`a lt 6b+4`
`3blt 8`
Which of the following consists of all the a-values that satisfy the system of inequaities above ?

Options

A. `a lt 20`
B. `a lt 16 `
C. `a lt 12`
D. `a lt 8/3`

Explanation

To solve the given system of inequalities:

1. **Write down the inequalities:**
\[ a < 6b + 4 \]
\[ 3b < 8 \]

2. **Solve the second inequality for \( b \):**
\[ 3b < 8 \]
Divide both sides by 3:
\[ b < \frac{8}{3} \]

3. **Substitute \( b < \frac{8}{3} \) into the first inequality:**
We know that \( b < \frac{8}{3} \). Now, substitute this into the inequality for \( a \):
\[ a < 6b + 4 \]
Since \( b < \frac{8}{3} \), we can find the upper limit for \( a \):
\[ a < 6\left(\frac{8}{3}\right) + 4 \]

4. **Calculate \( 6 \times \frac{8}{3} \):**
\[ 6 \times \frac{8}{3} = \frac{48}{3} = 16 \]

5. **Now substitute back into the inequality for \( a \):**
\[ a < 16 + 4 \]
\[ a < 20 \]

6. **Conclusion:**
The solution for \( a \) is:
\[ a < 20 \]

Thus, the set of all \( a \)-values that satisfy the system of inequalities is \( a < 20 \).