CBSE Class 10 Pair of Linear Equations in Two Variables Sure Shot Questions Set 02

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PRACTICE QUESTIONS
CLASS X : CHAPTER - 3
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
CONDITIONS FOR SOLVING LINEAR EQUATIONS

1. Find the value of k, so that the following system of equations has no solution:
3x - y - 5 = 0; 6x - 2y - k = 0.

Answer:
1. Write the equations as 3x - y - 5 = 0 and 6x - 2y - k = 0, so a1 = 3, b1 = -1, c1 = -5 and a2 = 6, b2 = -2, c2 = -k.
2. For no solution, \( \frac{3}{6} = \frac{-1}{-2} \neq \frac{-5}{-k} \).
3. Here \( \frac{3}{6} = \frac{1}{2} \) and \( \frac{-1}{-2} = \frac{1}{2} \), so the first two ratios are already equal.
4. The third ratio must be different: \( \frac{5}{k} \) ≠ \( \frac{1}{2} \), so k ≠ 10.
5. Hence, the system has no solution for every real value of k except k = 10 (for k = 10 the lines coincide).

Teacher's Note:
a) For no solution, the lines must be parallel: a1/a2 = b1/b2 but not equal to c1/c2.
b) After finding k, always test the third ratio; if it is also equal the lines coincide.

2. Find the value of k, so that the following system of equations has a non-zero solution:
3x + 5y = 0; kx + 10y = 0.

Answer:
1. The system is homogeneous, so x = 0, y = 0 is always a solution.
2. A non-zero solution exists only when the system has infinitely many solutions, that is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \).
3. \( \frac{3}{k} = \frac{5}{10} \), so 5k = 30.
4. k = 6
5. Hence, k = 6. For this value the second equation is 6x + 10y = 0, which is twice the first equation.

Teacher's Note:
a) A homogeneous system has a non-zero solution only when its two lines coincide.
b) Here both equations pass through the origin, so the ratio c1/c2 is automatically equal to the others.

Find the value of k, so that the following system of equations has no solution:

3. 3x + y = 1; (2k - 1)x + (k - 1)y = (2k - 1).

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 3, b1 = 1, c1 = -1 and a2 = 2k - 1, b2 = k - 1, c2 = 1 - 2k.
2. For no solution, \( \frac{3}{2 k - 1} = \frac{1}{k - 1} \) and \( \frac{1}{k - 1} \neq \frac{-1}{1 - 2 k} \).
3. From \( \frac{3}{2 k - 1} = \frac{1}{k - 1} \), a1b2 = a2b1, so 3k - 3 = 2k - 1, which gives k = 2.
4. For k = 2, \( \frac{b_1}{b_2} = \frac{1}{1} \) = 1 and \( \frac{c_1}{c_2} = \frac{-1}{-3} \) = \( \frac{1}{3} \). They are not equal, so this value is valid.
5. Hence, k = 2.

Teacher's Note:
a) For no solution, the lines must be parallel: a1/a2 = b1/b2 but not equal to c1/c2.
b) After finding k, always test the third ratio; if it is also equal the lines coincide.

4. 3x + y = 1; (2k - 1)x + (k - 1)y = (2k + 1).

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 3, b1 = 1, c1 = -1 and a2 = 2k - 1, b2 = k - 1, c2 = -2k - 1.
2. For no solution, \( \frac{3}{2 k - 1} = \frac{1}{k - 1} \) and \( \frac{1}{k - 1} \neq \frac{-1}{- 2 k - 1} \).
3. From \( \frac{3}{2 k - 1} = \frac{1}{k - 1} \), a1b2 = a2b1, so 3k - 3 = 2k - 1, which gives k = 2.
4. For k = 2, \( \frac{b_1}{b_2} = \frac{1}{1} \) = 1 and \( \frac{c_1}{c_2} = \frac{-1}{-5} \) = \( \frac{1}{5} \). They are not equal, so this value is valid.
5. Hence, k = 2.

Teacher's Note:
a) Write both equations as ax + by + c = 0 before taking ratios; the constant must go to the left side.
b) Parallel lines give no solution; the third ratio c1/c2 must be different.
c) Reject any value of k that makes all three ratios equal.

5. x - 2y = 3; 3x + ky = 1.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = -2, c1 = -3 and a2 = 3, b2 = k, c2 = -1.
2. For no solution, \( \frac{1}{3} = \frac{-2}{k} \) and \( \frac{-2}{k} \neq \frac{-3}{-1} \).
3. From \( \frac{1}{3} = \frac{-2}{k} \), a1b2 = a2b1, so k = -6, which gives k = -6.
4. For k = -6, \( \frac{b_1}{b_2} = \frac{-2}{-6} \) = \( \frac{1}{3} \) and \( \frac{c_1}{c_2} = \frac{-3}{-1} \) = 3. They are not equal, so this value is valid.
5. Hence, k = -6.

Teacher's Note:
a) For no solution, the lines must be parallel: a1/a2 = b1/b2 but not equal to c1/c2.
b) After finding k, always test the third ratio; if it is also equal the lines coincide.

