Maharashtra Board Class 11 Maths Part 2 Chapter 8 Linear Inequations 8.1 Solutions

Std 11 Maths 2 Exercise 8.1 Solutions Commerce Maths

Question 1. Write the inequations that represent the interval and state whether the interval is bounded or unbounded:
(i) \([-4, \frac{7}{3}]\)
Answer:
\([-4, \frac{7}{3}]\)
Here, x takes values between -4 and \(\frac{7}{3}\) including -4 and \(\frac{7}{3}\).
\(\therefore\) the required inequation is \(-4 \le x \le \frac{7}{3}\)
\(\therefore\) it is a bounded (closed) interval.
In simple words: This interval means x can be any number from -4 to \(\frac{7}{3}\), including both -4 and \(\frac{7}{3}\). Because both endpoints are included and it has a definite start and end, it's a bounded, or closed, interval.

🎯 Exam Tip: Remember that square brackets [ ] indicate a closed interval (endpoints included), while parentheses ( ) indicate an open interval (endpoints excluded). Bounded intervals have finite endpoints.

(ii) \((0, 0.9]\)
Answer:
\((0, 0.9]\)
Here, x takes values between 0 and 0.9, including 0.9 and excluding 0.
\(\therefore\) the required inequation is \(0 < x \le 0.9\)
\(\therefore\) it is a bounded (semi-right closed) interval.
In simple words: This interval allows x to be any number greater than 0 but less than or equal to 0.9. It's bounded because it has clear start and end points, but "semi-right closed" because only the right endpoint (0.9) is included.

🎯 Exam Tip: Intervals with one square bracket and one parenthesis are called semi-open or semi-closed. Pay attention to which side is open/closed to correctly write the inequality and classify the interval type.

(iii) \((-\infty, \infty)\)
Answer:
\((-\infty, \infty)\)
Here, x takes values between \(-\infty\) and \(\infty\)
\(\therefore\) the required inequation is \(-\infty < x < \infty\)
\(\therefore\) it is an unbounded (open) interval.
In simple words: This interval represents all real numbers, meaning x can be any value without any upper or lower limits. Since it extends infinitely in both directions, it is considered unbounded and open.

🎯 Exam Tip: Infinity symbols (\(\infty\)) always accompany parentheses, as infinity is not a number that can be included. Any interval involving infinity is always unbounded.

(iv) \([5, \infty)\)
Answer:
\([5, \infty)\)
Here, x takes values between 5 and \(\infty\) including 5.
\(\therefore\) the required inequation is \(5 \le x < \infty\)
\(\therefore\) it is an unbounded (semi-left closed) interval.
In simple words: This interval means x can be any number that is 5 or greater, extending infinitely. It is unbounded because it goes to infinity, and "semi-left closed" because 5 is included.

🎯 Exam Tip: When writing inequalities for unbounded intervals, always use the correct comparison operator (\(\le\), \(\ge\)) for the finite endpoint if it's included, and strict inequality (<, >) for infinity.

(v) \((-11, -2)\)
Answer:
\((-11, -2)\)
Here, x takes values between -11 and -2
\(\therefore\) the required inequation is \(-11 < x < -2\)
\(\therefore\) it is a bounded (open) interval.
In simple words: This interval means x can be any number strictly between -11 and -2, but not including -11 or -2. It's a bounded interval because it has definite start and end points, and it's open because neither endpoint is included.

🎯 Exam Tip: An open interval means the values approach the endpoints but never actually reach them. Both parentheses signify an open interval and strict inequalities.

(vi) \((-\infty, 3)\)
Answer:
\((-\infty, 3)\)
Here, x takes values between \(-\infty\) and 3
\(\therefore\) the required inequation is \(-\infty < x < 3\)
\(\therefore\) it is an unbounded (open) interval.
In simple words: This interval includes all real numbers less than 3, extending infinitely in the negative direction. It's unbounded because it goes to negative infinity, and open because 3 is not included.

🎯 Exam Tip: Intervals extending to negative infinity represent values "less than" a specific number. The open parenthesis at 3 means x cannot equal 3.

Question 2. Solve the following inequations
(i) \(3x - 36 > 0\)
Answer:
\(3x - 36 > 0\)
Adding 36 both sides, we get
\(3x - 36 + 36 > 0 + 36\)
\(\therefore 3x > 36\)
Dividing both sides by 3, we get
\(\frac{3x}{3} > \frac{36}{3}\)
\(\therefore x > 12\)
\(\therefore\) x takes all real values more than 12
\(\therefore\) Solution set = \((12, \infty)\)
In simple words: To solve this, isolate x by adding 36 to both sides, then divide by 3. The solution is all numbers strictly greater than 12.

