Verified Answer :

  • For Problem 22, the system \(7y > 2y + 10\) and \(2y > 5y - 15\) yields the interval \(\boxed{2 < y < 5}\).
  • For Problem 24, the compound inequality \(4x + 6 > 10 - x\) or \(9 - 3x \geq -1\) simplifies to \(x > \frac{4}{5}\) or \(x \leq \frac{10}{3}\), which includes \(\boxed{\text{all real numbers}}\).

Explanation

  1. Analysis of the inequalities

The problem requires solving two separate algebraic tasks involving linear inequalities.

For the first task (Problem 22), we have a system of two linear inequalities in \(y\):

  1. \(7y > 2y + 10\)
  2. \(2y > 5y - 15\)

To solve a system of inequalities, we must find the values of \(y\) that satisfy both conditions simultaneously, which corresponds to the intersection of the individual solution sets.

For the second task (Problem 24), we have a compound inequality in \(x\):

  1. \(4x + 6 > 10 - x\)
  2. \(9 - 3x \geq -1\)

To solve a compound inequality involving "OR", we must find the values of \(x\) that satisfy at least one of the conditions, which corresponds to the union of the individual solution sets.

  1. Solving the system in Problem 22

We solve each inequality in the system \(7y > 2y + 10\) and \(2y > 5y - 15\) separately.

First, consider \(7y > 2y + 10\):
Subtract \(2y\) from both sides:
\(5y > 10\)
Divide by \(5\):
\(y > 2\)

Next, consider \(2y > 5y - 15\):
Subtract \(5y\) from both sides:
\(-3y > -15\)
Divide by \(-3\), remembering that the inequality sign must be reversed when dividing by a negative number:
\(y < 5\)

The solution for the system is the intersection of \(y > 2\) and \(y < 5\), which is:
\(2 < y < 5\)

  1. Solving the compound inequality in Problem 24

We solve each inequality in the compound statement \(4x + 6 > 10 - x\) or \(9 - 3x \geq -1\) separately.

First, consider \(4x + 6 > 10 - x\):
Add \(x\) to both sides:
\(5x + 6 > 10\)
Subtract \(6\) from both sides:
\(5x > 4\)
Divide by \(5\):
\(x > \frac{4}{5}\)

Next, consider \(9 - 3x \geq -1\):
Subtract \(9\) from both sides:
\(-3x \geq -10\)
Divide by \(-3\), remembering that the inequality sign must be reversed:
\(x \leq \frac{10}{3}\)

The solution for the compound inequality is the union of \(x > \frac{4}{5}\) and \(x \leq \frac{10}{3}\). Since every real number is either greater than \(\frac{4}{5}\) or less than or equal to \(\frac{10}{3}\), these two intervals cover the entire real line.
The solution set is all real numbers, denoted as \(\mathbb{R}\) or \((-\infty, \infty)\).

  1. Final conclusions

The solution for the system in Problem 22 is \(\boxed{2 < y < 5}\).
The solution for the compound inequality in Problem 24 is \(\boxed{\text{all real numbers}}\).

Examples

Linear inequalities appear frequently in resource management and budgeting. For example, if a store needs to maintain a inventory level \(y\) for a perishable product, where the lower bound ensures sufficient stock (\(y > 2\) units) and the upper bound ensures freshness (\(y < 5\) units), the acceptable range is \(2 < y < 5\). Compound inequalities are used when there are multiple acceptable states, such as setting thresholds for automated equipment where any value \(x\) outside of a specific "danger zone" is safe to operate, covering the entire range of potential values.

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