Maharashtra Board Class 9 Maths Chapter 2 Set 2.5 Algebra Standard Part 1 Real Numbers Solutions

Question 1. Find the value.
i. \( | 15 - 2 | \)
ii. \( | 4 - 9 | \)
iii. \( | 7 | \times | -4 | \)
Answer:
i. \( |15 - 2| = |13| = 13 \)
ii. \( |4 - 9| = |-5| = 5 \)
iii. \( |7| \times |-4| = 7 \times 4 = 28 \)
The absolute value of any real number represents its distance from zero on the number line and is always non-negative.
In simple words: Absolute value bars make any number inside them positive. So, even if you get a negative number like -5, it becomes positive 5.

🎯 Exam Tip: Always perform the arithmetic operations inside the absolute value bars first before converting the final result to a positive number.

Question 2. Solve.
(i) \( |3x - 5| = 1 \)
(ii) \( |7 - 2x| = 5 \)
(iii) \( \left| \frac{8 - x}{2} \right| = 5 \)
(iv) \( \left| 5 + \frac{x}{4} \right| = 5 \)
Answer:
(i) \( |3x - 5| = 1 \)
\( \implies 3x - 5 = 1 \) or \( 3x - 5 = -1 \)
\( \implies 3x = 1 + 5 \) or \( 3x = -1 + 5 \)
\( \implies 3x = 6 \) or \( 3x = 4 \)
\( \implies x = \frac{6}{3} \) or \( x = \frac{4}{3} \)
\( \implies x = 2 \) or \( x = \frac{4}{3} \)

(ii) \( |7 - 2x| = 5 \)
\( \implies 7 - 2x = 5 \) or \( 7 - 2x = -5 \)
\( \implies 7 - 5 = 2x \) or \( 7 + 5 = 2x \)
\( \implies 2x = 2 \) or \( 2x = 12 \)
\( \implies x = \frac{2}{2} \) or \( x = \frac{12}{2} \)
\( \implies x = 1 \) or \( x = 6 \)

(iii) \( \left| \frac{8 - x}{2} \right| = 5 \)
\( \implies \frac{8 - x}{2} = 5 \) or \( \frac{8 - x}{2} = -5 \)
\( \implies 8 - x = 10 \) or \( 8 - x = -10 \) [Multiplying both sides by 2]
\( \implies 8 - 10 = x \) or \( 8 + 10 = x \)
\( \implies x = -2 \) or \( x = 18 \)

(iv) \( \left| 5 + \frac{x}{4} \right| = 5 \)
\( \implies 5 + \frac{x}{4} = 5 \) or \( 5 + \frac{x}{4} = -5 \)
\( \implies \frac{x}{4} = 5 - 5 \) or \( \frac{x}{4} = -5 - 5 \)
\( \implies \frac{x}{4} = 0 \) or \( \frac{x}{4} = -10 \)
\( \implies x = 0 \) or \( x = -40 \)
In simple words: To solve absolute value equations, we split them into two separate equations: one where the expression inside equals the positive value, and one where it equals the negative value. Then, we solve each equation individually to find the two possible answers.

🎯 Exam Tip: Always remember that \( |x| = a \) (where \( a \ge 0 \)) splits into two cases: \( x = a \) or \( x = -a \). Don't forget to solve both cases to get full marks!

Question 2. Solve:
(iv) \( \left| 5 + \frac{x}{4} \right| = 5 \)
Answer:
\( \left| 5 + \frac{x}{4} \right| = 5 \)
\( \therefore 5 + \frac{x}{4} = 5 \) or \( 5 + \frac{x}{4} = -5 \)
\( \therefore \frac{x}{4} = 5 - 5 \) or \( \frac{x}{4} = -5 - 5 \)
\( \therefore \frac{x}{4} = 0 \) or \( \frac{x}{4} = -10 \)
\( \therefore x = 0 \) or \( x = -40 \) [Multiplying both sides by 4]
In simple words: Since the absolute value of a number is its distance from zero, the expression inside the bars can be either positive 5 or negative 5. We solve both cases separately to find the two possible values for x.

🎯 Exam Tip: Always split absolute value equations into two separate equations (one positive, one negative) right from the first step to avoid losing half of your marks.