a Write two inequalities: one that represents cost and one that represents the number of pounds. Then graph each inequality using the boundary line and a test point.
B
b Choose any point from the solution to the system.
C
c Is (4,1) a solution to the system?
A
aSystem of Inequalities: 4x+3y ≤ 21 x+y ≥ 3
Graph:
B
bExample Point: (3,2) Interpretation: The point (3,2) means that if we buy 3 pounds of blueberries and 2 pounds of strawberries we will have at least 3 pounds of fruit and spend no more than 21 dollars.
C
c Yes, see solution.
Practice makes perfect
a The system of inequalities for this exercise will have two inequalities. One will represent cost and the other will represent the number of pounds. For both we will let x represent the number of pounds of blueberries purchased and y represent the number of pounds of strawberries purchased.
Inequality I: It is given that one pound of blueberries costs 4 dollars, one pound of strawberries costs 3 dollars, and we can spend at most 21 dollars. The phrase at most is represented by the symbol ≤. Let's write this inequality.
4x+3y ≤ 21
Inequality II: It is given that we need at least 3 pounds of fruit. The phrase at least is represented by ≥. Now, let's write the second inequality.
x+y ≥ 3
With these two inequalities, we can represent this situation as a system.
4x+3y ≤ 21 & (I) x+y ≥ 3 & (II)
Writing in Slope-Intercept Form
To graph the system, it is helpful if we rewrite the inequalities in slope-intercept form. For Inequality (II) this only requires subtracting x from both sides.
x+y ≥ 3 ⇒ y ≥ - x+3
For Inequality (I) there are more steps.
We can now write our system as follows.
y ≤ - 43x+7 & (I) y ≥ - x+3 & (II)
Graphing the System
To graph a system of inequalities graph each inequality separately. The solution to the system is the intersection of the individual solution sets.
Graphing Inequality (I)
To graph the inequality we will first draw the boundary line, which is represented as the equation as follows.
y= - 43x+7
In this form - 43 is the slope and 7 is the y-intercept. Since the symbol is ≤, the line should be solid.
To determine if we should shade the region to the left or the right of the boundary line we can use a test point. For simplicity we will use (0,0). If we substitute (0,0) into the inequality and it makes a true statement, that means (0,0) is a solution of the inequality. Thus, we shade the region containing (0,0). If not, we shade the other region.
Since 0 is less than 7, (0,0) is a solution to the inequality. Thus, we shade the region to the left of the boundary line.
Graphing Inequality (II)
To graph Inequality (II) we will add to the graph from above. Let's consider the equation of the boundary line.
y = - x+3.
We see it has a slope of - 1 and a y-intercept of 3. We will determine which region to shade (either above or below the line) by using (0,0) as a test point.
As 0 is not greater than 3, we will shade the region that does not contain the point (0,0). In other words, we will shade above the boundary line. Adding this to the graph from above, we have the following.
Solution to the System
The solution to the system is the intersection of both shaded regions. The graph below shows the solution to the system.
b The shaded region from the last graph in Part A shows the solution to the system. Any point within that region solves the system because it satisfies both inequalities simultaneously.
We can arbitrarily choose any point for this exercise. We will choose (3,2).
Recall that x represents the number of pounds of blueberries and y represents the number of pounds of strawberries. The point (3,2) means that if we buy 3 pounds of blueberries and 2 pounds of strawberries we will have at least 3 pounds of fruit and spend no more than 21 dollars.
c If it is possible to buy 4 pounds of blueberries and 1 pound of strawberries and satisfy the restraints of both inequalities, the point (4,1) will be a solution to the system. One way we can determine if this is true is by seeing if (4,1) lies in the shaded region of the graph.
From the graph we can see that (4,1) is a solution. Therefore, it is possible to buy 4 pounds of blueberries and 1 pound of strawberries.
