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velikii [3]
3 years ago
7

Mike and his friends bought cheese wafers for $2 per packet and chocolate wafers for $1 per packet at a carnival. They spent a t

otal of $25 to buy a total of 20 packets of wafers of the two varieties.
Part A: Write a system of equations that can be solved to find the number of packets of cheese wafers and the number of packets of chocolate wafers that Mike and his friends bought at the carnival. Define the variables used in the equations. (5 points)

Part B: How many packets of chocolate wafers and cheese wafers did they buy? Explain how you got the answer and why you selected a particular method to get the answer. (5 points)
Mathematics
1 answer:
trapecia [35]3 years ago
4 0
A. 2x+y=25; x+y=20
x represents cheese wafers and y represents chocolate wafers.

B. I will choose elimination since there is y and y. I can multiply one equation by -1 to get y and -y, which cancels out.
2x+y=25
-x-y=-20

Add equations
x=5

Plug x in
5+y=20
y=15

Final answer: 5 cheese wafers, 15 chocolate wafers
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Step-by-step explanation:

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The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter
Natasha2012 [34]

Answer:

(1\frac{1}{2},1) - False

(4,6) - True

(18,12) -- False

(0,0) -- True

(2\frac{1}{2},3\frac{3}{4}) -- True

Step-by-step explanation:

The points are

(1\frac{1}{2},1) , (4,6), (18,12), (0,0) and (2\frac{1}{2},3\frac{3}{4}) ---- missing from the question

Given

y = 1\frac{1}{2}x

Required

Determine if each of the points would be on y = 1\frac{1}{2}x

To do this, we simply substitute the value of x and of each point in y = 1\frac{1}{2}x.

(a) (1\frac{1}{2},1)

In this case;

x = 1\frac{1}{2} and y = 1

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 1\frac{1}{2}

y = \frac{3}{2} * \frac{3}{2}

y = \frac{9}{4}

y = 2\frac{1}{4}

<em>The point </em>(1\frac{1}{2},1)<em>  won't be on the graph because the corresponding value of y for </em>x = 1\frac{1}{2}<em> is </em>y = 2\frac{1}{4}<em></em>

(b) (4,6)

In this case;

x = 4

y = 6

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 4

y = \frac{3}{2} * 4

y = \frac{3* 4}{2}

y = \frac{12}{2}

y = 6

<em>The point </em>(4,6)<em>  would be on the graph because the corresponding value of y for </em>x = 4 is y = 6

(c) (18,12)

In this case:

x = 18;y = 12

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 18

y = \frac{3}{2} * 18

y = \frac{3* 18}{2}

y = \frac{54}{2}

y = 27

<em>The point </em>(18,12)<em>  wouldn't be on the graph because the corresponding value of y for </em>x = 18<em> is </em>y = 12<em></em>

(d) (0,0)

In this case;

x =0; y = 0

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 0

y = 0

<em>The point </em>(0,0)<em>  would be on the graph because the corresponding value of y for </em>x = 0 is y = 0

(e) (2\frac{1}{2},3\frac{3}{4})

In this case:

x = 2\frac{1}{2}; y = 3\frac{3}{4}

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 2\frac{1}{2}

y = \frac{3}{2} * \frac{5}{2}

y = \frac{15}{4}

y = 3\frac{3}{4}

<em>The point </em>(2\frac{1}{2},3\frac{3}{4}) <em>  would be on the graph because the corresponding value of y for </em>x = 2\frac{1}{2} is y = 3\frac{3}{4}

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