Chapter 7 · Practice sets 3 & 4Textbook pages 137–140

Practice Sets 3 & 4 – Animated Solutions ചിന്തിക്കാം

Ten questions that put the multiplication and division rules to work: checking the distributive law, putting negative values into expressions, finding integer points, and dividing with negative numbers. Try each in your notebook first, use the hint if you are stuck, then watch the animation.

📖 Tools you will use (pages 5, 7 and 9 of this guide)

Practice set 3 · Question 1 – checking the distributive law

Q1Take several positive and negative numbers as x, y, z and compute (x + y)z and xz + yz. Is (x + y)z = xz + yz true in every case?

x, y, z ആയി പല അധിസംഖ്യകളും ന്യൂനസംഖ്യകളും എടുത്ത്, (x + y)z ഉം xz + yz ഉം കണക്കാക്കുക. എല്ലാറ്റിലും (x + y)z = xz + yz എന്ന സമവാക്യം ശരിയാകുന്നുണ്ടോ എന്നു പരിശോധിക്കുക.

Compute each side separately, carefully with signs. Try a few mixes, e.g. x = 3, y = −5, z = −2.

🎬 Solution animation
🧪 Try your own x, y, z
Answer: Yes, in every case. For example x = 3, y = −5, z = −2: (3 + (−5)) × (−2) = (−2) × (−2) = 4 and 3 × (−2) + (−5) × (−2) = −6 + 10 = 4. The multiplication rules were chosen precisely so that this law keeps working.

Question 2 – (x + y)(u + v) with negatives

Q2Prove that the identity (x + y)(u + v) = xu + xv + yu + yv is also true when −x, −y, −u, −v are put in place of x, y, u, v.

−x + (−y) = −(x + y). Then use (−a)(−b) = ab on both sides.

🎬 Solution animation
Answer: Left side: (−x + (−y))(−u + (−v)) = (−(x + y))(−(u + v)) = (x + y)(u + v). Right side: (−x)(−u) + (−x)(−v) + (−y)(−u) + (−y)(−v) = xu + xv + yu + yv. Both equal the original, so the identity still holds.

Question 3 – putting negative values into equations

Q3Find y for the given x: (i) y = x², x = −1 (ii) y = x² + 3x + 2, x = −1 (iii) y = x² + 3x + 2, x = −2 (iv) y = (x + 1)(x + 2), x = −1 (v) y = (x + 1)(x + 2), x = −2.

(−1)² = 1, (−2)² = 4, 3 × (−1) = −3, 3 × (−2) = −6. A product with a zero factor is 0.

🎬 Solution animation
Answer: (i) 1 (ii) 1 − 3 + 2 = 0 (iii) 4 − 6 + 2 = 0 (iv) 0 × 1 = 0 (v) (−1) × 0 = 0. Parts (ii)–(v) agree because x² + 3x + 2 = (x + 1)(x + 2) for all numbers.

Question 4 – why x² + 4x + 4 is never negative

Q4In y = x² + 4x + 4 take many positive and negative values for x and calculate y. Why is y always positive or zero, whatever x is?

Compare with (x + 2)² = x² + 4x + 4.

🎬 Solution animation
🎚️ Slide x
Answer: x = −5, −4, −3, −2, −1, 0, 1, 2 give y = 9, 4, 1, 0, 1, 4, 9, 16. Since x² + 4x + 4 = (x + 2)², y is a square, and the square of any number – positive or negative – is positive (or 0 when x = −2).

Question 5 – integer points

Q5Natural numbers, their negatives and zero together are called integers. (i) How many pairs of integers satisfy x² + y² = 25? (ii) How many satisfy x² − y² = 25?

Which squares add to 25? (0 + 25, 9 + 16.) Each number can be positive or negative. For (ii) write x² − y² = (x − y)(x + y) = 25.

🎬 Solution animation
🔎 Which equation?

0 = x² + y² = 25 (12 points), 1 = x² − y² = 25 (6 points). Hover the coloured points for their coordinates.

Answer: (i) 12 pairs: (±5, 0), (0, ±5), (±3, ±4), (±4, ±3). (ii) 6 pairs: (±5, 0) and (±13, ±12), since (x − y)(x + y) = 25 = 1 × 25 = 5 × 5 = 25 × 1 and the same with negatives.

Questions 6 and 7 – a long product

Q61 × 2 × 3 × 4 × 5 = 120. What is (−1) × (−2) × (−3) × (−4) × (−5)?

Q7What is (1 × 2 × 3 × 4 × 5) + [(−1) × (−2) × (−3) × (−4) × (−5)]?

Multiply two at a time and watch the sign flip with each new negative factor.

🎬 Solution animation
Answer: Q6: −120 (five negative factors – an odd number – make the product negative). Q7: 120 + (−120) = 0.

Practice set 4 · Question 1 – y = 1/x

Q1In the equation y = 1/x, find y when x is 2, −2, ½, −½.

y = 1/x എന്ന സമവാക്യത്തിൽ, x ആയി 2, −2, ½, −½ എന്നീ സംഖ്യകളെടുത്താൽ y ആയി കിട്ടുന്ന സംഖ്യകൾ കണക്കാക്കുക.

1 ÷ ½ = 2. A positive divided by a negative is negative.

🎬 Solution animation
Answer: x = 2 ⇒ y = ½; x = −2 ⇒ y = −½; x = ½ ⇒ y = 2; x = −½ ⇒ y = −2.

Question 2 – y = 1/(x − 1) + 1/(x + 1)

Q2In y = 1/(x − 1) + 1/(x + 1), find y when x = −2 and when x = −½.

For x = −2: x − 1 = −3 and x + 1 = −1. For x = −½: x − 1 = −1½ = −3/2 and x + 1 = ½.

🎬 Solution animation
Answer: x = −2: 1/(−3) + 1/(−1) = −⅓ − 1 = −4/3. x = −½: 1/(−3/2) + 1/(½) = −⅔ + 2 = 4/3.

Question 3 – z = x/y − y/x

Q3In z = x/y − y/x, find z for (i) x = 10, y = −5 (ii) x = −10, y = 5 (iii) x = −10, y = −5.

Find x/y and y/x first, with signs; then subtract.

🎬 Solution animation
Answer: (i) −2 − (−½) = −3/2 (ii) −2 − (−½) = −3/2 (iii) 2 − ½ = 3/2.
🌍 Why these matter

Putting negative values into formulas is everyday work for scientists and engineers: temperatures below zero in a heat formula, a velocity downward in a motion formula, a loss in a profit formula. The rules of this chapter make every formula work for every value.

Check yourself

Next: Real-world applications ▶