6. x + 2y = 5; 3x + ky + 15 = 0.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = 2, c1 = -5 and a2 = 3, b2 = k, c2 = 15.
2. For no solution, \( \frac{1}{3} = \frac{2}{k} \) and \( \frac{2}{k} \neq \frac{-5}{15} \).
3. From \( \frac{1}{3} = \frac{2}{k} \), a1b2 = a2b1, so k = 6, which gives k = 6.
4. For k = 6, \( \frac{b_1}{b_2} = \frac{2}{6} \) = \( \frac{1}{3} \) and \( \frac{c_1}{c_2} = \frac{-5}{15} \) = \( - \frac{1}{3} \). They are not equal, so this value is valid.
5. Hence, k = 6.

Teacher's Note:
a) Write both equations as ax + by + c = 0 before taking ratios; the constant must go to the left side.
b) Parallel lines give no solution; the third ratio c1/c2 must be different.
c) Reject any value of k that makes all three ratios equal.

7. kx + 2y = 5; 3x - 4y = 10.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = k, b1 = 2, c1 = -5 and a2 = 3, b2 = -4, c2 = -10.
2. For no solution, \( \frac{k}{3} = \frac{2}{-4} \) and \( \frac{2}{-4} \neq \frac{-5}{-10} \).
3. From \( \frac{k}{3} = \frac{2}{-4} \), a1b2 = a2b1, so -4k = 6, which gives k = -3/2.
4. For k = -3/2, \( \frac{b_1}{b_2} = \frac{2}{-4} \) = \( - \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-5}{-10} \) = \( \frac{1}{2} \). They are not equal, so this value is valid.
5. Hence, k = -3/2.

Teacher's Note:
a) For no solution, the lines must be parallel: a1/a2 = b1/b2 but not equal to c1/c2.
b) After finding k, always test the third ratio; if it is also equal the lines coincide.

8. x + 2y = 3; 5x + ky + 7 = 0.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = 2, c1 = -3 and a2 = 5, b2 = k, c2 = 7.
2. For no solution, \( \frac{1}{5} = \frac{2}{k} \) and \( \frac{2}{k} \neq \frac{-3}{7} \).
3. From \( \frac{1}{5} = \frac{2}{k} \), a1b2 = a2b1, so k = 10, which gives k = 10.
4. For k = 10, \( \frac{b_1}{b_2} = \frac{2}{10} \) = \( \frac{1}{5} \) and \( \frac{c_1}{c_2} = \frac{-3}{7} \) = \( - \frac{3}{7} \). They are not equal, so this value is valid.
5. Hence, k = 10.

Teacher's Note:
a) Write both equations as ax + by + c = 0 before taking ratios; the constant must go to the left side.
b) Parallel lines give no solution; the third ratio c1/c2 must be different.
c) Reject any value of k that makes all three ratios equal.

9. 8x + 5y = 9; kx + 10y = 15.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 8, b1 = 5, c1 = -9 and a2 = k, b2 = 10, c2 = -15.
2. For no solution, \( \frac{8}{k} = \frac{5}{10} \) and \( \frac{5}{10} \neq \frac{-9}{-15} \).
3. From \( \frac{8}{k} = \frac{5}{10} \), a1b2 = a2b1, so 80 = 5k, which gives k = 16.
4. For k = 16, \( \frac{b_1}{b_2} = \frac{5}{10} \) = \( \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-9}{-15} \) = \( \frac{3}{5} \). They are not equal, so this value is valid.
5. Hence, k = 16.

Teacher's Note:
a) For no solution, the lines must be parallel: a1/a2 = b1/b2 but not equal to c1/c2.
b) After finding k, always test the third ratio; if it is also equal the lines coincide.

10. (3k + 1)x + 3y - 2 = 0; (k² + 1)x + (k - 2)y - 5 = 0.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 3k + 1, b1 = 3, c1 = -2 and a2 = k² + 1, b2 = k - 2, c2 = -5.
2. For no solution, \( \frac{3 k + 1}{k^{2} + 1} = \frac{3}{k - 2} \) and \( \frac{3}{k - 2} \neq \frac{-2}{-5} \).
3. From \( \frac{3 k + 1}{k^{2} + 1} = \frac{3}{k - 2} \), a1b2 = a2b1, so 3k² - 5k - 2 = 3k² + 3, which gives k = -1.
4. For k = -1, \( \frac{b_1}{b_2} = \frac{3}{-3} \) = -1 and \( \frac{c_1}{c_2} = \frac{-2}{-5} \) = \( \frac{2}{5} \). They are not equal, so this value is valid.
5. Hence, k = -1.

Teacher's Note:
a) Write both equations as ax + by + c = 0 before taking ratios; the constant must go to the left side.
b) Parallel lines give no solution; the third ratio c1/c2 must be different.
c) Reject any value of k that makes all three ratios equal.

11. kx + 3y = 3; 12x + ky = 6.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = k, b1 = 3, c1 = -3 and a2 = 12, b2 = k, c2 = -6.
2. For no solution, \( \frac{k}{12} = \frac{3}{k} \) and \( \frac{3}{k} \neq \frac{-3}{-6} \).
3. From \( \frac{k}{12} = \frac{3}{k} \), a1b2 = a2b1, so k² = 36, which gives k² - 36 = 0, so k = -6 or k = 6.
4. For k = -6, \( \frac{b_1}{b_2} = \frac{3}{-6} \) = \( - \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-3}{-6} \) = \( \frac{1}{2} \). They are not equal, so this value is valid.
5. For k = 6, \( \frac{b_1}{b_2} = \frac{3}{6} \) = \( \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-3}{-6} \) = \( \frac{1}{2} \). They are equal, so the lines coincide and this value is rejected.
6. Hence, k = -6.

