Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.2

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.2

Question 1. Solve for x
(i) \( |3 - x| < 7 \)
(ii) \( |4x - 5| \ge -2 \)
(iii) \( |3 - \frac{3}{4}x| \le \frac{1}{4} \)
(iv) \( |x| - 10 < -3 \)
Answer:
(i) We need to solve the inequality \( |3 - x| < 7 \).
When \( |A| < k \), it means \( -k < A < k \). So, we can write:
\( -7 < 3 - x < 7 \)
First, subtract 3 from all parts of the inequality:
\( -7 - 3 < -x < 7 - 3 \)
\( -10 < -x < 4 \)
Now, multiply the entire inequality by -1. Remember to reverse the inequality signs when multiplying by a negative number:
\( 10 > x > -4 \)
We can rewrite this in the standard ascending order:
\( -4 < x < 10 \)
The solution set in interval notation is \( x \in (-4, 10) \). This type of inequality is often used to define a range of acceptable values in real-world problems.


(ii) We need to solve the inequality \( |4x - 5| \ge -2 \).
The absolute value of any real number is always non-negative. This means \( |4x - 5| \) will always be greater than or equal to 0.
Since any non-negative number is always greater than or equal to a negative number (-2 in this case), the inequality \( |4x - 5| \ge -2 \) is true for all real numbers.
So, the solution set is \( x \in R \) (all real numbers). Absolute value functions ensure distance is always positive, making it impossible to be less than zero.

(iii) We need to solve the inequality \( |3 - \frac{3}{4}x| \le \frac{1}{4} \).
When \( |A| \le k \), it means \( -k \le A \le k \). So, we can write:
\( -\frac{1}{4} \le 3 - \frac{3}{4}x \le \frac{1}{4} \)
First, subtract 3 from all parts:
\( -\frac{1}{4} - 3 \le -\frac{3}{4}x \le \frac{1}{4} - 3 \)
\( -\frac{1}{4} - \frac{12}{4} \le -\frac{3}{4}x \le \frac{1}{4} - \frac{12}{4} \)
\( -\frac{13}{4} \le -\frac{3}{4}x \le -\frac{11}{4} \)
Now, multiply the entire inequality by \( -\frac{4}{3} \). Remember to reverse the inequality signs when multiplying by a negative number:
\( (-\frac{13}{4})(-\frac{4}{3}) \ge x \ge (-\frac{11}{4})(-\frac{4}{3}) \)
\( \frac{13}{3} \ge x \ge \frac{11}{3} \)
Rewrite this in ascending order:
\( \frac{11}{3} \le x \le \frac{13}{3} \)
The solution set in interval notation is \( x \in [\frac{11}{3}, \frac{13}{3}] \). This interval shows all values of x between and including the two fractions.


(iv) We need to solve the inequality \( |x| - 10 < -3 \).
First, add 10 to both sides of the inequality:
\( |x| < -3 + 10 \)
\( |x| < 7 \)
When \( |A| < k \), it means \( -k < A < k \). So, we can write:
\( -7 < x < 7 \)
The solution set in interval notation is \( x \in (-7, 7) \). This represents all numbers between -7 and 7, but not including -7 or 7.
In simple words: For each part, we isolate the absolute value term and then use the rules of absolute inequalities to find the range of x. Remember to flip the inequality sign if you multiply or divide by a negative number.

🎯 Exam Tip: Always remember that multiplying or dividing an inequality by a negative number requires reversing the inequality sign. This is a common mistake in algebra.

Question 2. Solve \( \frac{1}{|2x-1|} < 6 \) and express the solution using the interval notation.
Answer:
Given the inequality \( \frac{1}{|2x-1|} < 6 \).
First, we must ensure the denominator is not zero. So, \( |2x-1| \neq 0 \), which means \( 2x-1 \neq 0 \), so \( x \neq \frac{1}{2} \).
Since \( |2x-1| \) is always positive, we can multiply both sides by \( |2x-1| \) without changing the inequality direction:
\( 1 < 6|2x-1| \)
Now, divide by 6:
\( \frac{1}{6} < |2x-1| \)
This can be written as \( |2x-1| > \frac{1}{6} \).
For an inequality of the form \( |A| > k \), the solution is \( A < -k \) or \( A > k \).
So, we have two separate inequalities to solve:
Case 1: \( 2x-1 < -\frac{1}{6} \)
\( \implies 2x < 1 - \frac{1}{6} \)
\( \implies 2x < \frac{6-1}{6} \)
\( \implies 2x < \frac{5}{6} \)
\( \implies x < \frac{5}{12} \)
Case 2: \( 2x-1 > \frac{1}{6} \)
\( \implies 2x > 1 + \frac{1}{6} \)
\( \implies 2x > \frac{6+1}{6} \)
\( \implies 2x > \frac{7}{6} \)
\( \implies x > \frac{7}{12} \)
Combining these two cases, the solution set is \( x \in (-\infty, \frac{5}{12}) \cup (\frac{7}{12}, \infty) \). This type of solution gives two separate regions on the number line where the inequality holds true.
In simple words: First, flip the fraction and the number 6 to solve it easier, making sure not to divide by zero. Then, split the problem into two parts using "less than negative" or "greater than positive" rules for absolute values. Combine these two sets of answers for the final solution.

