Question

1. $8b - 6\geq7b + 1$ or $2b+2 > 9b - 5$
2. $-5 - 2r<10 - 3r

Explanation:

Step 1: Solve the first - part of the compound inequality in problem 13

We have \(-5 - 2r<10 - 3r\). Add \(3r\) to both sides: \(-5 - 2r+3r<10 - 3r+3r\), which simplifies to \(-5 + r<10\). Then add 5 to both sides: \(r<10 + 5\), so \(r<15\).

Step 2: Solve the second - part of the compound inequality in problem 13

We have \(10 - 3r

Step 3: Combine the results

Combining \(r < 15\) and \(r>5\), the solution is \(5 < r<15\).

Step 4: Solve the first inequality in problem 11

For \(8b-6\geq7b + 1\), subtract \(7b\) from both sides: \(8b-7b-6\geq7b-7b + 1\), which gives \(b-6\geq1\). Add 6 to both sides: \(b\geq1 + 6\), so \(b\geq7\).

Step 5: Solve the second inequality in problem 11

For \(2b + 2>9b-5\), subtract \(2b\) from both sides: \(2b-2b + 2>9b-2b-5\), which gives \(2>7b-5\). Add 5 to both sides: \(2 + 5>7b-5 + 5\), so \(7>7b\). Divide both sides by 7: \(\frac{7}{7}>\frac{7b}{7}\), and we get \(1>b\) or \(b < 1\).

Step 6: Combine the results for problem 11

The solution for \(8b-6\geq7b + 1\) or \(2b + 2>9b-5\) is \(b<1\) or \(b\geq7\).

Answer:

For problem 11: \(b<1\) or \(b\geq7\); For problem 13: \(5 < r<15\)