Teacher's Note:
a) For no solution, the lines must be parallel: a1/a2 = b1/b2 but not equal to c1/c2.
b) After finding k, always test the third ratio; if it is also equal the lines coincide.

Find the value of k, so that the following system of equations has a unique solution:

12. x - 2y = 3; 3x + ky = 1.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = -2, c1 = -3 and a2 = 3, b2 = k, c2 = -1.
2. For a unique solution, \( \frac{1}{3} \) ≠ \( \frac{-2}{k} \).
3. Cross-multiplying, a1b2 ≠ a2b1, so k ≠ -6.
4. This gives k + 6 ≠ 0, so k ≠ -6.
5. Hence, the system has a unique solution for every real value of k except k = -6.

Teacher's Note:
a) Cross-multiply the ratio a1/a2 and b1/b2 and keep the inequality sign.
b) State clearly that every other real value of k gives a unique solution.
c) Do not use the c1/c2 ratio here; it is not needed.

13. x + 2y = 5; 3x + ky + 15 = 0.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = 2, c1 = -5 and a2 = 3, b2 = k, c2 = 15.
2. For a unique solution, \( \frac{1}{3} \) ≠ \( \frac{2}{k} \).
3. Cross-multiplying, a1b2 ≠ a2b1, so k ≠ 6.
4. This gives k - 6 ≠ 0, so k ≠ 6.
5. Hence, the system has a unique solution for every real value of k except k = 6.

Teacher's Note:
a) Unique solution means the lines intersect, so a1/a2 is not equal to b1/b2.
b) The answer is a "not equal to" condition, not a single value of k.

14. kx + 2y = 5; 3x - 4y = 10.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = k, b1 = 2, c1 = -5 and a2 = 3, b2 = -4, c2 = -10.
2. For a unique solution, \( \frac{k}{3} \) ≠ \( \frac{2}{-4} \).
3. Cross-multiplying, a1b2 ≠ a2b1, so -4k ≠ 6.
4. This gives -4k - 6 ≠ 0, so k ≠ -3/2.
5. Hence, the system has a unique solution for every real value of k except k = -3/2.

Teacher's Note:
a) Cross-multiply the ratio a1/a2 and b1/b2 and keep the inequality sign.
b) State clearly that every other real value of k gives a unique solution.
c) Do not use the c1/c2 ratio here; it is not needed.

15. x + 2y = 3; 5x + ky + 7 = 0.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = 2, c1 = -3 and a2 = 5, b2 = k, c2 = 7.
2. For a unique solution, \( \frac{1}{5} \) ≠ \( \frac{2}{k} \).
3. Cross-multiplying, a1b2 ≠ a2b1, so k ≠ 10.
4. This gives k - 10 ≠ 0, so k ≠ 10.
5. Hence, the system has a unique solution for every real value of k except k = 10.

Teacher's Note:
a) Unique solution means the lines intersect, so a1/a2 is not equal to b1/b2.
b) The answer is a "not equal to" condition, not a single value of k.

16. 8x + 5y = 9; kx + 10y = 15.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 8, b1 = 5, c1 = -9 and a2 = k, b2 = 10, c2 = -15.
2. For a unique solution, \( \frac{8}{k} \) ≠ \( \frac{5}{10} \).
3. Cross-multiplying, a1b2 ≠ a2b1, so 80 ≠ 5k.
4. This gives 80 - 5k ≠ 0, so k ≠ 16.
5. Hence, the system has a unique solution for every real value of k except k = 16.

Teacher's Note:
a) Cross-multiply the ratio a1/a2 and b1/b2 and keep the inequality sign.
b) State clearly that every other real value of k gives a unique solution.
c) Do not use the c1/c2 ratio here; it is not needed.

17. kx + 3y = (k - 3); 12x + ky = k.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = k, b1 = 3, c1 = 3 - k and a2 = 12, b2 = k, c2 = -k.
2. For a unique solution, \( \frac{k}{12} \) ≠ \( \frac{3}{k} \).
3. Cross-multiplying, a1b2 ≠ a2b1, so k² ≠ 36.
4. This gives k² - 36 ≠ 0, so k ≠ -6 and k ≠ 6.
5. Hence, the system has a unique solution for every real value of k except k = -6 and k = 6.

Teacher's Note:
a) Unique solution means the lines intersect, so a1/a2 is not equal to b1/b2.
b) The answer is a "not equal to" condition, not a single value of k.

18. kx + 2y = 5; 3x + y = 1.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = k, b1 = 2, c1 = -5 and a2 = 3, b2 = 1, c2 = -1.
2. For a unique solution, \( \frac{k}{3} \) ≠ \( \frac{2}{1} \).
3. Cross-multiplying, a1b2 ≠ a2b1, so k ≠ 6.
4. This gives k - 6 ≠ 0, so k ≠ 6.
5. Hence, the system has a unique solution for every real value of k except k = 6.