🎯 Exam Tip: When dealing with fractional inequalities, always first consider the domain restrictions, especially to prevent division by zero, before solving the inequality.

Question 3. Solve \( -3 |x| + 5 \le -2 \) and graph the solution set in a number line.
Answer:
Given the inequality \( -3 |x| + 5 \le -2 \).
First, isolate the absolute value term by subtracting 5 from both sides:
\( -3 |x| \le -2 - 5 \)
\( -3 |x| \le -7 \)
Now, divide both sides by -3. Remember to reverse the inequality sign because you are dividing by a negative number:
\( |x| \ge \frac{-7}{-3} \)
\( |x| \ge \frac{7}{3} \)
For an inequality of the form \( |A| \ge k \), the solution is \( A \le -k \) or \( A \ge k \).
So, we have two separate inequalities:
\( x \le -\frac{7}{3} \) or \( x \ge \frac{7}{3} \)
The solution set in interval notation is \( x \in (-\infty, -\frac{7}{3}] \cup [\frac{7}{3}, \infty) \). These are the numbers that are at least \( \frac{7}{3} \) units away from zero in either direction.


In simple words: To solve this, first get the absolute value part by itself. When you divide by a negative number, flip the inequality sign. Then, remember that "greater than or equal to" for an absolute value means the answer is outside of a certain range, creating two separate solution intervals.

🎯 Exam Tip: When graphing a solution set, use solid circles for "inclusive" inequalities (≤ or ≥) and open circles for "exclusive" inequalities (< or >).

Question 4. Solve \( 2|x + 1| - 6 \le 7 \) and graph the solution set in a number line.
Answer:
Given the inequality \( 2|x + 1| - 6 \le 7 \).
First, isolate the absolute value term by adding 6 to both sides:
\( 2|x + 1| \le 7 + 6 \)
\( 2|x + 1| \le 13 \)
Now, divide both sides by 2:
\( |x + 1| \le \frac{13}{2} \)
For an inequality of the form \( |A| \le k \), the solution is \( -k \le A \le k \).
So, we can write:
\( -\frac{13}{2} \le x + 1 \le \frac{13}{2} \)
Now, subtract 1 from all parts of the inequality:
\( -\frac{13}{2} - 1 \le x \le \frac{13}{2} - 1 \)
\( -\frac{13}{2} - \frac{2}{2} \le x \le \frac{13}{2} - \frac{2}{2} \)
\( -\frac{15}{2} \le x \le \frac{11}{2} \)
The solution set in interval notation is \( x \in [-\frac{15}{2}, \frac{11}{2}] \). This represents a continuous range of numbers, including the endpoints.


In simple words: First, get the part with the absolute value alone on one side. Then, use the rule that for "less than or equal to" inequalities, the solution is between the negative and positive values of the number on the other side. Finally, solve for x in all three parts.

🎯 Exam Tip: Remember to solve for 'x' by applying operations to all three parts of the compound inequality simultaneously to maintain balance.

Question 5. Solve \( \frac{1}{5} |10x - 2| < 1 \)
Answer:
Given the inequality \( \frac{1}{5} |10x - 2| < 1 \).
First, multiply both sides by 5 to isolate the absolute value term:
\( |10x - 2| < 1 \times 5 \)
\( |10x - 2| < 5 \)
For an inequality of the form \( |A| < k \), the solution is \( -k < A < k \).
So, we can write:
\( -5 < 10x - 2 < 5 \)
Now, add 2 to all parts of the inequality:
\( -5 + 2 < 10x < 5 + 2 \)
\( -3 < 10x < 7 \)
Finally, divide all parts by 10 (a positive number, so no change in inequality signs):
\( -\frac{3}{10} < x < \frac{7}{10} \)
The solution set in interval notation is \( x \in (-\frac{3}{10}, \frac{7}{10}) \). This means x can be any number strictly between these two fractions.


In simple words: First, multiply to get the absolute value part alone. Then, for "less than" inequalities, the solution is between the negative and positive versions of the number on the other side. Solve this combined inequality to find the range for x.

🎯 Exam Tip: When an inequality has a fraction outside the absolute value, clear the fraction first to simplify the problem before applying the absolute value rules.

Question 6. Solve \( |5x - 12| < -2 \)
Answer:
Given the inequality \( |5x - 12| < -2 \).
By the definition of an absolute value (also called modulus function), the absolute value of any real number is always non-negative. This means \( |5x - 12| \) will always be greater than or equal to 0.
A value that is always greater than or equal to 0 can never be less than -2 (a negative number).
Therefore, there is no real number x that can satisfy this inequality.
The solution set is the empty set, which can be written as \( \emptyset \) or {}. This type of inequality has no possible answers.
In simple words: The absolute value of any number is always positive or zero. It can never be a negative number. So, it is impossible for something that is always positive or zero to be less than a negative number like -2. Therefore, there is no solution.

🎯 Exam Tip: Always remember the fundamental property of absolute values: they are always non-negative (\( |a| \ge 0 \)). This simple rule can often help you quickly solve or dismiss certain absolute value inequalities.