Question

directions: graph each linear inequality. 1. $yleq-\frac{5}{4}x + 2$ 2. $4x - y < - 1$ 3. in football, a kicker can score one point for the extra - point on a touchdown or three points for a field goal. logan is a kicker for his high school football team. the record number of points scored by a kicker in a single game is 24. logan is hoping to beat this record in his next game. a) write a linear inequality to represent this situation, then graph. b) using your graph, give two possible combinations of extra - point kicks and field goals logan could score to beat the current record. directions: graph each system of linear inequalities. clearly indicate the solution region. 4. $x + 2ygeq - 12$ $y < 6x+7$ 5. $2x - 5y < - 20$ $x > - 3$

Explanation:

Step1: Define variables for problem 3

Let $x$ be the number of extra - point kicks and $y$ be the number of field goals. Since an extra - point kick is 1 point and a field goal is 3 points, and Logan wants to beat the record of 24 points, the linear inequality is $x + 3y>24$.

Step2: Graph the boundary line for $x + 3y>24$

First, graph the boundary line $x + 3y=24$. Rewrite it in slope - intercept form $y=-\frac{1}{3}x + 8$. The $y$ - intercept is 8 and the slope is $-\frac{1}{3}$. Since the inequality is $x + 3y>24$, the boundary line is dashed (because the points on the line are not part of the solution) and we shade the region above the line.

Step3: Find two possible combinations for problem 3b

We can choose points in the shaded region. For example, when $x = 0$, from the inequality $x+3y>24$, we have $3y>24$ or $y > 8$. Let $y = 9$, so one combination is $(0,9)$. When $y = 0$, we have $x>24$, let $x = 25$, so another combination is $(25,0)$.

Step4: Graph $y\leq-\frac{5}{4}x + 2$ for problem 1

The $y$ - intercept is 2 and the slope is $-\frac{5}{4}$. Since the inequality is $\leq$, the boundary line is solid and we shade the region below the line.

Step5: Graph $4x-y<-1$ for problem 2

Rewrite it as $y>4x + 1$. The $y$ - intercept is 1 and the slope is 4. Since the inequality is $>$, the boundary line is dashed and we shade the region above the line.

Step6: Graph $x + 2y\geq-12$ and $y<6x + 7$ for problem 4

For $x + 2y\geq-12$, rewrite as $y\geq-\frac{1}{2}x-6$. The boundary line is solid and we shade above it. For $y<6x + 7$, the boundary line is dashed and we shade below it. The solution region is the intersection of the two shaded regions.

Step7: Graph $2x-5y<-20$ and $x>-3$ for problem 5

Rewrite $2x-5y<-20$ as $y>\frac{2}{5}x + 4$. The boundary line is dashed and we shade above it. For $x>-3$, the boundary line is $x=-3$ (dashed) and we shade to the right of it. The solution region is the intersection of the two shaded regions.

Answer:

1. For $y\leq-\frac{5}{4}x + 2$, graph a solid line $y =-\frac{5}{4}x+2$ and shade below it.
2. For $4x - y<-1$, graph a dashed line $y = 4x + 1$ and shade above it.

3a. The linear inequality is $x + 3y>24$. Graph a dashed line $y=-\frac{1}{3}x + 8$ and shade above it.
3b. Two possible combinations are $(0,9)$ and $(25,0)$.

1. Graph a solid line $y=-\frac{1}{2}x - 6$ and shade above it, graph a dashed line $y=6x + 7$ and shade below it. The solution region is their intersection.
2. Graph a dashed line $y=\frac{2}{5}x + 4$ and shade above it, graph a dashed line $x=-3$ and shade to the right of it. The solution region is their intersection.