Teacher's Note:
a) Cross-multiply the ratio a1/a2 and b1/b2 and keep the inequality sign.
b) State clearly that every other real value of k gives a unique solution.
c) Do not use the c1/c2 ratio here; it is not needed.

19. x - 2y = 3; 3x + ky = 1.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = -2, c1 = -3 and a2 = 3, b2 = k, c2 = -1.
2. For a unique solution, \( \frac{1}{3} \) ≠ \( \frac{-2}{k} \).
3. Cross-multiplying, a1b2 ≠ a2b1, so k ≠ -6.
4. This gives k + 6 ≠ 0, so k ≠ -6.
5. Hence, the system has a unique solution for every real value of k except k = -6.

Teacher's Note:
a) Unique solution means the lines intersect, so a1/a2 is not equal to b1/b2.
b) The answer is a "not equal to" condition, not a single value of k.

20. 4x - 5y = k; 2x - 3y = 12.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 4, b1 = -5, c1 = -k and a2 = 2, b2 = -3, c2 = -12.
2. For a unique solution, \( \frac{4}{2} \) ≠ \( \frac{-5}{-3} \).
3. Here a1b2 = -12 and a2b1 = -10, which are not equal, so a1b2 - a2b1 = -2, which is never zero.
4. Hence, the system has a unique solution for every real value of k.

Teacher's Note:
a) Cross-multiply the ratio a1/a2 and b1/b2 and keep the inequality sign.
b) State clearly that every other real value of k gives a unique solution.
c) Do not use the c1/c2 ratio here; it is not needed.

For what value of k, the following pair of linear equations has infinite number of solutions:

21. kx + 3y = (2k + 1); 2(k + 1)x + 9y = (7k + 1).

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = k, b1 = 3, c1 = -2k - 1 and a2 = 2k + 2, b2 = 9, c2 = -7k - 1.
2. For infinite solutions, \( \frac{k}{2 k + 2} = \frac{3}{9} = \frac{- 2 k - 1}{- 7 k - 1} \).
3. From \( \frac{k}{2 k + 2} = \frac{3}{9} \): 9k = 6k + 6, so 3k - 6 = 0, giving k = 2.
4. From \( \frac{3}{9} = \frac{- 2 k - 1}{- 7 k - 1} \): -21k - 3 = -18k - 9, so 6 - 3k = 0, giving k = 2.
5. The value common to both results is k = 2, which satisfies all three ratios.
6. Hence, k = 2.

Teacher's Note:
a) Infinite solutions mean coincident lines: a1/a2 = b1/b2 = c1/c2.
b) Solve two of the ratio equations and check the answer in the third.

22. 2x + 3y = 2; (k + 2)x + (2k + 1)y = 2(k - 1).

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 2, b1 = 3, c1 = -2 and a2 = k + 2, b2 = 2k + 1, c2 = 2 - 2k.
2. For infinite solutions, \( \frac{2}{k + 2} = \frac{3}{2 k + 1} = \frac{-2}{2 - 2 k} \).
3. From \( \frac{2}{k + 2} = \frac{3}{2 k + 1} \): 4k + 2 = 3k + 6, so k - 4 = 0, giving k = 4.
4. From \( \frac{3}{2 k + 1} = \frac{-2}{2 - 2 k} \): 6 - 6k = -4k - 2, so 8 - 2k = 0, giving k = 4.
5. The value common to both results is k = 4, which satisfies all three ratios.
6. Hence, k = 4.

Teacher's Note:
a) Take the three ratios in pairs and solve for k from each pair.
b) The value of k that satisfies all the pairs is the answer; reject the rest.
c) Verify by putting k back into the original equations: they become multiples of each other.

23. x + (2k - 1)y = 4; kx + 6y = k + 6.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = 2k - 1, c1 = -4 and a2 = k, b2 = 6, c2 = -k - 6.
2. For infinite solutions, \( \frac{1}{k} = \frac{2 k - 1}{6} = \frac{-4}{- k - 6} \).
3. From \( \frac{1}{k} = \frac{2 k - 1}{6} \): 6 = k(2k - 1), so -2k² + k + 6 = 0, giving k = -3/2, 2.
4. From \( \frac{2 k - 1}{6} = \frac{-4}{- k - 6} \): -2k² - 11k + 6 = -24, so -2k² - 11k + 30 = 0, giving k = -15/2, 2.
5. The value common to both results is k = 2, which satisfies all three ratios.
6. Hence, k = 2.

Teacher's Note:
a) Infinite solutions mean coincident lines: a1/a2 = b1/b2 = c1/c2.
b) Solve two of the ratio equations and check the answer in the third.

24. (k - 1)x - y = 5; (k + 1)x + (1 - k)y = (3k + 1).