🎯 Exam Tip: Remember that when you add or subtract a number from both sides of an inequality, the inequality sign does not change. The goal is to isolate the variable.

(ii) \(7x - 25 \le -4\)
Answer:
\(7x-25 \le -4\)
Adding 25 on both sides, we get
\(7x - 25 + 25 \le -4 + 25\)
\(\therefore 7x \le 21\)
Dividing both sides by 7, we get
\(x \le 3\)
\(\therefore\) x takes all real values less or equal to 3.
\(\therefore\) Solution Set = \((-\infty, 3]\)
In simple words: To find the solution, first add 25 to both sides, then divide by 7. This shows that x must be 3 or any number smaller than 3.

🎯 Exam Tip: Division by a positive number does not change the inequality direction. The solution set includes 3, so a square bracket is used for the upper bound.

(iii) \(0 < \frac{x-5}{4} < 3\)
Answer:
\(0 < \frac{x-5}{4} < 3\)
Multiplying by 4 throughout, we get
\(0 \times 4 < x-5 < 3 \times 4\)
\(0 < x-5 < 12\)
Adding 5 on both sides, we get
\(0 + 5 < x-5 + 5 < 12 + 5\)
\(5 < x < 17\)
x takes all real values between 5 and 17.
\(\therefore\) Solution set = \((5, 17)\)
In simple words: To solve this compound inequality, multiply all parts by 4, then add 5 to all parts. This isolates x, showing it lies strictly between 5 and 17.

🎯 Exam Tip: When solving compound inequalities, perform operations on all parts simultaneously to maintain equivalence. The solution represents an open interval since strict inequalities are used.

(iv) \(|7x - 4| < 10\)
Answer:
\(|7x-4| < 10\)
This implies \(-10 < 7x - 4 < 10\) (using the property: \(|x| < k\) is same as \(-k < x < k\))
Adding 4 on both sides, we get
\(-10 + 4 < 7x - 4 + 4 < 10 + 4\)
\(-6 < 7x < 14\)
Dividing both sides by 7, we get
\(\frac{-6}{7} < \frac{7x}{7} < \frac{14}{7}\)
\(\therefore -\frac{6}{7} < x < 2\)
\(\therefore\) x takes all real values between \(-\frac{6}{7}\) and 2.
\(\therefore\) Solution set = \((-\frac{6}{7}, 2)\)
In simple words: An absolute value inequality like \(|A| < B\) can be rewritten as \(-B < A < B\). Apply this, then solve the resulting compound inequality by adding 4 and then dividing by 7 across all parts to find the range for x.

🎯 Exam Tip: Always convert absolute value inequalities into compound inequalities first. Remember that dividing by a positive number maintains the inequality direction.

Question 3. Sketch the graph which represents the solution set for the following inequations:
(i) \(x > 5\)
Answer:
\(x > 5\)
Here, x takes all real values that are greater than 5.
\(\therefore\) Solution set represents the unbounded (open) interval \((5, \infty)\)
\(\therefore\) the required graph of the solution set is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -5 से 9 तक पूर्णांकों को दर्शाती है। 5 पर एक खुला वृत्त (hollow circle) है, और 5 से दाईं ओर एक तीर के साथ रेखा विस्तारित है, जो दर्शाती है कि x का मान 5 से अधिक है (x > 5)।
In simple words: The graph for \(x > 5\) is a number line with an open circle at 5 and a line extending indefinitely to the right, indicating all numbers strictly greater than 5.

🎯 Exam Tip: For strict inequalities (> or <), use an open circle (or hollow dot) at the critical point on the number line. The arrow indicates the direction of all possible values for x.

(ii) \(x \ge 5\)
Answer:
\(x \ge 5\)
Here, x takes all real values that are greater than or equal to 5
\(\therefore\) Solution set represents the unbounded (semi-left closed) interval \([5, \infty)\)
\(\therefore\) the required graph of the solution set is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -4 से 9 तक पूर्णांकों को दर्शाती है। 5 पर एक ठोस वृत्त (solid circle) है, और 5 से दाईं ओर एक तीर के साथ रेखा विस्तारित है, जो दर्शाती है कि x का मान 5 के बराबर या उससे अधिक है (x ≥ 5)।
In simple words: The graph for \(x \ge 5\) is a number line with a solid circle at 5 and a line extending indefinitely to the right, showing all numbers greater than or equal to 5.

🎯 Exam Tip: For inequalities including equality (\(\ge\) or \(\le\)), use a solid circle (or closed dot) at the critical point on the number line. This indicates that the endpoint is part of the solution set.

