# Understanding the Elimination Method (Grade 10)

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## Understanding the Elimination Method

1.

You have a system of two linear equations in two variables with unique solution [math](a,b)[/math]. Are you able to change one equation to a new equation while keeping the same unique solution? If so, how many of these new equations exist?

- No.
- Yes, one.
- Yes, four.
- Yes, an infinite number.

2.

Which of the following is an equivalent system of equations to the one given?

[math]3x+5y=5[/math]

[math]2x+y=8[/math]

[math]3x+5y=5[/math]

[math]2x+y=8[/math]

- [math]-7x = -35, \ \ 2x+y=8[/math]
- [math]3x+5y=5, \ \ -13x = -17 [/math]
- [math]3x+5y=5, \ \ 2x=8[/math]
- [math]y=11, \ \ 2x+y=8[/math]

3.

Which of the following systems of equations is equivalent to the one given? There may be more than one correct answer.

[math]4x+3y=-2[/math]

[math]2x-2y=6[/math]

[math]4x+3y=-2[/math]

[math]2x-2y=6[/math]

- [math]4x+3y=-2, \ \ \ -7/2y = 7 [/math]
- [math]4x+3y=-2, \ \ \ 2x = -8[/math]
- [math]7y = -14, \ \ \ 2x-2y=6[/math]
- [math]x = -11, \ \ \ 2x- 2 = 6[/math]

4.

Which of the following systems of equations is equivalent to the one given? There may be more than one correct answer.

[math]2x+3y=6[/math]

[math]4x+2y=-4[/math]

[math]2x+3y=6[/math]

[math]4x+2y=-4[/math]

- [math]3y=8, \ \ \ \ 4x+2y=-4[/math]
- [math]-2x=10, \ \ \ \ 4x+2y=-4[/math]
- [math]6x+5y=2, \ \ \ \ 4x+2y=-4[/math]
- [math]2x+3y=6, \ \ \ \ 4y=16[/math]

5.

Which of the following are correct steps you could apply to the given system of equations (where [math]e_1, e_2[/math] stand for equations 1 and 2), such that the resulting system of equations has the same answer and has eliminated one variable in one of the equations? There may be more than one correct answer.

[math]e_1: \ \ 2x+4y=-6[/math]

[math]e_2: \ \ 4x-2y=18[/math]

[math]e_1: \ \ 2x+4y=-6[/math]

[math]e_2: \ \ 4x-2y=18[/math]

- Replace equation 2 with [math]e_2 - 2e_1[/math].
- Replace equation 1 with [math]e_1 - 2e_2[/math].
- Replace equation 1 with [math]e_1 - 1/2e_2[/math]
- Replace equation 2 with [math]e_2 -1/2e_1[/math]

6.

Which of the following are correct steps you could apply to the given system of equations (where [math]e_1, e_2[/math] stand for equations 1 and 2), such that the resulting system of equations has the same answer and has eliminated one variable in one of the equations? There may be more than one correct answer.

[math]e_1: \ \ 2x+2y=6[/math]

[math]e_2: \ \ 5x+3y=5[/math]

[math]e_1: \ \ 2x+2y=6[/math]

[math]e_2: \ \ 5x+3y=5[/math]

- Replace equation 2 with [math]e_2 - 6/5 e_1[/math].
- Replace equation 1 with [math]e_1 - 2/3e_2[/math].
- Replace equation 2 with [math]e_2 - 2/5e_1[/math].
- Replace equation 1 with [math]e_1 - 5/2e_2[/math].

7.

For the given system of equations, the following work was done to create another system of equations with the same solution. Is this work correct? If not, choose the correct reason why.

Our given system is as follows.

[math]3x-3y=-18[/math]

[math]-5x+2y=9[/math]

We will multiply the first equation by [math]2/3[/math]. Now we can add this with the second equation.

[math] ul{:(,2x-2y=,-12),(+,-5x+2y=,9):}[/math]

[math]{:( \ \ \ ,-3x \ \ \ \ \ \ \ \ \ =, -3):}[/math]

This results in the following new system of equations.

[math]3x-3y=-18[/math]

[math]-3x=-3[/math]

Our given system is as follows.

[math]3x-3y=-18[/math]

[math]-5x+2y=9[/math]

We will multiply the first equation by [math]2/3[/math]. Now we can add this with the second equation.

[math] ul{:(,2x-2y=,-12),(+,-5x+2y=,9):}[/math]

[math]{:( \ \ \ ,-3x \ \ \ \ \ \ \ \ \ =, -3):}[/math]

This results in the following new system of equations.

[math]3x-3y=-18[/math]

[math]-3x=-3[/math]

- This is correct.
- You cannot add equations.
- [math]2/3[/math] is not the correct scalar to multiply by.
- The equations were added in the wrong order.

8.

For the given system of equations, the following work was done to create another system of equations with the same solution. Is this work correct? If not, choose the correct reason why.

We start with our given system of equations.

[math]4x-3y=8[/math]

[math]2x+6y=-11[/math]

We will multiply the first equation by 2. Now we can add this with the second equation.

[math] ul{:(,4x-6y=,8),(+,2x+6y=,-11):}[/math]

[math]{:( \ \ \ ,6x \ \ \ \ \ \ \ \ \ =, -3):}[/math]

This results in the following new system of equations.

[math]4x-3y=8[/math]

[math]6x=-3[/math]

We start with our given system of equations.

[math]4x-3y=8[/math]

[math]2x+6y=-11[/math]

We will multiply the first equation by 2. Now we can add this with the second equation.

[math] ul{:(,4x-6y=,8),(+,2x+6y=,-11):}[/math]

[math]{:( \ \ \ ,6x \ \ \ \ \ \ \ \ \ =, -3):}[/math]

This results in the following new system of equations.

[math]4x-3y=8[/math]

[math]6x=-3[/math]

- This is correct.
- The first equation has to be multiplied by [math]1/2[/math].
- The equations should be subtracted.
- Only the second term of the first equation was multiplied by 2.

9.

For the given system of equations, the following work was done to create another system of equations with the same solution. Is this work correct? If not, choose the correct reason why.

Our given system is as follows.

[math]5x+4y=1[/math]

[math]-3x-2y=1[/math]

We will multiply the second equation by [math]-1[/math]. Now we can subtract this from the first equation.

[math] ul{:(,5x+4y=,1),(-,3x+2y=,-1):}[/math]

[math]{:( \ \ \ ,2x+2y =, 2):}[/math]

This results in the following new system of equations.

[math]2x+2y=2[/math]

[math]-3x-2y=1[/math]

Our given system is as follows.

[math]5x+4y=1[/math]

[math]-3x-2y=1[/math]

We will multiply the second equation by [math]-1[/math]. Now we can subtract this from the first equation.

[math] ul{:(,5x+4y=,1),(-,3x+2y=,-1):}[/math]

[math]{:( \ \ \ ,2x+2y =, 2):}[/math]

This results in the following new system of equations.

[math]2x+2y=2[/math]

[math]-3x-2y=1[/math]

- This is correct.
- You must add the equations, not subtract.
- No variable has been eliminated, therefore the new system of equations is invalid.
- You have to multiply by a scalar other than [math]pm1.[/math]

10.

Prove that, given a system of linear equations with a unique solution [math](x_0,y_0)[/math], replacing one equation with the sum of that equation and a multiple of the other equation will result in a system of equations with the same solution [math](x_0,y_0)[/math].

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