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = k - 1, b1 = -1, c1 = -5 and a2 = k + 1, b2 = 1 - k, c2 = -3k - 1.
2. For infinite solutions, \( \frac{k - 1}{k + 1} = \frac{-1}{1 - k} = \frac{-5}{- 3 k - 1} \).
3. From \( \frac{k - 1}{k + 1} = \frac{-1}{1 - k} \): -k² + 2k - 1 = -k - 1, so k(3 - k) = 0, giving k = 0, 3.
4. From \( \frac{-1}{1 - k} = \frac{-5}{- 3 k - 1} \): 3k + 1 = 5k - 5, so 6 - 2k = 0, giving k = 3.
5. The value common to both results is k = 3, which satisfies all three ratios.
6. Hence, k = 3.

Teacher's Note:
a) Take the three ratios in pairs and solve for k from each pair.
b) The value of k that satisfies all the pairs is the answer; reject the rest.
c) Verify by putting k back into the original equations: they become multiples of each other.

25. x + (k + 1)y = 5; (k + 1)x + 9y = (8k - 1).

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 1, b1 = k + 1, c1 = -5 and a2 = k + 1, b2 = 9, c2 = 1 - 8k.
2. For infinite solutions, \( \frac{1}{k + 1} = \frac{k + 1}{9} = \frac{-5}{1 - 8 k} \).
3. From \( \frac{1}{k + 1} = \frac{k + 1}{9} \): 9 = k² + 2k + 1, so -k² - 2k + 8 = 0, giving k = -4, 2.
4. From \( \frac{k + 1}{9} = \frac{-5}{1 - 8 k} \): -8k² - 7k + 1 = -45, so -8k² - 7k + 46 = 0, giving k = -23/8, 2.
5. The value common to both results is k = 2, which satisfies all three ratios.
6. Hence, k = 2.

Teacher's Note:
a) Infinite solutions mean coincident lines: a1/a2 = b1/b2 = c1/c2.
b) Solve two of the ratio equations and check the answer in the third.

26. 2x + 3y = 7; (k - 1)x + (k + 2)y = 3k.

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 2, b1 = 3, c1 = -7 and a2 = k - 1, b2 = k + 2, c2 = -3k.
2. For infinite solutions, \( \frac{2}{k - 1} = \frac{3}{k + 2} = \frac{-7}{- 3 k} \).
3. From \( \frac{2}{k - 1} = \frac{3}{k + 2} \): 2k + 4 = 3k - 3, so 7 - k = 0, giving k = 7.
4. From \( \frac{3}{k + 2} = \frac{-7}{- 3 k} \): -9k = -7k - 14, so 14 - 2k = 0, giving k = 7.
5. The value common to both results is k = 7, which satisfies all three ratios.
6. Hence, k = 7.

Teacher's Note:
a) Take the three ratios in pairs and solve for k from each pair.
b) The value of k that satisfies all the pairs is the answer; reject the rest.
c) Verify by putting k back into the original equations: they become multiples of each other.

27. 2x + (k - 2)y = k; 6x + (2k - 1)y = (2k + 5).

Answer:
1. The system is a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, with a1 = 2, b1 = k - 2, c1 = -k and a2 = 6, b2 = 2k - 1, c2 = -2k - 5.
2. For infinite solutions, \( \frac{2}{6} = \frac{k - 2}{2 k - 1} = \frac{- k}{- 2 k - 5} \).
3. From \( \frac{2}{6} = \frac{k - 2}{2 k - 1} \): 4k - 2 = 6k - 12, so 10 - 2k = 0, giving k = 5.
4. From \( \frac{k - 2}{2 k - 1} = \frac{- k}{- 2 k - 5} \): -2k² - k + 10 = k(1 - 2k), so 10 - 2k = 0, giving k = 5.
5. The value common to both results is k = 5, which satisfies all three ratios.
6. Hence, k = 5.

Teacher's Note:
a) Infinite solutions mean coincident lines: a1/a2 = b1/b2 = c1/c2.
b) Solve two of the ratio equations and check the answer in the third.

Find the value of a and b for which each of the following systems of linear equations has a infinite number of solutions:

28. (a - 1)x + 3y = 2; 6x + (1 - 2b)y = 6.

Answer:
1. Here a1 = a - 1, b1 = 3, c1 = -2 and a2 = 6, b2 = 1 - 2b, c2 = -6.
2. For infinite solutions, \( \frac{a - 1}{6} = \frac{3}{1 - 2b} = \frac{-2}{-6} \).
3. Since \( \frac{-2}{-6} = \frac{1}{3} \), \( \frac{a - 1}{6} = \frac{1}{3} \), so a - 1 = 2 and a = 3.
4. Also \( \frac{3}{1 - 2b} = \frac{1}{3} \), so 1 - 2b = 9, which gives b = -4.
5. Hence, a = 3 and b = -4.

Teacher's Note:
a) Equate each of the first two ratios with the constant ratio, which is known here.
b) Two unknowns need two equations; here each ratio gives one value directly.

29. 2x - 3y = 7; (a + b)x - (a + b - 3)y = 4a + b.