(iii) \(x < 3\)
Answer:
\(x < 3\)
Here, x takes all real values that are less than 3.
\(\therefore\) Solution set represents the unbounded (open) interval \((-\infty, 3)\)
\(\therefore\) the required graph of the solution set is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -9 से 5 तक पूर्णांकों को दर्शाती है। 3 पर एक खुला वृत्त (hollow circle) है, और 3 से बाईं ओर एक तीर के साथ रेखा विस्तारित है, जो दर्शाती है कि x का मान 3 से कम है (x < 3)।
In simple words: The graph for \(x < 3\) is a number line with an open circle at 3 and a line extending indefinitely to the left, representing all numbers strictly less than 3.

🎯 Exam Tip: When graphing, ensure the direction of the arrow matches the inequality. "<" means values to the left, ">" means values to the right.

(iv) \(x \le 3\)
Answer:
\(x \le 3\)
Here, x takes all real values less than and including 3
\(\therefore\) Solution set represents the unbounded (semi-right closed) interval \((-\infty, 3]\)
\(\therefore\) the required graph of the solution set is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -4 से 4 तक पूर्णांकों को दर्शाती है। 3 पर एक ठोस वृत्त (solid circle) है, और 3 से बाईं ओर एक तीर के साथ रेखा विस्तारित है, जो दर्शाती है कि x का मान 3 के बराबर या उससे कम है (x ≤ 3)।
In simple words: The graph for \(x \le 3\) is a number line with a solid circle at 3 and a line extending indefinitely to the left, showing all numbers less than or equal to 3.

🎯 Exam Tip: For "less than or equal to" inequalities, the solid dot and leftward arrow correctly depict the solution set on a number line.

(v) \(-4 < x < 3\)
Answer:
\(-4 < x < 3\)
Here, x takes all real values between -4 and 3.
\(\therefore\) Solution set represents the bounded (open) interval \((-4, 3)\)
\(\therefore\) the required graph of the solution set is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -6 से 7 तक पूर्णांकों को दर्शाती है। -4 और 3 पर खुले वृत्त (hollow circles) हैं, और उनके बीच एक रेखा खंड है, जो दर्शाता है कि x का मान -4 से अधिक और 3 से कम है (-4 < x < 3)।
In simple words: The graph for \(-4 < x < 3\) is a number line with open circles at -4 and 3, connected by a line segment, showing all numbers strictly between these two values.

🎯 Exam Tip: Compound inequalities like this are represented by a segment on the number line. Open circles at both ends signify that the interval is open and the endpoints are not included.

(vi) \(-2 \le x < 2.5\)
Answer:
\(-2 \le x < 2.5\)
Here, x takes all values between -2 and 2.5 including -2.
\(\therefore\) Solution set represents the bounded (semi-left closed) interval \([-2, 2.5)\)
\(\therefore\) the required graph of the solution set is as follows.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -5 से 6 तक पूर्णांकों को दर्शाती है। -2 पर एक ठोस वृत्त (solid circle) और 2.5 पर एक खुला वृत्त (hollow circle) है, और उनके बीच एक रेखा खंड है, जो दर्शाता है कि x का मान -2 के बराबर या उससे अधिक और 2.5 से कम है (-2 ≤ x < 2.5)।
In simple words: The graph for \(-2 \le x < 2.5\) is a number line with a solid circle at -2 and an open circle at 2.5, connected by a line segment. This shows that -2 is included, but 2.5 is not.

🎯 Exam Tip: For semi-closed intervals, one endpoint uses a solid dot (included) and the other an open dot (excluded). Always plot decimal points accurately between integers.

(vii) \(-3 \le x \le 1\)
Answer:
\(-3 \le x \le 1\)
Here, x takes all real values between -3 and 1 including -3 and 1
\(\therefore\) Solution set represents the bounded (closed) interval \([-3, 1]\)
\(\therefore\) the required graph of the solution set is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -4 से 4 तक पूर्णांकों को दर्शाती है। -3 और 1 पर ठोस वृत्त (solid circles) हैं, और उनके बीच एक रेखा खंड है, जो दर्शाता है कि x का मान -3 के बराबर या उससे अधिक और 1 के बराबर या उससे कम है (-3 ≤ x ≤ 1)।
In simple words: The graph for \(-3 \le x \le 1\) is a number line with solid circles at -3 and 1, connected by a line segment, meaning all numbers between -3 and 1, inclusive, are part of the solution.

🎯 Exam Tip: A closed interval like this uses solid dots at both endpoints and a connecting line segment, indicating that all values from -3 to 1, including -3 and 1, satisfy the inequality.

