a Let x be the number of girls and y be the number of boys. Write a system of inequalities by using the given information. To graph the system, begin by determining the boundary lines of the inequalities.
B
b The possible combinations are the points with whole number coordinates that are in the shaded region.
C
c The possible combinations are the points with whole number coordinates that are in the shaded region.
A
a
B
b See solution.
C
c See solution.
Practice makes perfect
a Let x be the number of girls and y be the number of boys. Using the given information, we can write three inequalities that represent the situation. To write them, we will make an organized table.
Verbal Expression
Algebraic Expression
The number of players is less than or equal to 15.
x+ y≤ 15
The number of players is greater than or equal to 10.
x+ y≥ 10
Number of girls is greater than number of boys.
x > y
Now we have three inequalities to write a system.
x+y≤ 15 & (I) x+y≥ 10 & (II) x > y & (III)
We will graph the system starting from Inequality I. We will first determine the boundary line of it.
&Inequality I && Boundary Line I
&x+y ≤ 15 &&x+y = 15
The boundary line is in standard form. Therefore, it would be a better choice to find its intercepts to graph it. We will substitute y= 0 for the x-intercept and x= 0 for the y-intercept.
x+y=15
Operation
x-intercept
y-intercept
Substitution
x+ 0=15
0+y=15
Calculation
x=15
y=15
Point
(15,0)
(0,15)
Now we can plot the intercepts and connect them with a line segment. Notice that the number of boys and girls cannot be negative, so the line will be bound by the axes. The boundary line will also be solid because of the non-strict inequality.
Next, we will decide which region we should shade. Therefore, we will test the point (0,0).
Since the point satisfies the inequality, region that contains the point will be shaded. Let's do it!
We will graph Inequality II proceeding in the same way. First, we will determine its boundary line.
&Inequality II && Boundary Line II
&x+y ≥ 10 &&x+y = 10
Next, we will find its intercepts.
x+y=10
Operation
x-intercept
y-intercept
Substitution
x+ 0=10
0+y=10
Calculation
x=10
y=10
Point
(10,0)
(0,10)
Now we will decide the region that will be shaded.
The point did not satisfy the inequality, so we will shade the region that does not contain the point.
Finally, we will graph Inequality III.
&Inequality III && Boundary Line III
&x ≥ y &&x = y
The boundary line of Inequality III is the identity line. The line will be dashed because of the strict inequality. The inequality also states that the points with x-coordinates greater than their y-coordinates are included in the solution set. Thus, below the boundary line will be shaded.
The overlapping section of the graph above is the solution set of the system.
b The possible combinations are the points with whole number coordinates that are in the shaded region. Let's show them!
Now we can list the combinations.
(6,4), (6,5), (7,3), (7,4), (7,5), (7,6),
(8,2), (8,3), (8,4), (8,5), (8,6), (8,7),
(9,1), (9,2), (9,3), (9,4), (9,5), (9,6),
(10,0), (10,1), (10,2), (10,3), (10,4), (10,5),
(11,0), (11,1), (11,2), (11,3), (11,4), (12,0),
(12,1), (12,2), (12,3), (13,0), (13,1), (13,2),
(14,0), (14,1), (15,0)
c The possible combinations are the points with whole number coordinates that are in the shaded region. Therefore, there is finite number of possibilities.
