Exercise 29 Page 261

• 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 $\le .$ Let's write this inequality.

• Inequality II: It is given that we need at least $3$ pounds of fruit. The phrase at least is represented by $\ge .$ Now, let's write the second inequality.

With these two inequalities, we can represent this situation as a system.

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. For Inequality (I) there are more steps. We can now write our system as follows.

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. In this form $\text{-}\frac{4}{3}$ is the slope and $7$ is the $y\text{-}$intercept. Since the symbol is $\le ,$ 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. We see it has a slope of $\text{-}1$ and a $y\text{-}$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.