(viii) \(|x| < 4\)
Answer:
\(|x| < 4 \implies -4 < x < 4\)
Here, x takes all real values between -4 and 4.
\(\therefore\) Solution set represents bounded (open) interval \((-4, 4)\)
\(\therefore\) the required graph of the solution set is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -5 से 5 तक पूर्णांकों को दर्शाती है। -4 और 4 पर खुले वृत्त (hollow circles) हैं, और उनके बीच एक रेखा खंड है, जो दर्शाता है कि x का निरपेक्ष मान 4 से कम है (|x| < 4)।
In simple words: The inequality \(|x| < 4\) means that x is between -4 and 4. The graph shows open circles at -4 and 4, with a line connecting them, indicating all numbers strictly between these two values.

🎯 Exam Tip: Remember that \(|x| < a\) translates to \(-a < x < a\). This represents an open interval centered at zero on the number line.

(ix) \(|x| \ge 3.5\)
Answer:
\(|x| \ge 3.5 \implies x \ge 3.5\) or \(x \le -3.5\)
Here, x takes values greater than or equal to 3.5 or it takes values less than or equal to -3.5
\(\therefore\) Solution set represents the unbounded (semi-left closed) interval \([3.5, \infty)\) or the unbounded (semi-right closed) interval \((-\infty, -3.5]\)
\(\therefore x \in (-\infty, -3.5] \cup [3.5, \infty)\)
\(\therefore\) the required graph of the solution set is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संख्या रेखा -5 से 5 तक पूर्णांकों को दर्शाती है। -3.5 और 3.5 पर ठोस वृत्त (solid circles) हैं। -3.5 से बाईं ओर एक तीर के साथ रेखा विस्तारित है और 3.5 से दाईं ओर एक तीर के साथ रेखा विस्तारित है, जो दर्शाता है कि x का निरपेक्ष मान 3.5 के बराबर या उससे अधिक है (|x| ≥ 3.5)।
In simple words: The inequality \(|x| \ge 3.5\) means x is either greater than or equal to 3.5, or less than or equal to -3.5. The graph shows solid circles at -3.5 and 3.5, with lines extending outwards from both points.

🎯 Exam Tip: For absolute value inequalities of the form \(|x| \ge a\), the solution is always a union of two separate intervals: \(x \le -a\) or \(x \ge a\). Remember to use solid dots for inclusive inequalities.

Question 4. Solve the inequations:
(i) \(5x + 7 > 4 - 2x\)
Answer:
\(5x + 7 > 4 - 2x\)
Adding 2x on both sides, we get
\(5x + 2x + 7 > 4 - 2x + 2x\)
\(7x + 7 > 4\)
Subtracting 7 from both sides, we get
\(7x + 7 - 7 > 4 - 7\)
\(7x > -3\)
Dividing by 7 on both sides, we get
\(x > -\frac{3}{7}\)
i.e., x takes all real values greater than \(-\frac{3}{7}\)
\(\therefore\) the solution set is \((-\frac{3}{7}, \infty)\)
In simple words: Combine x terms on one side and constant terms on the other. Add 2x to both sides, then subtract 7, and finally divide by 7 to find that x must be greater than \(-\frac{3}{7}\).

🎯 Exam Tip: When moving terms across the inequality sign, remember to change their operation (addition becomes subtraction, multiplication becomes division). Be careful with negative signs during division/multiplication.

(ii) \(3x + 1 \ge 6x - 4\)
Answer:
\(3x + 1 \ge 6x - 4\)
Subtracting 3x from both sides, we get
\(3x - 3x + 1 \ge 6x - 3x - 4\)
\(1 \ge 3x - 4\)
Adding 4 on both sides, we get
\(1 + 4 \ge 3x - 4 + 4\)
\(5 \ge 3x\)
Dividing by 3 on both sides, we get
\(\frac{5}{3} \ge \frac{3x}{3}\)
\(\frac{5}{3} \ge x\)
i.e., \(x \le \frac{5}{3}\)
i.e., x takes all real values less than or equal to \(\frac{5}{3}\).
\(\therefore\) the solution set is \((-\infty, \frac{5}{3}]\)
In simple words: Move the x terms to one side and constants to the other. Subtract 3x, then add 4, and divide by 3 to find that x is less than or equal to \(\frac{5}{3}\).

🎯 Exam Tip: It's often helpful to rearrange the inequality so the variable has a positive coefficient. Reading \(5 \ge 3x\) as \(3x \le 5\) can help prevent errors in writing the final interval.

(iii) \(4 - 2x < 3(3 - x)\)
Answer:
\(4 - 2x < 3(3 - x)\)
\(\therefore 4 - 2x < 9 - 3x\)
Adding 3x on both sides, we get
\(4 - 2x + 3x < 9 - 3x + 3x\)
\(4 + x < 9\)
Subtracting 4 from both sides, we get
\(4 + x - 4 < 9 - 4\)
\(x < 5\)
i.e., x takes all real values less than 5
\(\therefore\) the solution set is \((-\infty, 5)\)
In simple words: First, distribute the 3 on the right side. Then, gather all x terms on one side and constants on the other, ensuring the inequality sign is maintained, to find that x must be less than 5.