Answer:
1. Here a1 = 2, b1 = -3, c1 = -7 and a2 = a + b, b2 = -(a + b - 3), c2 = -(4a + b).
2. For infinite solutions, \( \frac{2}{a + b} = \frac{-3}{-(a + b - 3)} = \frac{-7}{-(4a + b)} \).
3. From \( \frac{2}{a + b} = \frac{3}{a + b - 3} \): 2(a + b - 3) = 3(a + b), so a + b = -6.
4. Then \( \frac{2}{a + b} = \frac{2}{-6} = -\frac{1}{3} \).
5. From \( \frac{2}{a + b} = \frac{7}{4a + b} \): 4a + b = -21.
6. Subtracting a + b = -6 from 4a + b = -21 gives 3a = -15, so a = -5 and b = -1.
7. Hence, a = -5 and b = -1.

Teacher's Note:
a) Notice that the first ratio only has a + b, so solve for a + b before finding a and b.
b) Check the answer: the second equation becomes -6x + 9y = -21, which is -3 times the first equation.
c) Do not forget the sign of the term - (a + b - 3)y when reading b2.

30. 2x + 3y = 7; (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1.

Answer:
1. Here a1 = 2, b1 = 3, c1 = -7 and a2 = a + b + 1, b2 = a + 2b + 2, c2 = -(4a + 4b + 1).
2. For infinite solutions, \( \frac{2}{a + b + 1} = \frac{3}{a + 2b + 2} = \frac{7}{4(a + b) + 1} \).
3. From the first two ratios: 2(a + 2b + 2) = 3(a + b + 1), so a - b = 1.
4. From the first and third ratios: 2(4a + 4b + 1) = 7(a + b + 1), so a + b = 5.
5. Adding a - b = 1 and a + b = 5 gives 2a = 6, so a = 3.
6. Then b = 5 - 3 = 2.
7. Hence, a = 3 and b = 2.

Teacher's Note:
a) Cross-multiply two pairs of ratios to get two linear equations in a and b.
b) Solving by addition is quick because the equations are a - b = 1 and a + b = 5.
c) Check the third ratio before writing the final answer.

31. 2x + 3y = 7; a(x + y) - b(x - y) = 3a + b - 2.

Answer:
1. Rewrite the second equation as (a - b)x + (a + b)y = 3a + b - 2, so a2 = a - b, b2 = a + b, c2 = -(3a + b - 2).
2. For infinite solutions, \( \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3a + b - 2} \).
3. From the first two ratios: 2(a + b) = 3(a - b), so a = 5b.
4. From the second and third ratios: 3(3a + b - 2) = 7(a + b), so 2a - 4b = 6, that is a - 2b = 3.
5. Putting a = 5b gives 5b - 2b = 3, so b = 1 and a = 5.
6. Hence, a = 5 and b = 1.

Teacher's Note:
a) First expand the brackets and collect the coefficients of x and y.
b) A common error is to leave b(x - y) unexpanded, which gives wrong signs.
c) Verify: with a = 5, b = 1 the second equation is 4x + 6y = 14, twice the first equation.

32. (2a - 1)x + 3y = 5; 3x + (b - 1)y = 2.

Answer:
1. Here a1 = 2a - 1, b1 = 3, c1 = -5 and a2 = 3, b2 = b - 1, c2 = -2.
2. For infinite solutions, \( \frac{2a - 1}{3} = \frac{3}{b - 1} = \frac{5}{2} \).
3. \( \frac{2a - 1}{3} = \frac{5}{2} \), so 2(2a - 1) = 15, 4a = 17 and a = \( \frac{17}{4} \).
4. \( \frac{3}{b - 1} = \frac{5}{2} \), so 5(b - 1) = 6 and b = \( \frac{11}{5} \).
5. Hence, a = \( \frac{17}{4} \) and b = \( \frac{11}{5} \).

Teacher's Note:
a) Use the known ratio of constants, 5/2, with each of the other ratios.
b) Answers need not be integers; keep them as fractions.

33. Find the value of k, so that the following system of equations has a non-zero solution:
5x - 3y = 0; 2x + ky = 0.

Answer:
1. The system is homogeneous, so a non-zero solution exists only if the lines coincide.
2. \( \frac{5}{2} = \frac{-3}{k} \), so 5k = -6.
3. k = \( -\frac{6}{5} \)
4. Hence, k = \( -\frac{6}{5} \).

Teacher's Note:
a) A non-zero solution of a homogeneous pair needs a1/a2 = b1/b2.
b) Watch the negative sign of b1 when cross-multiplying.

Show that the following system of the equations has a unique solution and hence find the solution of the given system of equations.

34. \( \frac{x}{3} + \frac{y}{2} = 3 \); x - 2y = 2

Answer:
1. Multiplying the first equation by 6, 2x + 3y = 18, so a1 = 2, b1 = 3 and a2 = 1, b2 = -2 for x - 2y = 2.
2. \( \frac{a_1}{a_2} = \frac{2}{1} = 2 \) and \( \frac{b_1}{b_2} = \frac{3}{-2} \).
3. Since the two ratios are not equal, the system has a unique solution.
4. From x - 2y = 2, x = 2 + 2y.
5. Substituting in 2x + 3y = 18: 2(2 + 2y) + 3y = 18, so 4 + 7y = 18 and y = 2.
6. Then x = 2 + 2(2) = 6.
7. Hence, x = 6 and y = 2.

Teacher's Note:
a) Clear the fractions first by multiplying by the LCM, 6.
b) Always substitute the answer back in both original equations to verify: 6/3 + 2/2 = 3 and 6 - 4 = 2.

