Chapter 7 · Practice set 2Textbook page 132

Practice Set 2 – Animated Solutions ചിന്തിക്കാം

Three questions that use the new subtraction rules: x − (−y) = x + y and x − y = −(y − x). The last one is a little puzzle you can prove with algebra.

📖 Tools you will use

x + (−y) = x − y and x − (−y) = x + y.
x − y = −(y − x) for any numbers x, y.
Consecutive numbers: if the first is a, the next ones are a + 1, a + 2, a + 3.

Question 1 · x − (y − z) and (x − y) + z

Q1. Take several positive and negative numbers as x, y, z and compute x − (y − z) and (x − y) + z. Are they always equal?

x, y, z ഇവ പല അധിസംഖ്യകളായും ന്യൂനസംഖ്യകളായും എടുത്ത്, x − (y − z) ഉം (x − y) + z ഉം കണക്കാക്കുക. രണ്ടും ഒരേ സംഖ്യയാണോ?

Work out the bracket first. Remember: subtracting a negative number is adding.

🎬 Solution animation

🎚️ Try any numbers

x y z

Answer: always equal. For example x = 5, y = −3, z = 2: x − (y − z) = 5 − (−5) = 10 and (x − y) + z = 8 + 2 = 10. Removing a bracket with a minus in front changes the sign of everything inside: x − (y − z) = x − y + z.

Question 2 · Pairs with x − y = 1 and x − y = −1

Q2. Find two pairs x, y of each kind: (i) x positive, y negative, x − y = 1 (ii) x negative, y positive, x − y = −1.

If y is negative, x − y = x + (size of y). Two positive numbers add to 1 – how big can each be?

🎬 Solution animation

Answer: (i) x = ½, y = −½ (½ − (−½) = ½ + ½ = 1) and x = 0.3, y = −0.7. (ii) x = −½, y = ½ (−½ − ½ = −1) and x = −0.2, y = 0.8. No pairs of whole numbers work: x − y is the sum of two positive sizes that make 1, so both must be fractions.

💡 Why only fractions

With y negative, x − y = x + |y|: a positive number plus another positive number. If both were whole numbers the smallest possible sum would be 1 + 1 = 2, already more than 1. So both parts must be fractions between 0 and 1.

Question 3 · The a − b − c + d puzzle

Q3. (i) Take a, b, c, d as four consecutive natural numbers, or their negatives, and compute a − b − c + d. (ii) Explain with algebra why it is always 0. (iii) What about a + b − c − d? (iv) And a − b + c − d?

Write b = a + 1, c = a + 2, d = a + 3 and simplify.

🎬 Solution animation – the algebra

🎚️ Choose the first number

a use the negatives (−a, −a−1, …)

Answer: (i) 1 − 2 − 3 + 4 = 0, 5 − 6 − 7 + 8 = 0, −1 − (−2) − (−3) + (−4) = 0. (ii) a − (a + 1) − (a + 2) + (a + 3) = 0: the a's cancel and −1 − 2 + 3 = 0. (iii) a + b − c − d = −1 + … = −4 for naturals (4 for their negatives). (iv) a − b + c − d = −2 (2 for negatives).

🌍 Why this matters

A pattern seen in a few examples is only a guess; the algebra with a, a + 1, a + 2, a + 3 proves it for every starting number at once – including negatives.

Check yourself

🧭 Summary

x − (y − z) = (x − y) + z = x − y + z for all numbers.
x positive, y negative, x − y = 1 needs fractions (½ and −½).
For four consecutive numbers: a − b − c + d = 0, a + b − c − d = −4, a − b + c − d = −2 (signs flip for negatives).

Next: Speed: multiplication and division of negative numbers ▶