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The sum of two numbers is 22. Three times one number increased by five is 35.(4 points)

a) What are the equations you would use? (Response area) and (Response area)

b) What are the two numbers? (Response area) and (response area)

responses
x + y = 22
3x +5 = 35
x + 8 = 35
3x = 35
12
8
x - y = 22
14
10

11 answers

a) The equations you would use are:

  1. \( x + y = 22 \) (This equation represents the sum of the two numbers.)
  2. \( 3x + 5 = 35 \) (This equation represents three times one number increased by five.)

b) To find the two numbers, we can solve the equations.

  1. From the second equation, we can solve for \( x \): \[ 3x + 5 = 35 \] Subtract 5 from both sides: \[ 3x = 30 \] Now divide by 3: \[ x = 10 \]

  2. Substituting \( x = 10 \) back into the first equation to find \( y \): \[ x + y = 22 \ 10 + y = 22 \] Subtract 10 from both sides: \[ y = 12 \]

Thus the two numbers are:

  1. \( x = 10 \)
  2. \( y = 12 \)

Final responses:

  • \( x = 10 \)
  • \( y = 12 \)
use the responses provided and answer all questions

Sure! I'll use the responses you provided.

a) The equations you would use are:

  1. \( x + y = 22 \)
  2. \( 3x + 5 = 35 \)

b) To find the two numbers using the provided responses:

  1. From \( 3x + 5 = 35 \):

    • Subtract 5 from both sides: \[ 3x = 30 \]
    • Divide by 3: \[ x = 10 \]

  2. Substitute \( x = 10 \) into \( x + y = 22 \): \[ 10 + y = 22 \]

    • Subtract 10 from both sides: \[ y = 12 \]

So, using the provided responses, the two numbers are:

  • First number: 10
  • Second number: 12

Final answers:

  • a) \( x + y = 22 \) and \( 3x + 5 = 35 \)
  • b) 10 and 12
oelle currently has 18 rocks in her collection and gains 4 each week. Lewis currently has 30 rocks in his collection and gains 3 each week. Set up a system of equations to show how many rocks each has in their collection.(6 points)
a) Joelle has y = (Response area)x + (Response area)

b) Lewis has y = (Response area)x + (Response area)

c) After how many weeks will Joelle and Lewis have the same number of rocks in their collections? (Response area) weeks

d) How many rocks will Joelle and Lewis have when the amount of rocks in their collection is equal? They will each have (Response area) rocks in their collection.

response
4
65
12
70
3
2
18
30
66

To set up the equations for Joelle and Lewis, we can use the following format:

Let \( x \) represent the number of weeks.

a) Joelle has:

  • Initial amount: 18 rocks
  • Gains: 4 rocks per week

The equation for Joelle's rocks can be expressed as: \[ y = 4x + 18 \]

b) Lewis has:

  • Initial amount: 30 rocks
  • Gains: 3 rocks per week

The equation for Lewis's rocks can be expressed as: \[ y = 3x + 30 \]

c) To find how many weeks it will take for Joelle and Lewis to have the same number of rocks, we set the two equations equal to each other: \[ 4x + 18 = 3x + 30 \]

Now, solve for \( x \):

  1. Subtract \( 3x \) from both sides: \[ 4x - 3x + 18 = 30 \] \[ x + 18 = 30 \]

  2. Subtract 18 from both sides: \[ x = 12 \]

So, Joelle and Lewis will have the same number of rocks after 12 weeks.

d) To determine how many rocks they will each have at that time, substitute \( x = 12 \) back into either equation. We'll use Joelle's equation: \[ y = 4(12) + 18 \] \[ y = 48 + 18 \] \[ y = 66 \]

So, when Joelle and Lewis have the same number of rocks, they will each have 66 rocks in their collection.

Final responses:

  • a) Joelle has \( y = 4x + 18 \)
  • b) Lewis has \( y = 3x + 30 \)
  • c) \( 12 \) weeks
  • d) They will each have \( 66 \) rocks in their collection.
Not including tax, a total of 19 pens and markers cost $11.50. The pens cost $0.25 each, and the markers cost $0.75 each. Write the system of equations that could be used to solve for the number of pens, p, and the number of markers, m, bought.(4 points)
p + m = (Response area)

(Response area) p + (Response area) m = (response area)

responses
1
0.25
19
30.5
0.75
1.50
11.50

To set up the system of equations for the problem, we will define \( p \) as the number of pens and \( m \) as the number of markers.