35. 3x + 5y = 12; 5x + 3y = 4.

Answer:
1. Here a1 = 3, b1 = 5 and a2 = 5, b2 = 3.
2. \( \frac{a_1}{a_2} = \frac{3}{5} \) and \( \frac{b_1}{b_2} = \frac{5}{3} \), which are not equal, so the system has a unique solution.
3. Adding the equations: 8x + 8y = 16, so x + y = 2.
4. Subtracting the second from the first: -2x + 2y = 8, so y - x = 4.
5. Adding x + y = 2 and y - x = 4 gives 2y = 6, so y = 3.
6. Then x = 2 - 3 = -1.
7. Hence, x = -1 and y = 3.

Teacher's Note:
a) For equations with swapped coefficients, adding and subtracting is quicker than substitution.
b) Verify: 3(-1) + 5(3) = 12 and 5(-1) + 3(3) = 4.
c) Unique solution means the lines meet at exactly one point, here (-1, 3).

Very Short Answer 

Question. How many solutions does the pair of equations \( y = 0 \) and \( y = -5 \) have?
Answer: Sol. \( y = 0 \) and \( y = -5 \) are Parallel lines, hence no solution.

Question. If \( ax + by = a^2 - b^2 \) and \( bx + ay = 0 \), find the value of \( (x + y) \).
Answer: Sol. \( ax + by = a^2 - b^2 \)
\( ay + bx = 0 \)
\( a(x + y) + b(x + y) = a^2 - b^2 \) ...[adding
\( (x + y)(a + b) = a^2 - b^2 \)
\( (x + y) = \frac{(a + b)(a - b)}{(a + b)} \)
\( \therefore x + y = a - b \)

Short Answer-II 

Question. Solve the following pair of equations for x and y:
\( \frac{a^2}{x} - \frac{b^2}{y} = 0; \frac{a^2b}{x} + \frac{b^2a}{y} = a + b, x \neq 0; y \neq 0 \)
Answer: Sol. \( \frac{a^2}{x} - \frac{b^2}{y} = 0 \) ...(i)
\( \frac{a^2b}{x} + \frac{b^2a}{y} = a + b \) ...(ii)
Multiplying (i) by \( a \) and (ii) by 1,
\( \frac{a^3}{x} - \frac{b^2a}{y} = 0 \)
\( \frac{a^2b}{x} + \frac{b^2a}{y} = a + b \)
\( \frac{a^3 + a^2b}{x} = a + b \) ...[By adding
\( \frac{a^2(a + b)}{x} = \frac{a + b}{1} \)
\( x(a + b) = a^2(a + b) \therefore x = a^2 \)
Putting the value of \( x \) in (i), we get \( \frac{a^2}{a^2} - \frac{b^2}{y} = 0 \Rightarrow 1 = \frac{b^2}{y} \Rightarrow y = b^2 \)

Question. Solve for x and y:
\( \frac{10}{x + y} + \frac{2}{x - y} = 4; \frac{15}{x + y} - \frac{5}{x - y} = -2; x + y \neq 0, x - y \neq 0 \)
Answer: Sol. \( \frac{10}{x + y} + \frac{2}{x - y} = 4 \) ...(i)
\( \frac{15}{x + y} - \frac{5}{x - y} = -2 \) ...(ii)
Multiplying (i) by 5 and (ii) by 2, we get
\( \frac{50}{x + y} + \frac{10}{x - y} = 20 \)
\( \frac{30}{x + y} - \frac{10}{x - y} = -4 \)
\( \frac{80}{x + y} = 16 \) ...[By adding
\( 16(x + y) = 80 \Rightarrow x + y = 5 \) ...(iii)
Putting the value of \( (x + y) \) in (i), we get \( \frac{10}{5} + \frac{2}{x - y} = 4 \Rightarrow 2 + \frac{2}{x - y} = 4 \)
\( \frac{2}{x - y} = 4 - 2 = 2 \Rightarrow 2(x - y) = 2 \therefore x - y = 1 \) ...(iv)
Solving (iii) and (iv), we get \( 2x = 6 \Rightarrow x = 3 \)
Putting the value of \( x \) in (iii), we get \( 3 + y = 5 \Rightarrow y = 2 \)
\( \therefore x = 3, y = 2 \)

Question. Solve the following pair of linear equations for x and y:
\( 141x + 93y = 189; \)
\( 93x + 141y = 45 \)
Answer: Sol. \( 141x + 93y = 189 \)
\( 93x + 141y = 45 \)
\( 234x + 234y = 234 \) ...[By adding
\( \Rightarrow x + y = 1 \) ...(i) (\( \div \) by 234)
Again
\( 141x + 93y = 189 \)
\( 93x + 141y = 45 \)
\( 48x - 48y = 144 \) ...[By subtracting
\( \Rightarrow x - y = 3 \) ...(ii) (\( \div \) by 48)
By adding (i) and (ii), we get \( 2x = 4 \Rightarrow x = 2 \)
Putting the value of \( x \) in (i), we get \( 2 + y = 1 \Rightarrow y = -1 \)
\( \therefore x = 2, y = -1 \)