🎯 Exam Tip: Always simplify both sides of the inequality first, especially by distributing multiplication over parentheses, before attempting to isolate the variable.

(iv) \(\frac{3}{4}x - 6 \le x - 7\)
Answer:
\(\frac{3}{4}x - 6 \le x - 7\)
Multiplying by 4 on both sides, we get
\(4(\frac{3}{4}x - 6) \le 4(x - 7)\)
\(3x - 24 \le 4x - 28\)
Subtracting 3x from both sides, we get
\(3x - 3x - 24 \le 4x - 3x - 28\)
\(-24 \le x - 28\)
Adding 28 on both the sides, we get
\(-24 + 28 \le x - 28 + 28\)
\(\therefore 4 \le x\) i.e., \(x \ge 4\)
i.e., x takes all real values greater or equal to 4.
\(\therefore\) the solution set is \([4, \infty)\)
In simple words: Eliminate the fraction by multiplying by 4. Then, rearrange terms to get x on one side and constants on the other, yielding that x must be greater than or equal to 4.

🎯 Exam Tip: Multiplying an inequality by the LCM of denominators helps clear fractions, making calculations simpler. Always distribute the multiplier to *every* term on both sides.

(v) \(-8 \le -(3x - 5) < 13\)
Answer:
\(-8 \le -(3x - 5) < 13\)
First, simplify the middle term: \(-(3x - 5) = -3x + 5\)
So, \(-8 \le -3x + 5 < 13\)
Subtracting 5 from all parts:
\(-8 - 5 \le -3x + 5 - 5 < 13 - 5\)
\(-13 \le -3x < 8\)
Dividing by -3 throughout (so inequality sign changes for both parts):
\(\frac{-13}{-3} \ge \frac{-3x}{-3} > \frac{8}{-3}\)
\(\frac{13}{3} \ge x > -\frac{8}{3}\)
Rearranging to standard form:
\(-\frac{8}{3} < x \le \frac{13}{3}\)
i.e., x takes all real values between \(-\frac{8}{3}\) and \(\frac{13}{3}\) including \(\frac{13}{3}\).
\(\therefore\) the solution set is \((-\frac{8}{3}, \frac{13}{3}]\)
In simple words: Distribute the negative sign, then subtract 5 from all parts of the compound inequality. Finally, divide all parts by -3, remembering to reverse both inequality signs.

🎯 Exam Tip: The critical rule for inequalities is to reverse the direction of the inequality sign whenever you multiply or divide by a negative number. This is a common point of error.

(vi) \(-1 < 3 - \frac{x}{5} \le 1\)
Answer:
\(-1 < 3 - \frac{x}{5} \le 1\)
Subtracting 3 from both sides, we get
\(-1 - 3 < 3 - \frac{x}{5} - 3 \le 1 - 3\)
\(-4 < -\frac{x}{5} \le -2\)
Multiplying by -1 throughout (so inequality sign changes):
\((-4) \times (-1) > (-\frac{x}{5}) \times (-1) \ge (-2) \times (-1)\)
\(4 > \frac{x}{5} \ge 2\)
Rearranging:
\(2 \le \frac{x}{5} < 4\)
Multiplying by 5 on both sides, we get
\(5 \times 2 \le 5 \times \frac{x}{5} < 5 \times 4\)
\(10 \le x < 20\)
i.e., x takes all real values between 10 and 20.
\(\therefore\) the solution set is \([10, 20)\)
In simple words: First, subtract 3 from all parts. Then, multiply all parts by -1, remembering to flip the inequality signs. Finally, multiply all parts by 5 to isolate x.

🎯 Exam Tip: Handle fractions by multiplying by the denominator. Always double-check the direction of inequality signs, especially after multiplying or dividing by negative numbers.

(vii) \(2|4 - 5x| \ge 9\)
Answer:
\(2|4 - 5x| \ge 9\)
Divide by 2:
\(\therefore |4 - 5x| \ge \frac{9}{2}\)
This implies \(4 - 5x \le -\frac{9}{2}\) or \(4 - 5x \ge \frac{9}{2}\) (using the property: \(|x| \ge a\) implies \(x \le -a\) or \(x \ge a\))

**Case 1:** \(4 - 5x \le -\frac{9}{2}\)
Subtracting 4 from both sides, we get
\(-5x \le -\frac{9}{2} - 4\)
\(-5x \le -\frac{9}{2} - \frac{8}{2}\)
\(-5x \le -\frac{17}{2}\)
Divide by -5 (so inequality sign changes)
\(x \ge \frac{-17/2}{-5}\)
\(x \ge \frac{17}{10}\)

