Exercise 3.2 answers: Pair of Linear Equations in Two Variables

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(i) x + y = 14

1. From x − y = 4, x = y + 4.
2. Put into x + y = 14: (y + 4) + y = 14, so 2y = 10, y = 5.
3. Then x = 5 + 4 = 9.

Answerx = 9, y = 5.

(ii) s − t = 3

1. From s − t = 3, s = t + 3.
2. Clear the fractions in s/3 + t/2 = 6 by multiplying by 6: 2s + 3t = 36.
3. Put s = t + 3 in: 2(t + 3) + 3t = 36, so 5t = 30, t = 6.
4. Then s = 6 + 3 = 9.

Answers = 9, t = 6.

(iii) 3x − y = 3

1. From 3x − y = 3, y = 3x − 3.
2. Put into 9x − 3y = 9: 9x − 3(3x − 3) = 9, so 9x − 9x + 9 = 9, which gives 9 = 9.
3. Every letter has cancelled, and what is left is true — the second equation is just the first one multiplied by 3.

AnswerThere is no single pair — every point on the line 3x − y = 3 is a solution (infinitely many solutions).

(iv) 0.2x + 0.3y = 1.3

1. From 0.2x + 0.3y = 1.3, x = (1.3 − 0.3y)/0.2 = 6.5 − 1.5y.
2. Put into 0.4x + 0.5y = 2.3: 0.4(6.5 − 1.5y) + 0.5y = 2.3, so 2.6 − 0.6y + 0.5y = 2.3, giving −0.1y = −0.3, y = 3.
3. Then x = 6.5 − 1.5(3) = 6.5 − 4.5 = 2.

Answerx = 2, y = 3.

(v) √2 x + √3 y = 0

1. From √2 x + √3 y = 0, x = −(√3/√2) y.
2. Put into √3 x − √8 y = 0: √3 × (−(√3/√2) y) − √8 y = 0, so −(3/√2) y − √8 y = 0.
3. Both terms have the same sign and y multiplies a non-zero number, so y = 0.
4. Then x = 0 as well.

Answerx = 0, y = 0.

(vi) 3x/2 − 5y/3 = −2

1. From x/3 + y/2 = 13/6, multiply by 6: 2x + 3y = 13, so x = (13 − 3y)/2.
2. Put into (3x/2) − (5y/3) = −2: multiply everything by 6 first to clear fractions: 9x − 10y = −12.
3. Substitute x = (13 − 3y)/2: 9(13 − 3y)/2 − 10y = −12. Multiply by 2: 9(13 − 3y) − 20y = −24, so 117 − 27y − 20y = −24, giving −47y = −141, y = 3.
4. Then x = (13 − 9)/2 = 2.

Answerx = 2, y = 3.

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1. Both equations already have the same x-term (2x), so subtract the second from the first: (2x + 3y) − (2x − 4y) = 11 − (−24), giving 7y = 35, so y = 5.
2. Put y = 5 into 2x + 3y = 11: 2x + 15 = 11, so 2x = −4, x = −2.
3. Now use y = mx + 3 with x = −2, y = 5: 5 = m(−2) + 3, so 2 = −2m, m = −1.

Answerx = −2, y = 5, and m = −1.

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(i) difference between two numbers is 26 and one number is three times

1. Let the larger number be x and the smaller be y.
2. The difference is 26: x − y = 26.
3. One number is three times the other: x = 3y.
4. Substitute x = 3y into x − y = 26: 3y − y = 26, so 2y = 26, y = 13.
5. Then x = 3 × 13 = 39.
6. (If negative numbers were allowed, −13 and −39 would also fit, since −13 − (−39) = 26 and −39 = 3 × (−13). The answer expected here uses positive numbers.)

AnswerThe two numbers are 39 and 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees.

1. Let the larger angle be x° and the smaller be y°.
2. Supplementary angles add to 180°: x + y = 180.
3. The larger exceeds the smaller by 18°: x − y = 18, so x = y + 18.
4. Substitute into x + y = 180: (y + 18) + y = 180, so 2y = 162, y = 81.
5. Then x = 81 + 18 = 99.

AnswerThe larger angle is 99° and the smaller is 81°.

(iii) The coach of a cricket team buys 7 bats and 6 balls

1. Let one bat cost ₹x and one ball cost ₹y.
2. 7 bats and 6 balls cost ₹3800: 7x + 6y = 3800.
3. 3 bats and 5 balls cost ₹1750: 3x + 5y = 1750, so x = (1750 − 5y)/3.
4. Substitute into 7x + 6y = 3800: 7(1750 − 5y)/3 + 6y = 3800. Multiply by 3: 12250 − 35y + 18y = 11400, so 17y = 850, y = 50.
5. Then x = (1750 − 5 × 50)/3 = 1500/3 = 500.

AnswerOne bat costs ₹500 and one ball costs ₹50.

(iv) The taxi charges in a city consist of a fixed charge

1. Let the fixed charge be ₹x and the charge per km be ₹y.
2. For 10 km: x + 10y = 105, so x = 105 − 10y.
3. For 15 km: x + 15y = 155.
4. Substitute: (105 − 10y) + 15y = 155, so 5y = 50, y = 10.
5. Then x = 105 − 10 × 10 = 5.
6. For 25 km: x + 25y = 5 + 25 × 10 = 255.

AnswerThe fixed charge is ₹5, the charge per km is ₹10, and a 25 km journey costs ₹255.

(v) A fraction becomes 9/11, if 2 is added to both …

1. Let the fraction be x/y.
2. (x + 2)/(y + 2) = 9/11 gives 11(x + 2) = 9(y + 2), that is 11x − 9y = −4.
3. (x + 3)/(y + 3) = 5/6 gives 6(x + 3) = 5(y + 3), that is 6x − 5y = −3, so y = (6x + 3)/5.
4. Substitute into 11x − 9y = −4: 11x − 9(6x + 3)/5 = −4. Multiply by 5: 55x − 54x − 27 = −20, so x = 7.
5. Then y = (6 × 7 + 3)/5 = 45/5 = 9.

AnswerThe fraction is 7/9.

(vi) Five years hence, the age of Jacob will be three times

1. Let Jacob's present age be x years and his son's be y years.
2. Five years hence: x + 5 = 3(y + 5), so x = 3y + 10.
3. Five years ago: x − 5 = 7(y − 5), so x − 7y = −30.
4. Substitute x = 3y + 10: 3y + 10 − 7y = −30, so −4y = −40, y = 10.
5. Then x = 3 × 10 + 10 = 40.

AnswerJacob is 40 years old and his son is 10 years old.

Watch this explained “Two sentences about ages”, 7:41 into Substitution: rewriting one unknown so only the other survives