SOLUTION: For every calendar that is sold at a fundraising banquet, $8 goes to charity. For every ticket that is sold, $25 goes to charity. The organizers’ goal is to raise at least $6000.

Question 1170937: For every calendar that is sold at a fundraising banquet, $8 goes to charity. For every ticket that is sold, $25 goes to charity. The organizers’ goal is to raise at least $6000. The organizers need to know how many calendars and tickets must be sold to meet their goal.
a) define the variables and write a linear inequality to represent the situation.
b) graph the linear inequality to help determine whether each of the following points is in the solution set. The first coordinate is the number of calendars and the second is the number of tickets.
i)(400,100) ii) (500,100) iii) (100,200)

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your inequality is:

8x + 25y >= 6000

at the point (400,100), x = 400 and y = 100.
8x + 25y >= 6000 becomes:
8*400 + 25*100 >= 6000 which becomes:
3200 + 2500 >= 6000 which becomes:
5700 >= 6000 which is false.

at the point (500,100), x = 500 and y = 100.
8x + 25y >= 6000 becomes:
8*500 + 25*100 >= 6000 which becomes:
4000 + 2500 >= 6000 which becomes:
6500 >= 6000 which is true.

at the point (100,200), x = 100 and y = 200.
8x + 25y >= 6000 becomes:
8*100 + 25*200 >= 6000 which becomes:
800 + 5000 >= 6000 which becomes:
5800 >= 6000 which is false.

the only point that satisfies the inequality is the point (500,100).
the other two points don't satisfy the inequality.

graph the equation of 8x + 25y = 6000
any point below that line does not satisfy the inequality.
any point on or above that line satisfies the inequality.

i graphed the line and plotted the points (400,100), (500,100), (100,200).
only the point (500,100) satisfied the inequality.
the other two points didn't.

this was confirmed by the algebraic solutions that were determined above.

here's the graph.

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