**Case 2:** \(4 - 5x \ge \frac{9}{2}\)
Subtracting 4 from both sides, we get
\(-5x \ge \frac{9}{2} - 4\)
\(-5x \ge \frac{9}{2} - \frac{8}{2}\)
\(-5x \ge \frac{1}{2}\)
Divide by -5 (so inequality sign changes)
\(x \le \frac{1/2}{-5}\)
\(x \le -\frac{1}{10}\)

\(\therefore\) x takes all real values less than or equal to \(-\frac{1}{10}\) or it takes all real values greater or equal to \(\frac{17}{10}\).
\(\therefore\) the solution set is \((-\infty, -\frac{1}{10}] \cup [\frac{17}{10}, \infty)\)
In simple words: First, divide by 2. Then, convert the absolute value inequality into two separate inequalities: one where \(4 - 5x\) is less than or equal to the negative value, and another where it's greater than or equal to the positive value. Solve each inequality separately, remembering to flip the sign when dividing by a negative number.

🎯 Exam Tip: Absolute value inequalities with "greater than or equal to" (\(\ge\)) signs result in two separate, disconnected intervals (a union), unlike "less than" inequalities which form a single interval. Be meticulous with sign changes.

(viii) \(|2x + 7| \le 25\)
Answer:
\(|2x + 7| \le 25\)
This implies \(-25 \le 2x + 7 \le 25\) (using the property: \(|x| \le a\) implies \(-a \le x \le a\))
Subtracting 7 from both sides, we get
\(-25 - 7 \le 2x + 7 - 7 \le 25 - 7\)
\(-32 \le 2x \le 18\)
Dividing by 2 on both sides, we get
\(\frac{-32}{2} \le \frac{2x}{2} \le \frac{18}{2}\)
\(-16 \le x \le 9\)
\(\therefore\) x can take all real values between -16 and 9 including -16 and 9.
\(\therefore\) the solution set is \([-16, 9]\)
In simple words: Convert the absolute value inequality into a compound inequality. Subtract 7 from all parts, then divide all parts by 2 to find the range of x values that satisfy the condition.

🎯 Exam Tip: Absolute value inequalities with "less than or equal to" (\(\le\)) signs result in a single, closed interval. Ensure all operations are applied consistently across all three parts of the compound inequality.

(ix) \(|2x + 3| > 1\)
Answer:
\(|2x + 3| > 1\)
Dividing by 2 on both sides, we get
\(|x + \frac{3}{2}| > \frac{1}{2}\)
This implies \(x + \frac{3}{2} < -\frac{1}{2}\) or \(x + \frac{3}{2} > \frac{1}{2}\) (using the property: \(|y| > a\) implies \(y < -a\) or \(y > a\))

**Case 1:** \(x + \frac{3}{2} < -\frac{1}{2}\)
Subtracting \(\frac{3}{2}\) from both sides, we get
\(x < -\frac{1}{2} - \frac{3}{2}\)
\(x < -\frac{4}{2}\)
\(x < -2\)

**Case 2:** \(x + \frac{3}{2} > \frac{1}{2}\)
Subtracting \(\frac{3}{2}\) from both sides, we get
\(x > \frac{1}{2} - \frac{3}{2}\)
\(x > -\frac{2}{2}\)
\(x > -1\)

\(\therefore\) x can take all real values less than -2 or it can take values greater than -1.
\(\therefore\) Solution set is \((-\infty, -2) \cup (-1, \infty)\)
In simple words: First, simplify the absolute value by dividing by 2. Then, split the inequality into two separate conditions: one where the expression inside the absolute value is less than the negative value, and one where it's greater than the positive value. Solve each for x.

🎯 Exam Tip: For \(|Ax + B| > C\), always separate into \(Ax + B < -C\) or \(Ax + B > C\). Remember to express the final answer as a union of the two resulting intervals.

(x) \(\frac{x+5}{x-3} < 0\)
Answer:
\(\frac{x+5}{x-3} < 0\)
Since \(\frac{a}{b} < 0\), when a > 0 and b < 0 or a < 0 and b > 0
**Case I:** \(x + 5 > 0\) and \(x - 3 < 0\)
\(\therefore x > -5\) and \(x < 3\)
The intersection of these two conditions is \(-5 < x < 3\)
\(\therefore\) solution set = \((-5, 3)\)

**Case II:** \(x + 5 < 0\) and \(x - 3 > 0\)
\(\therefore x < -5\) and \(x > 3\)
This condition is not possible, as x cannot be simultaneously less than -5 and greater than 3.
\(\therefore\) solution set = \(\Phi\) (empty set)

\(\therefore\) solution set of the given inequation is \((-5, 3)\)
In simple words: A fraction is negative if its numerator and denominator have opposite signs. Consider two cases: numerator positive and denominator negative, or vice-versa. Find the range of x for each case, and combine valid solutions.

