Question

Quiz
Question 1 (1 point)

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Solve the following by elimination:
x-2y=12
5x+3y=-44


Question 1 options:

(-8, -4)


No Solution


Infinite Solutions


(-4, -8)

Question 2 (1 point)

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Solve the following by elimination:
4x+3y=-1
5x+4y=1

Question 2 options:

(-7, 9)


(9, -7)


Infinite Solutions


No Solution

Question 3 (1 point)

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Solve the following by elimination:
2x+5y=14
4x+2y=-4

Question 3 options:

(4, -3)


(-3, 4)


Infinite Solutions


No Solution

Question 4 (1 point)

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Solve the following by elimination:


3x+y=2
6x+2y=4
Question 4 options:

(6, 8)


Infinite Solutions


(0, 0)


No Solution

Question 5 (1 point)

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Solve the following by elimination:
2x+y=-2
5x+3y=-8

Question 5 options:

(-6, 2)


No Solution


(2, -6)


Infinite Solutions


0 of 5 questions saved
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Answer 1

Let's solve each of the questions one by one using the elimination method.

### Question 1
Given:
1. \( x - 2y = 12 \) (Equation 1)
2. \( 5x + 3y = -44 \) (Equation 2)

To eliminate \( x \), multiply Equation 1 by 5:
- \( 5x - 10y = 60 \)

Now, subtract Equation 2 from this:
- \( (5x - 10y) - (5x + 3y) = 60 - (-44) \)
- \( -10y - 3y = 60 + 44 \)
- \( -13y = 104 \)
- \( y = -8 \)

Substituting \( y = -8 \) back into Equation 1 to find \( x \):
- \( x - 2(-8) = 12 \)
- \( x + 16 = 12 \)
- \( x = 12 - 16 \)
- \( x = -4 \)

So the solution is \( (-4, -8) \).

### Question 2
Given:
1. \( 4x + 3y = -1 \) (Equation 1)
2. \( 5x + 4y = 1 \) (Equation 2)

To eliminate \( y \), multiply Equation 1 by 4 and Equation 2 by 3:
- \( 16x + 12y = -4 \)
- \( 15x + 12y = 3 \)

Now, subtract Equation 2 from this:
- \( (16x + 12y) - (15x + 12y) = -4 - 3 \)
- \( x = -7 \)

Substituting \( x = -7 \) back into Equation 1 to find \( y \):
- \( 4(-7) + 3y = -1 \)
- \( -28 + 3y = -1 \)
- \( 3y = 27 \)
- \( y = 9 \)

So the solution is \( (-7, 9) \).

### Question 3
Given:
1. \( 2x + 5y = 14 \) (Equation 1)
2. \( 4x + 2y = -4 \) (Equation 2)

To eliminate \( x \), multiply Equation 1 by 2:
- \( 4x + 10y = 28 \)

Now, subtract Equation 2 from this:
- \( (4x + 10y) - (4x + 2y) = 28 - (-4) \)
- \( 10y - 2y = 28 + 4 \)
- \( 8y = 32 \)
- \( y = 4 \)

Substituting \( y = 4 \) back into Equation 1 to find \( x \):
- \( 2x + 5(4) = 14 \)
- \( 2x + 20 = 14 \)
- \( 2x = 14 - 20 \)
- \( 2x = -6 \)
- \( x = -3 \)

So the solution is \( (-3, 4) \).

### Question 4
Given:
1. \( 3x + y = 2 \) (Equation 1)
2. \( 6x + 2y = 4 \) (Equation 2)

Notice that Equation 2 is just double Equation 1. This means both equations represent the same line.

Thus, there are infinite solutions.

### Question 5
Given:
1. \( 2x + y = -2 \) (Equation 1)
2. \( 5x + 3y = -8 \) (Equation 2)

To eliminate \( y \), multiply Equation 1 by 3:
- \( 6x + 3y = -6 \)

Now, subtract Equation 2 from this:
- \( (6x + 3y) - (5x + 3y) = -6 - (-8) \)
- \( x = 2 \)

Substituting \( x = 2 \) back into Equation 1 to find \( y \):
- \( 2(2) + y = -2 \)
- \( 4 + y = -2 \)
- \( y = -6 \)

So the solution is \( (2, -6) \).

### Final Answers:
1. (-4, -8)
2. (-7, 9)
3. (-3, 4)
4. Infinite Solutions
5. (2, -6)

Feel free to ask if you have any additional questions!