently asked questions about solving systems of linear equations.

Q: What is a system of linear equations?

A: 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.

Q: How do I know which method to use to solve a system of linear equations?

A: There are several methods to solve a system of linear equations, including the method of substitution, the method of elimination, and the method of graphing. The method of substitution and elimination is a more efficient method for solving systems of linear equations.

Q: What is the difference between the method of substitution and the method of elimination?

A: The method of substitution involves substituting the value of one variable into the other equation to solve for the other variable. The method of elimination involves adding or subtracting the equations to eliminate one of the variables.

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

A: To multiply the equations by necessary multiples, you need to multiply the coefficients of the variables in both equations by the same number. For example, if you have the equation 2x + 3y = 7, you can multiply it by 2 to get 4x + 6y = 14.

Q: How do I subtract the second equation from the first equation?

A: To subtract the second equation from the first equation, you need to subtract the coefficients of the variables in both equations. For example, if you have the equations 4x + 6y = 14 and 2x + 3y = 7, you can subtract the second equation from the first equation to get 2x + 3y = 7.

Q: How do I solve for y?

A: To solve for y, you need to isolate the variable y in one of the equations. For example, if you have the equation 2x + 3y = 7, you can subtract 2x from both sides to get 3y = -2x + 7.

Q: How do I substitute the value of y into one of the original equations?

A: To substitute the value of y into one of the original equations, you need to replace the variable y with its value in the equation. For example, if you have the equation 2x + 3y = 7 and y = 2, you can substitute y = 2 into the equation to get 2x + 3(2) = 7.

Q: What are some common mistakes to avoid when solving systems of linear equations?

A: Some common mistakes to avoid when solving systems of linear equations include:

  • Not multiplying the equations by necessary multiples
  • Not subtracting the second equation from the first equation
  • Not solving for y
  • Not substituting the value of y into one of the original equations

Q: How do I check my answer?

A: To check your answer, you need to plug the values of x and y back into the original equations to make sure they are true.

Conclusion

In conclusion, solving a system of linear equations is a fundamental concept in mathematics. By following the steps outlined in this article, you can solve systems of linear equations with ease. Remember to multiply the equations by necessary multiples, subtract the second equation from the first equation, solve for y, and substitute the value of y into one of the original equations.

Final Answer

The final answer is:

  • x=5x = 5
  • y=0y = 0

Discussion

This system of linear equations can be solved using other methods such as the method of substitution or the method of elimination. However, the method of substitution and elimination is a more efficient method for solving systems of linear equations.

Example Problems

Here are some example problems that can be solved using the method of substitution and elimination:

  • Solve the system of equations: 2x+3y=74x−2y=12\begin{array}{l} 2x + 3y = 7 \\ 4x - 2y = 12 \end{array}
  • Solve the system of equations: x+2y=53x−4y=11\begin{array}{l} x + 2y = 5 \\ 3x - 4y = 11 \end{array}

Tips and Tricks

Here are some tips and tricks for solving systems of linear equations:

  • Make sure to multiply the equations by necessary multiples to make the coefficients of x in both equations equal.
  • Subtract the second equation from the first equation to eliminate the variable x.
  • Solve for y and substitute the value of y into one of the original equations to solve for x.
  • Use the method of substitution and elimination to solve systems of linear equations.

Conclusion

In conclusion, solving a system of linear equations is a fundamental concept in mathematics. By following the steps outlined in this article, you can solve systems of linear equations with ease. Remember to multiply the equations by necessary multiples, subtract the second equation from the first equation, solve for y, and substitute the value of y into one of the original equations.