🎯 Exam Tip: When solving rational inequalities, always consider the signs of both the numerator and the denominator. A number line or sign chart can be very useful to visualize the intervals where the expression is positive or negative.

(xi) \(\frac{x-2}{x+5} > 0\)
Answer:
\(\frac{x-2}{x+5} > 0\)
Since \(\frac{a}{b} > 0\), when a > 0 and b > 0 or a < 0 and b < 0

**Case I:** \(x - 2 > 0\) and \(x + 5 > 0\)
\(\therefore x > 2\) and \(x > -5\)
The intersection of these two conditions is \(x > 2\)
\(\therefore\) solution set = \((2, \infty)\)

**Case II:** \(x - 2 < 0\) and \(x + 5 < 0\)
\(\therefore x < 2\) and \(x < -5\)
The intersection of these two conditions is \(x < -5\)
\(\therefore\) solution set = \((-\infty, -5)\)

\(\therefore\) the solution set of the given inequation is \((-\infty, -5) \cup (2, \infty)\)
In simple words: A fraction is positive if its numerator and denominator have the same sign. Consider two scenarios: both positive or both negative. Solve for x in each case and combine the valid ranges using a union (\(\cup\)) symbol.

🎯 Exam Tip: For rational inequalities greater than zero, the solution set will typically be a union of two disjoint intervals. Remember that the denominator cannot be zero, so \(x \ne -5\) in this example.

Question 5. Rajiv obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Answer:
Let \(x_1, x_2, x_3\) denote the marks in 1st, 2nd and 3rd unit test respectively. Then
\(\frac{x_1+x_2+x_3}{3} \ge 60\)
\(\therefore \frac{70+75+x_3}{3} \ge 60\)
\(\therefore \frac{145 + x_3}{3} \ge 60\)
Multiply both sides by 3:
\(145 + x_3 \ge 3(60)\)
\(145 + x_3 \ge 180\)
Subtracting 145 from both sides, we get
\(x_3 \ge 180 - 145\)
\(\therefore x_3 \ge 35\)
Rajiv must obtain a minimum of 35 marks to maintain an average of at least 60 marks.
In simple words: To find the minimum marks needed, set up an inequality where the average of the three test scores is greater than or equal to 60. Solve this inequality for the third test score.

🎯 Exam Tip: For "at least" problems, the inequality will be "greater than or equal to" (\(\ge\)). Average is sum of scores divided by number of scores. Be careful with calculations to find the exact minimum value.

Question 6. To receive Grade 'A' in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in the first four examinations are 87, 92, 94, and 95, find the minimum marks that Sunita must obtain in the fifth examination to get a grade 'A' in the course.
Answer:
Let \(x_1, x_2, x_3, x_4, x_5\) denote the marks in five examinations. Then
\(\frac{x_1+x_2+x_3+x_4+x_5}{5} \ge 90\)
\(\therefore \frac{87+92+94+95+x_5}{5} \ge 90\)
\(\therefore \frac{368 + x_5}{5} \ge 90\)
Multiply both sides by 5:
\(368 + x_5 \ge 5 \times 90\)
\(368 + x_5 \ge 450\)
Subtracting 368 from both sides, we get
\(x_5 \ge 450 - 368\)
\(\therefore x_5 \ge 82\)
Sunita must obtain a minimum of 82 marks in the 5th examination to get a grade of A.
In simple words: To find the minimum marks for an 'A' grade, set up an inequality where the average of five test scores is at least 90. Substitute the known scores and solve for the unknown fifth score.

🎯 Exam Tip: Clearly define your variables. The phrase "average of 90 marks or more" translates directly to \(\ge 90\). Ensure correct sum and division for the average calculation.

Question 7. Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.
Answer:
Let two consecutive positive odd integers be \(2n - 1, 2n + 1\) where \(n \ge 1 \in \mathbb{Z}\).
Given that both integers are smaller than 10:
\(2n - 1 < 10\) and \(2n + 1 < 10\)
From \(2n - 1 < 10 \implies 2n < 11 \implies n < \frac{11}{2}\)
From \(2n + 1 < 10 \implies 2n < 9 \implies n < \frac{9}{2}\)
Combining these, we need \(n < \frac{9}{2}\) .......(i)

Also, their sum is more than 11:
\((2n - 1) + (2n + 1) > 11\)
\(4n > 11\)
\(\therefore n > \frac{11}{4}\) .......(ii)

From (i) and (ii), we have:
\(\frac{11}{4} < n < \frac{9}{2}\)
Since, n is an integer,
\(\frac{11}{4} = 2.75\) and \(\frac{9}{2} = 4.5\)
So, \(2.75 < n < 4.5\)
The integers n satisfying this are \(n = 3, 4\)