  1. The first equation represents the total number of items bought (pens and markers): \[ p + m = 19 \]

  2. The second equation represents the total cost of the pens and markers:

    • The cost of pens is \( 0.25p \) (each pen costs $0.25).
    • The cost of markers is \( 0.75m \) (each marker costs $0.75).
    • The total cost is $11.50.

    So, the second equation would be: \[ 0.25p + 0.75m = 11.50 \]

Thus, the system of equations is:

  1. \( p + m = 19 \)
  2. \( 0.25p + 0.75m = 11.50 \)

Final responses:

  • First equation: \( p + m = 19 \)
  • Second equation: \( 0.25p + 0.75m = 11.50 \)
oyce wants to go to the zoo with her friend over spring break. You can pay $45 to get into the zoo each time, or join a membership for $50 and then only pay $20 each visit to the zoo. How many times will Joyce need to visit the zoo to justify the expense of the membership fee?(1 point)
Responses

3
3

5
5

10
10

2

To determine how many times Joyce needs to visit the zoo to justify the expense of the membership, let's set up the cost for both options:

  1. Cost without membership: Each visit costs $45.

    • If Joyce visits the zoo \( x \) times, the total cost without membership is: \[ \text{Cost without membership} = 45x \]

  2. Cost with membership: The membership costs $50, and each visit costs $20 after that.

    • If Joyce visits the zoo \( x \) times, the total cost with membership is: \[ \text{Cost with membership} = 50 + 20x \]

Now we need to find out when the cost with membership becomes less than or equal to the cost without membership: \[ 50 + 20x \leq 45x \]

Rearranging this inequality:

  1. First, subtract \( 20x \) from both sides: \[ 50 \leq 45x - 20x \] \[ 50 \leq 25x \]

  2. Now, divide by 25: \[ 2 \leq x \] or: \[ x \geq 2 \]

This means Joyce needs to visit the zoo at least 2 times to justify the expense of the membership.

Since the responses include options of \( 2, 3, 5, \) and \( 10\), the correct response would be:

  • 2
Two systems of equations are shown:

4x+y=−1

x+y=2

and

8x+2y=−2

−4x+cy=−8

What would the value of c need to be to get the same solution for both systems?

(1 point)
Responses

2
2

4
4

1
1

-4
-4

-2
-2
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To determine what value of \( c \) would make the second system of equations have the same solution as the first, we first need to analyze the first system of equations.

First System of Equations

  1. \( 4x + y = -1 \)
  2. \( x + y = 2 \)

We can solve for the variables \( x \) and \( y \).

From the second equation, we can express \( y \): \[ y = 2 - x \]

Now we can substitute \( y \) into the first equation: \[ 4x + (2 - x) = -1 \ 4x + 2 - x = -1 \ 3x + 2 = -1 \ 3x = -1 - 2 \ 3x = -3 \ x = -1 \]

Now substitute \( x = -1 \) back into \( y = 2 - x \): \[ y = 2 - (-1) = 2 + 1 = 3 \]

So the solution to the first system is \( (x, y) = (-1, 3) \).

Second System of Equations

  1. \( 8x + 2y = -2 \)
  2. \( -4x + cy = -8 \)

Now, we check if the first equation of the second system is consistent with the solution obtained from the first system.

If we multiply the first equation of the first system by 2, we get: \[ 2(4x + y) = 2(-1) \ 8x + 2y = -2 \]

So, the first equation of both systems is indeed the same.

Now, let's analyze the second equation of the second system: To ensure the same solution happens for both systems, we can substitute \( x = -1 \) and \( y = 3 \) into the second equation and solve for \( c \): \[ -4(-1) + c(3) = -8 \ 4 + 3c = -8 \]

Now, subtract 4 from both sides: \[ 3c = -8 - 4 \ 3c = -12 \]

Now, divide by 3: \[ c = -4 \]

Thus, the value of \( c \) that will ensure both systems have the same solution is:

  • -4