Verified Answer :
- For Question 22, solve both inequalities to find \(y > 2\) and \(y < 5\), which combines to \(2 < y < 5\).
- For Question 24, solve both parts to find \(x > 4/5\) and \(x \leq 10/3\); their union covers the entire number line, so the solution is all real numbers \(x \in \mathbb{R}\).
- For Question 26, solve the two parts of the continued inequality to get \(x \leq 2\) and \(x > 1/2\), resulting in \(\frac{1}{2} < x \leq 2\).
- Final answers: \(\boxed{2 < y < 5}\); \(\boxed{x \in \mathbb{R}}\); \(\boxed{\frac{1}{2} < x \leq 2}\).
Explanation
- Analysis and Objective
We begin by analyzing each of the three inequality problems provided. These problems involve solving for a variable (either \(x\) or \(y\)) under various conditions such as conjunctions ('and') and disjunctions ('or').
- Question 22 asks for the range of \(y\) that satisfies both \(7y > 2y + 10\) and \(2y > 5y - 15\).
- Question 24 asks for the range of \(x\) that satisfies at least one of \(4x + 6 > 10 - x\) or \(9 - 3x \geq -1\).
- Question 26 asks for the range of \(x\) that satisfies the continued inequality \(6 - 3x \geq 4 - 2x > 5 - 4x\), which is interpreted as the conjunction of two separate inequalities.
Our objective is to solve each problem by isolating the variable and combining the individual solution sets according to the logical operators ('and' or 'or') specified in each case.
- Solving Question 22
For the first problem, we have the system:
- \(7y > 2y + 10\)
- \(2y > 5y - 15\)
Solving the first inequality:
Subtract \(2y\) from both sides:
\(5y > 10\)
Divide by \(5\):
\(y > 2\)
Solving the second inequality:
Subtract \(5y\) from both sides:
\(-3y > -15\)
Divide by \(-3\) (remembering to flip the inequality sign):
\(y < 5\)
The conjunction 'and' requires both conditions to be true simultaneously. Thus, the solution is the intersection of \(y > 2\) and \(y < 5\), which is:
\(2 < y < 5\)
- Solving Question 24
For the second problem, we have the system:
- \(4x + 6 > 10 - x\)
- \(9 - 3x \geq -1\)
Solving the first inequality:
Add \(x\) to both sides and subtract \(6\):
\(5x > 4\)
Divide by \(5\):
\(x > \frac{4}{5}\)
Solving the second inequality:
Subtract \(9\) from both sides:
\(-3x \geq -10\)
Divide by \(-3\) (flipping the inequality sign):
\(x \leq \frac{10}{3}\)
The disjunction 'or' requires at least one of the conditions to be true. We need to find the union of \(x > \frac{4}{5}\) and \(x \leq \frac{10}{3}\).
Since the set \(( \frac{4}{5}, \infty )\) and the set \(( -\infty, \frac{10}{3} ]\) cover all real numbers (because \(\frac{4}{5} < \frac{10}{3}\)), their union is the set of all real numbers:
\(x \in \mathbb{R}\) or all real numbers.
- Solving Question 26
For the third problem, we have the continued inequality:
\(6 - 3x \geq 4 - 2x > 5 - 4x\)
This is equivalent to the system of two inequalities:
- \(6 - 3x \geq 4 - 2x\)
- \(4 - 2x > 5 - 4x\)
Solving the first inequality:
\(6 - 4 \geq 3x - 2x\)
\(2 \geq x\) or \(x \leq 2\)
Solving the second inequality:
\(4x - 2x > 5 - 4\)
\(2x > 1\)
\(x > \frac{1}{2}\)
Since this is a continued inequality, it requires both conditions to be true simultaneously. Thus, the solution is the intersection of \(x \leq 2\) and \(x > \frac{1}{2}\):
\(\frac{1}{2} < x \leq 2\)
- Conclusion
We have solved each of the three inequality problems:
- Question 22: The solution is \(2 < y < 5\).
- Question 24: The solution is all real numbers, denoted as \((-\infty, \infty)\).
- Question 26: The solution is \(\frac{1}{2} < x \leq 2\).
The final answers are:
- Question 22: \(\boxed{2 < y < 5}\)
- Question 24: \(\boxed{x \in \mathbb{R}}\)
- Question 26: \(\boxed{\frac{1}{2} < x \leq 2}\)
Examples
Inequalities are fundamental in real-world scenarios where constraints or boundaries define feasible options. For instance, if you are designing a product, your budget might be represented as an inequality where your total costs must be less than or equal to a certain amount (\(C \leq \text{Budget}\)), while safety regulations might require your product weight to be greater than a minimum threshold (\(W > W_{min}\)). Solving compound inequalities, like in our examples, allows you to determine the 'safety zone' where all operational conditions are met, such as finding the temperature range in which a machine operates efficiently without overheating or underperforming.