Question

Quiz
Question 1 (1 point)

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Which of the following is a solution to the system of equations?
{2x+y=8
{x=5


Question 1 options:

Infinite Solutions


(5, -2)


No Solutions


(-2, 5)

Question 2 (1 point)

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Which of the following is a solution to the system of equations?
{y=-2x-1
{3x-4y=-40


Question 2 options:

(7, -4)


(-4, 7)


No Solutions


Infinite Solutions

Question 3 (1 point)

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Which of the following is a solution to the system of equations?
{5x+2y=4
{9x+2y=12


Question 3 options:

No Solutions


(2, -3)


(-3, 2)


Infinite Solutions

Question 4 (1 point)

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Which of the following is a solution to the system of equations?
y=2/3x -1
y=-x+4


Question 4 options:

(1, 3)


No Solutions


Infinite Solutions


(3, 1)

Question 5 (1 point)

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Which of the following is a solution to the system of equations?

3x-2y=-16
x+y=-7

Question 5 options:

(-6, -1)


Infinite Solutions


(-1, -6)


No Solutions


0 of 5 questions saved
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Answer 1

Let's solve each question one by one:

**Question 1:**
The system of equations is:
1. \(2x + y = 8\)
2. \(x = 5\)

Substituting \(x = 5\) into the first equation:
\[
2(5) + y = 8 \implies 10 + y = 8 \implies y = 8 - 10 = -2
\]
So the solution is \((5, -2)\).

**Answer:** (5, -2)

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**Question 2:**
The system of equations is:
1. \(y = -2x - 1\)
2. \(3x - 4y = -40\)

Substituting \(y = -2x - 1\) into the second equation:
\[
3x - 4(-2x - 1) = -40 \implies 3x + 8x + 4 = -40 \implies 11x + 4 = -40 \implies 11x = -44 \implies x = -4
\]
Now substituting \(x = -4\) back to find \(y\):
\[
y = -2(-4) - 1 = 8 - 1 = 7
\]
So the solution is \((-4, 7)\).

**Answer:** (-4, 7)

---

**Question 3:**
The system of equations is:
1. \(5x + 2y = 4\)
2. \(9x + 2y = 12\)

Subtract the first equation from the second:
\[
(9x + 2y) - (5x + 2y) = 12 - 4 \implies 4x = 8 \implies x = 2
\]
Now substituting \(x = 2\) back into the first equation to find \(y\):
\[
5(2) + 2y = 4 \implies 10 + 2y = 4 \implies 2y = 4 - 10 \implies 2y = -6 \implies y = -3
\]
So the solution is \((2, -3)\).

**Answer:** (2, -3)

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**Question 4:**
The system of equations is:
1. \(y = \frac{2}{3}x - 1\)
2. \(y = -x + 4\)

Setting the two equations equal to each other:
\[
\frac{2}{3}x - 1 = -x + 4
\]
Multiply by 3 to eliminate the fraction:
\[
2x - 3 = -3x + 12
\]
Combine like terms:
\[
2x + 3x = 12 + 3 \implies 5x = 15 \implies x = 3
\]
Substituting \(x = 3\) back to find \(y\):
\[
y = -3 + 4 = 1
\]
So the solution is \((3, 1)\).

**Answer:** (3, 1)

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**Question 5:**
The system of equations is:
1. \(3x - 2y = -16\)
2. \(x + y = -7\)

From the second equation, we can express \(y\) as:
\[
y = -7 - x
\]
Substituting in the first equation:
\[
3x - 2(-7 - x) = -16 \implies 3x + 14 + 2x = -16 \implies 5x + 14 = -16 \implies 5x = -30 \implies x = -6
\]
Now substituting \(x = -6\) back to find \(y\):
\[
y = -7 - (-6) = -7 + 6 = -1
\]
So the solution is \((-6, -1)\).

**Answer:** (-6, -1)

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**Final Answers:**
1. (5, -2)
2. (-4, 7)
3. (2, -3)
4. (3, 1)
5. (-6, -1)