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How can you algebraically write at least
and no more
?
Remember to instruct the calculator that you are dealing with an inequality.
Look for the overlapping areas.
Inequality (I): x+y≥ 10
Inequality (II): 16x+20y≤ 200
Graph:
Example Solutions: (0,10), (10,0), (11,11), and (12,0)
We know that x represents the number of student tickets, y represents the number of adult tickets, and we want to buy at least 10 tickets. The phrase at least
can algebraically be written as ≥. Therefore, the total number of bought tickets, x plus y, is greater than or equal to 10.
Before we graph the inequalities, we need to solve for y in terms of x.
x+y&≥ 10 &&⇔ y ≥ - x+10
16x+20y&≤ 200 &&⇔ y ≤ - 0.8x+10
To draw these inequalities, we press Y= on our calculator and write the right-hand side of the inequalities on the first two rows.
Next, to instruct the calculator that we are dealing with inequalities, we place the cursor before Y_1, press ENTER, and scroll to select the black triangle pointing up and to the right. This triangle shows we are dealing with a ≥ inequality. We do the same thing for the second inequality but choose the black triangle that points down and to the left.
Finally, we will re-size the standard viewing window. To do so, push the WINDOW button and change the settings to given size. Then push GRAPH.
The area that contains both patterns shows our solution set.
All the points in the overlapping patterns with whole number coordinates are part of the solution set. Both inequalities intercept the y-axis at (0,10) so that's one solution. Another solution is the x-intercept of the first inequality at (10,0). Two more solutions are (11,0) and (12,0).
| Number of Student Tickets, x | Number of Adult Tickets, y | Solution, (x,y) |
|---|---|---|
| 0 | 10 | (0,10) |
| 10 | 0 | (10,0) |
| 11 | 0 | (11,0) |
| 12 | 0 | (12,0) |
Please note that any whole-number coordinate pairs will represent possible combinations of student and adult tickets. These four are only examples.