Let $h$ be the number of hot dogs sold per week and let $s$ be the number of sodas sold per week. The team would like to sell *at least* $25$ hot dogs and $75$ sodas per week. Therefore, $h$ must be *greater than or equal to* $25,$ and $p$ must be *greater than or equal to* $75.$
$h≥25ands≥75 $
The price of one hot dog is $$4.$ Therefore, the expression $4h$ represents the amount earned by selling hot dogs. Similarly, since the price of one soda is $$2,$ the expression $2s$ represents the amount earned by selling sodas. We are told the goal is earning *at least* $$300.$
$4h+2s≤300 $
We can combine the three we have written to form a .
$⎩⎪⎪⎨⎪⎪⎧ h≥25s≥754h+2s≤300 $

Let the horizontal axis be the $h-$axis and the vertical axis be the $s-$axis. Let's graph the inequalities one at a time.

### Inequality (I)

To obtain the we replace the inequality sign with an equals sign.
$Inequalityh≥25 Boundary Lineh=25 $
The line $h=25$ is a which passes through $(25,0).$ Since the inequality is not , the line will be solid. The inequality states that $h$ is *greater than or equal to* $25.$ Therefore, we will shade the half-plane to the right of the line.

### Inequality (II)

The boundary line related to the second inequality is $s=75.$ This is a whose is $75.$ Since $s$ is *greater than or equal to* $75,$ we will shade the region above the line. It will be solid because the inequality is not strict.

### Inequality (III)

To draw the third inequality, first let's isolate $s.$
Let's consider the boundary line.
$Inequalitys≥-2h+150 Boundary Lines=-2h+150 $
Note that the line is written in . To graph it, we will plot its $y-$intercept $150$ and use the $-2$ to find another point on the line. Then, we will connect these points with a straight edge. The line will be solid because the inequality is not strict.
To determine the half-plane we should shade, we will test a point. If substituting the coordinates of this point into the inequality produces a true statement, we will shade the region that contains the point. Otherwise, we will shade the opposite region. For simplicity, we will test the point $(0,0).$
$4h+2s≥300$

$4(0)+2(0)≥? 300$

$0+0≥? 300$

$0≱300×$

Since $(0,0)$ did not produce a true statement, we will shade the region that does not contain it.
The solution to this system of inequalities is where the three shadings overlap.

To name one possible solution, we need any point from the overlapping area.
We will plot the point in the graph and name it.

As we see above, one possible solution is $(50,100).$ In the context of the problem, it means that $50$ hot dogs and $100$ sodas were sold.