Solve The System Of Equations Below Using Elimination. You Must Show All Work To Earn Full Credit. 2 X + 3 Y = 12 4 X − 3 Y = 6 \begin{array}{l} 2x + 3y = 12 \\ 4x - 3y = 6 \end{array} 2 X + 3 Y = 12 4 X − 3 Y = 6 ​ You May Type Your Work In The Box Below And Use The Math Icon For Math Functions, OR

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Introduction

In mathematics, a system of equations is a set of two or more equations that have the same variables. Solving a system of equations involves finding the values of the variables that satisfy all the equations in the system. There are several methods to solve systems of equations, including substitution, elimination, and graphing. In this article, we will focus on the elimination method, which involves adding or subtracting equations to eliminate one of the variables.

The Elimination Method

The elimination method involves adding or subtracting equations to eliminate one of the variables. To do this, we need to have the same coefficient for one of the variables in both equations. We can then add or subtract the equations to eliminate that variable.

Step 1: Multiply the Equations by Necessary Multiples

To eliminate one of the variables, we need to have the same coefficient for that variable in both equations. We can do this by multiplying one or both of the equations by necessary multiples.

Let's consider the system of equations:

2x+3y=124x−3y=6\begin{array}{l} 2x + 3y = 12 \\ 4x - 3y = 6 \end{array}

We can multiply the first equation by 2 and the second equation by 1 to get:

4x+6y=244x−3y=6\begin{array}{l} 4x + 6y = 24 \\ 4x - 3y = 6 \end{array}

Step 2: Add or Subtract the Equations

Now that we have the same coefficient for x in both equations, we can add or subtract the equations to eliminate x.

Let's add the two equations:

(4x+6y)+(4x−3y)=24+6(4x + 6y) + (4x - 3y) = 24 + 6

This simplifies to:

8x+3y=308x + 3y = 30

Step 3: Solve for One Variable

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

Let's solve for y in the equation:

8x+3y=308x + 3y = 30

We can subtract 8x from both sides to get:

3y=30−8x3y = 30 - 8x

Now, we can divide both sides by 3 to get:

y=30−8x3y = \frac{30 - 8x}{3}

Step 4: 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.

Let's substitute y into the first original equation:

2x+3y=122x + 3y = 12

Substituting y = (30 - 8x)/3, we get:

2x+3(30−8x3)=122x + 3(\frac{30 - 8x}{3}) = 12

Simplifying, we get:

2x+30−8x=122x + 30 - 8x = 12

Combine like terms:

−6x+30=12-6x + 30 = 12

Subtract 30 from both sides:

−6x=−18-6x = -18

Divide both sides by -6:

x=3x = 3

Step 5: Find the Value of y

Now that we have the value of x, we can find the value of y.

Substituting x = 3 into the equation y = (30 - 8x)/3, we get:

y=30−8(3)3y = \frac{30 - 8(3)}{3}

Simplifying, we get:

y=30−243y = \frac{30 - 24}{3}

y=63y = \frac{6}{3}

y=2y = 2

Conclusion

In this article, we used the elimination method to solve a system of equations. We multiplied the equations by necessary multiples, added or subtracted the equations to eliminate one of the variables, solved for one variable, substituted the value of the variable into one of the original equations, and found the value of the other variable. The final solution is x = 3 and y = 2.

Example Problems

  1. Solve the system of equations using the elimination method:

x+2y=63x−2y=2\begin{array}{l} x + 2y = 6 \\ 3x - 2y = 2 \end{array}

  1. Solve the system of equations using the elimination method:

2x+y=5x−2y=−3\begin{array}{l} 2x + y = 5 \\ x - 2y = -3 \end{array}

Tips and Tricks

  1. Make sure to multiply the equations by necessary multiples to have the same coefficient for one of the variables.
  2. Add or subtract the equations to eliminate one of the variables.
  3. Solve for one variable and substitute the value into one of the original equations.
  4. Make sure to check your solution by plugging the values back into the original equations.

Practice Problems

  1. Solve the system of equations using the elimination method:

x+3y=92x−3y=−3\begin{array}{l} x + 3y = 9 \\ 2x - 3y = -3 \end{array}

  1. Solve the system of equations using the elimination method:

2x+4y=12x−2y=−2\begin{array}{l} 2x + 4y = 12 \\ x - 2y = -2 \end{array}

Conclusion

Q: What is the elimination method?

A: The elimination method is a technique used to solve systems of equations by adding or subtracting equations to eliminate one of the variables.

Q: How do I know which variable to eliminate?

A: To determine which variable to eliminate, look for the coefficients of the variables in the equations. If the coefficients are the same, you can eliminate the variable by adding or subtracting the equations. If the coefficients are different, you can multiply one or both of the equations by necessary multiples to make the coefficients the same.

Q: What if I have two equations with the same variable but different coefficients?

A: If you have two equations with the same variable but different coefficients, you can multiply one or both of the equations by necessary multiples to make the coefficients the same. Then, you can add or subtract the equations to eliminate the variable.

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

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

Q: What if I get a contradiction when using the elimination method?

A: If you get a contradiction when using the elimination method, it means that the system of equations has no solution. This can happen if the equations are inconsistent or if there is no value that satisfies both equations.

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

A: Yes, you can use the elimination method to solve systems of equations with fractions or decimals. However, you may need to multiply the equations by necessary multiples to eliminate the fractions or decimals.

Q: What if I get a solution that is not an integer?

A: If you get a solution that is not an integer, it means that the solution is a decimal or a fraction. This is okay, as long as the solution satisfies both equations.

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

A: Yes, you can use the elimination method to solve systems of equations with absolute values. However, you may need to consider different cases and possibilities.

Q: What if I get a solution that is not unique?

A: If you get a solution that is not unique, it means that there are multiple solutions that satisfy both equations. This can happen if the equations are dependent or if there are multiple values that satisfy both equations.

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

A: Yes, you can use the elimination method to solve systems of equations with complex numbers. However, you may need to use complex arithmetic and consider different cases and possibilities.

Q: What if I get a solution that is not a real number?

A: If you get a solution that is not a real number, it means that the solution is a complex number. This is okay, as long as the solution satisfies both equations.

Conclusion

In conclusion, the elimination method is a powerful tool for solving systems of equations. By understanding the basics of the elimination method and practicing with different types of equations, you can become proficient in using the elimination method to solve systems of equations. Remember to always check your solution by plugging the values back into the original equations.

Practice Problems

  1. Solve the system of equations using the elimination method:

x+2y=63x−2y=2\begin{array}{l} x + 2y = 6 \\ 3x - 2y = 2 \end{array}

  1. Solve the system of equations using the elimination method:

2x+y=5x−2y=−3\begin{array}{l} 2x + y = 5 \\ x - 2y = -3 \end{array}

Tips and Tricks

  1. Make sure to multiply the equations by necessary multiples to have the same coefficient for one of the variables.
  2. Add or subtract the equations to eliminate one of the variables.
  3. Solve for one variable and substitute the value into one of the original equations.
  4. Make sure to check your solution by plugging the values back into the original equations.

Real-World Applications

  1. Solving systems of equations is used in many real-world applications, such as:
  • Physics: to solve problems involving motion and forces
  • Engineering: to design and optimize systems
  • Economics: to model and analyze economic systems
  • Computer Science: to solve problems involving algorithms and data structures

Conclusion

In conclusion, the elimination method is a powerful tool for solving systems of equations. By understanding the basics of the elimination method and practicing with different types of equations, you can become proficient in using the elimination method to solve systems of equations. Remember to always check your solution by plugging the values back into the original equations.