If \(n = 3\), the integers are \(2(3) - 1 = 5\), \(2(3) + 1 = 7\). The pair is (5, 7).
Both 5 and 7 are smaller than 10. Their sum \(5 + 7 = 12\), which is more than 11. (Valid)

If \(n = 4\), the integers are \(2(4) - 1 = 7\), \(2(4) + 1 = 9\). The pair is (7, 9).
Both 7 and 9 are smaller than 10. Their sum \(7 + 9 = 16\), which is more than 11. (Valid)

\(\therefore\) The pairs of positive consecutive odd integers are (5, 7) and (7, 9).
In simple words: Represent consecutive odd integers as \(2n-1\) and \(2n+1\). Set up inequalities for "smaller than 10" and "sum more than 11." Solve for n and find the integer values of n that satisfy both conditions to determine the pairs.

🎯 Exam Tip: When working with consecutive odd/even integers, represent them using expressions involving 'n' (e.g., \(2n-1, 2n+1\) for odd; \(2n, 2n+2\) for even). Ensure you satisfy all conditions of the problem simultaneously, often by finding the intersection of multiple inequalities for 'n'.

Question 8. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Answer:
Let \(2n, 2n + 2\) be two positive consecutive even integers where \(n \ge 1 \in \mathbb{Z}\).
Given that both integers are larger than 5:
\(2n > 5\) and \(2n + 2 > 5\)
From \(2n > 5 \implies n > \frac{5}{2}\)
From \(2n + 2 > 5 \implies 2n > 3 \implies n > \frac{3}{2}\)
Combining these, we need \(n > \frac{5}{2}\) .......(i)

Also, their sum is less than 23:
\((2n) + (2n + 2) < 23\)
\(4n + 2 < 23\)
\(4n < 21\)
\(\therefore n < \frac{21}{4}\) .......(ii)

From (i) and (ii), we have:
\(\frac{5}{2} < n < \frac{21}{4}\)
Since n is an integer,
\(\frac{5}{2} = 2.5\) and \(\frac{21}{4} = 5.25\)
So, \(2.5 < n < 5.25\)
The integers n satisfying this are \(n = 3, 4, 5\)

If \(n = 3\), the integers are \(2(3) = 6\), \(2(3) + 2 = 8\). The pair is (6, 8).
Both 6 and 8 are larger than 5. Their sum \(6 + 8 = 14\), which is less than 23. (Valid)

If \(n = 4\), the integers are \(2(4) = 8\), \(2(4) + 2 = 10\). The pair is (8, 10).
Both 8 and 10 are larger than 5. Their sum \(8 + 10 = 18\), which is less than 23. (Valid)

If \(n = 5\), the integers are \(2(5) = 10\), \(2(5) + 2 = 12\). The pair is (10, 12).
Both 10 and 12 are larger than 5. Their sum \(10 + 12 = 22\), which is less than 23. (Valid)

\(\therefore\) The pairs of positive even consecutive integers are (6, 8), (8, 10), (10, 12).
In simple words: Represent consecutive even integers as \(2n\) and \(2n+2\). Formulate inequalities based on "larger than 5" and "sum less than 23". Solve for n to find the possible integer values, then generate the corresponding pairs.

🎯 Exam Tip: Pay close attention to keywords like "positive", "consecutive", "even/odd", "larger/smaller than", and "sum less/more than" to correctly set up the variables and inequalities. Listing out the valid integer values for 'n' makes it easy to find the final pairs.

Question 9. The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum integer length of the shortest side.
Answer:
Let the shortest side be x.
Then longest side length = 2x
and third side length = x + 2
Perimeter = shortest side + longest side + third side
Perimeter = \(x + 2x + (x + 2)\)
Perimeter = \(4x + 2\)
Given, perimeter > 166
\(\therefore 4x + 2 > 166\)
Subtract 2 from both sides:
\(4x > 166 - 2\)
\(4x > 164\)
Divide by 4:
\(x > \frac{164}{4}\)
\(\therefore x > 41\)
Since x must be an integer, the minimum integer length of the shortest side is 42 cm.
In simple words: Assign the shortest side as 'x'. Express the other two sides and the perimeter in terms of 'x'. Set up an inequality for the perimeter being "more than 166 cm" and solve for 'x'. Since 'x' must be an integer, find the smallest integer satisfying the inequality.

🎯 Exam Tip: Clearly define variables for each side of the triangle. Ensure the perimeter formula is correct. When an inequality results in a non-integer value (e.g., \(x > 41\)), and an integer solution is required, choose the next whole number (e.g., 42 for >41, or 41 for \(\ge 41\)).