Thomas Edison was a newsboy during the Civil War. One day, he persuaded the editor to give him 300 copies of the paper instead of the usual 100. He went to the train station where people were eager for news of the war. He was able to sell the papers for 10¢ and 25¢ instead of the usual 5¢. If he wanted to earn at least $52, how many papers could he have sold for 10¢ and 25¢?
x = the number of 10¢ papers y = the number of 25¢ papers
The following system of inequalities can be used to represent the conditions
of this problem.
x + y ≤
300 He
can sell as many as 300 papers.
0.10x + 0.25y ≥ 52 He
wants to earn at least $52.
Since both x and y represent the number of
papers, neither can be a negative number. Thus, x ≤ 0 and y ≤ 0.
The solution of the system is the set of all ordered pairs that satisfy both
inequalities and lie in the first quadrant. The solution can be determined by
graphing each inequality in the same coordinate plane as shown below.
Recall that the graph of each inequality is called a half-plane.
· Any
point in the yellow region satisfies x + y
≤ 300.
· Any
point in the blue region satisfies 0.10x
+ 0.25y ≥ 52.
· Any
point in the green region satisfies both inequalities. The intersection of the
two half-planes represents the solution to the system of inequalities.
· The
graphs of x + y = 300 and 0.10x
+ 0.25y = 52 are
the boundaries of the region and are included in the graph of the system.
This solution is a region that contains the graphs of an
infinite number of ordered pairs. An example is (50, 225). This means that
Edison could have sold 50 papers at 10¢ and 225 papers at 25¢ to earn at least
$52.
Check:
Remember that the boundary lines of inequalities are only
included in the solution if the inequality symbol is greater than or equal to, ≥,
or less than or equal to, ≤.
Solve each system of inequalities by graphing. If the system
does not have a solution, write no solution.
1. y ≤ – 1
y > x – 3
Alternative
Solutions:
The solution is the
ordered pairs in the intersection
of the graphs of y ≤
– 1 and y > x – 3. The region is
shaded in green at the right. The
graphs of y
=
–1
and y = x – 3 are the boundaries
of this region. The graph of y
= x – 3
is a dashed line and is not
included in the solution of the system. Choose
a point and check the solution.
2.
2y < x – 2
–3x
+ 6y ≥ 12
Alternative Solutions:
The graphs of 2y = x – 2 and –3x + 6y = 12 are parallel lines. Check this by graphing or
by comparing the slopes.
Because the regions in the solution of 2y < x – 2 and –3x + 6y ≥ 12 have no points in common, the system of inequalities has
no solution.
Spending
Link
3. Luisa has $96 to spend on gifts for the holidays. She must buy at least 9
gifts. She plans to buy puzzles that cost $8 or $12. How many of each puzzle
can she buy?
Alternative
Solutions:
x = the number of $8 puzzles y = the number of $12
puzzles
The
following system of inequalities can be used to represent the conditions of
this problem.
x + y ≥ 9 She wants to buy at least 9 puzzles.
8x 12y ≤ 96 The
total cost must be no more than $96.
Because
the number of puzzles she can buy cannot be negative, both x ≥ 0
and y ≥ 0.
The
solutions are all of the ordered pairs in the intersection of the graphs of
these inequalities. Only the first quadrant is used because x ≥ 0
and y ≥ 0.
Any point
with whole-number coordinates in the region is a possible solution. For
example, since the point at (9, 1) is in the region, Luisa could buy 9 puzzles
for $8 and 1 for $12. The greatest number of gifts that she could buy would be
12 puzzles for $8 and no puzzles for $12.
A graphing calculator can be helpful in solving systems of
inequalities or in checking solutions.
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