Question

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Suppose the goal is to raise $600. One equation relating the calendar and poster sales to the $600 goal is 3c + 2p= 600. Suppose the company donating the calendars and posters said they would provide a total of $250 items. ​

Answer 1

Given equation:

`3c+2p=600`

where:

• c = number of calendars
• p = number of posters

Part (a)

If the company donating the calendars and posters said they would provide a total of 250 items then:

`c+p=250`

Part (b)

To graph:

• Rewrite both equations to make p the subject
• Find the c-intercept
• Find the p-intercept
• Connect the two points with a straight line

`3c+2p=600 \implies p=300-(3)/(2)c`

`\textsf{x-intercept}: \quad 300-(3)/(2)c=0 \implies c=200 \implies (200,0)`

`\textsf{y-intercept}: \quad 300-(3)/(2)(0)=300 \implies (0,300)`

`c+p=250 \implies p=250-c`

`\textsf{x-intercept}: \quad 250-c=0 \implies c=250 \implies (250,0)`

`\textsf{y-intercept}: \quad 250-0=250 \implies (0,250)`

Point of intersection (equate equations):

`\begin{aligned}300-(3)/(2)c &=250-c\\50 & =(1)/(2)c\\ c & =100\end{aligned}`

`p=250-100=150`

`\implies \left(100,150}\right)`

The point of intersection tells us that they need to sell 100 calendars and 150 posters to reach their goal of raising $600. Therefore, the company needs to provide 100 calendars and 150 posters (or 250 items in the ratio of calendars to posters of 2 : 3).

Answer 2

• 100 calendars and 150 posters

Explanation:

• Note it seem the "$" shouldn't be in front of 250 as it is the numbe of items

a) The second equation is

• c + p = 250

b) The graph of both equations is attached

The intersection point of the lines is (100, 150)

It indicates 100 calendars and 150 posters to be donated to meet both of targets $600 and 250 items.