Question

Below are two inequalities and the graphs of their lines without the shading. By imagining where the shading should be, identify which point would satisfy BOTH inequalities.
y, is greater than, minus, start fraction, 5, divided by, 6, end fraction, x, minus, 6
y>−
6
5
​
x−6
y, is greater than, one third, x, minus, 3
y>
3
1
​
x−3
x
y
Answer
Multiple Choice Answers
left parenthesis, 0, comma, minus, 5, right parenthesis(0,−5)
left parenthesis, minus, 7, comma, minus, 7, right parenthesis(−7,−7)
left parenthesis, minus, 9, comma, minus, 1, right parenthesis(−9,−1)
left parenthesis, 6, comma, 3, right parenthesis(6,3)

Answer 1

To determine which point satisfies both inequalities, we will first express the inequalities based on the given information:

1. **First Inequality:** \( y > -\frac{5}{6}x - 6 \)
2. **Second Inequality:** \( y > \frac{1}{3}x - 3 \)

Now we will evaluate each of the given points to see if they satisfy both inequalities.

### 1. Point (0, -5):
- For the first inequality:
\[
-5 > -\frac{5}{6}(0) - 6 \implies -5 > -6 \quad \text{(True)}
\]
- For the second inequality:
\[
-5 > \frac{1}{3}(0) - 3 \implies -5 > -3 \quad \text{(False)}
\]

### 2. Point (-7, -7):
- For the first inequality:
\[
-7 > -\frac{5}{6}(-7) - 6 \implies -7 > \frac{35}{6} - 6 \implies -7 > \frac{35}{6} - \frac{36}{6} \implies -7 > -\frac{1}{6} \quad \text{(False)}
\]
- For the second inequality:
\[
-7 > \frac{1}{3}(-7) - 3 \implies -7 > -\frac{7}{3} - 3 \implies -7 > -\frac{7}{3} - \frac{9}{3} \implies -7 > -\frac{16}{3} \quad \text{(True)}
\]

### 3. Point (-9, -1):
- For the first inequality:
\[
-1 > -\frac{5}{6}(-9) - 6 \implies -1 > \frac{45}{6} - 6 \implies -1 > \frac{45}{6} - \frac{36}{6} \implies -1 > \frac{9}{6} \quad \text{(False)}
\]
- For the second inequality:
\[
-1 > \frac{1}{3}(-9) - 3 \implies -1 > -3 - 3 \implies -1 > -6 \quad \text{(True)}
\]

### 4. Point (6, 3):
- For the first inequality:
\[
3 > -\frac{5}{6}(6) - 6 \implies 3 > -5 - 6 \implies 3 > -11 \quad \text{(True)}
\]
- For the second inequality:
\[
3 > \frac{1}{3}(6) - 3 \implies 3 > 2 - 3 \implies 3 > -1 \quad \text{(True)}
\]

After evaluating all the points, the only point that satisfies both inequalities is:

**Point (6, 3)**.

So, the answer is \( \boxed{(6, 3)} \).