Question. Solve the following pair of linear equations for x and y:
\( \frac{b}{a}x + \frac{a}{b}y = a^2 + b^2; x + y = 2ab \)
Answer: Sol. \( \frac{b}{a}x + \frac{a}{b}y = a^2 + b^2 \) ...(i)
\( x + y = 2ab \) ...(ii)
Multiplying (ii) by \( b/a \), we get
\( \frac{b}{a}x + \frac{b}{a}y = 2b^2 \)
Subtracting this from (i):
\( (\frac{a}{b} - \frac{b}{a})y = a^2 + b^2 - 2b^2 \)
\( (\frac{a^2 - b^2}{ab})y = (a^2 - b^2) \)
\( y = ab \)
Putting the value of \( y \) in (ii), we get \( x + ab = 2ab \Rightarrow x = ab \)
\( \therefore x = ab, y = ab \)

Question. Solve by elimination:
\( 3x = y + 5 \)
\( 5x - y = 11 \)
Answer: Sol. We have, \( 3x - y = 5 \) ...(i) and \( 5x - y = 11 \) ...(ii)
Subtracting (i) from (ii): \( 2x = 6 \Rightarrow x = 3 \)
Putting \( x = 3 \) in (i): \( 3(3) - y = 5 \Rightarrow 9 - 5 = y \therefore y = 4 \)
\( \therefore x = 3, y = 4 \)

Question. Solve by elimination:
\( 3x - y = 7 \)
\( 2x + 5y + 1 = 0 \)
Answer: Sol. \( 3x - y = 7 \) ...(i) and \( 2x + 5y = -1 \) ...(ii)
Multiplying equation (i) by 5: \( 15x - 5y = 35 \)
Adding this to (ii): \( 17x = 34 \Rightarrow x = 2 \)
Putting \( x = 2 \) in (i): \( 3(2) - y = 7 \Rightarrow 6 - 7 = y \therefore y = -1 \)
\( \therefore x = 2, y = -1 \)

Question. Solve for x and y:
\( 27x + 31y = 85; \)
\( 31x + 27y = 89 \)
Answer: Sol. Adding equations: \( 58x + 58y = 174 \Rightarrow x + y = 3 \) ...(i)
Subtracting equations: \( -4x + 4y = -4 \Rightarrow x - y = 1 \) ...(ii)
Adding (i) and (ii): \( 2x = 4 \Rightarrow x = 2 \)
Putting \( x = 2 \) in (i): \( 2 + y = 3 \Rightarrow y = 1 \)
\( \therefore x = 2, y = 1 \)

Question. Solve for x and y: \( \frac{x}{a} = \frac{y}{b}; ax + by = a^2 + b^2 \)
Answer: Sol. \( \frac{x}{a} - \frac{y}{b} = 0 \Rightarrow bx - ay = 0 \) ...(i)
\( ax + by = a^2 + b^2 \) ...(ii)
Multiplying (i) by \( b \) and (ii) by \( a \),
\( b^2x - aby = 0 \)
\( a^2x + aby = a(a^2 + b^2) \)
Adding: \( x(b^2 + a^2) = a(a^2 + b^2) \Rightarrow x = a \)
Putting \( x = a \) in (i): \( b(a) - ay = 0 \Rightarrow ay = ab \therefore y = b \)
\( \therefore x = a, y = b \)

Question. Solve: \( \frac{x}{a} + \frac{y}{b} = a + b; \frac{x}{a^2} + \frac{y}{b^2} = 2, a, b \neq 0 \)
Answer: Sol. \( \frac{bx + ay}{ab} = a + b \Rightarrow bx + ay = a^2b + ab^2 \) ...(i)
\( \frac{b^2x + a^2y}{a^2b^2} = 2 \Rightarrow b^2x + a^2y = 2a^2b^2 \) ...(ii)
Multiplying (i) by \( b \) and subtracting (ii) from it:
\( (b^2x + aby) - (b^2x + a^2y) = (a^2b^2 + ab^3) - 2a^2b^2 \)
\( ay(b - a) = ab^2(b - a) \therefore y = b^2 \)
Putting \( y = b^2 \) in (ii): \( \frac{x}{a^2} + 1 = 2 \Rightarrow \frac{x}{a^2} = 1 \therefore x = a^2 \)
\( \therefore x = a^2, y = b^2 \)

Question. Solve the following pair of equations:
\( 49x + 51y = 499 \)
\( 51x + 49y = 501 \)
Answer: Sol. Adding: \( 100x + 100y = 1000 \Rightarrow x + y = 10 \) ...(i)
Subtracting: \( -2x + 2y = -2 \Rightarrow -x + y = -1 \) ...(ii)
Solving (i) and (ii): \( 2y = 9 \Rightarrow y = \frac{9}{2} \)
Putting \( y = \frac{9}{2} \) in (i): \( x = 10 - \frac{9}{2} = \frac{11}{2} \)
\( \therefore x = \frac{11}{2}, y = \frac{9}{2} \)

Question. Find the two numbers whose sum is 75 and difference is 15.
Answer: Sol. Let the two numbers be \( x \) and \( y \).
\( x + y = 75 \) ...(i)
\( x - y = 15 \) ...(ii)
Adding: \( 2x = 90 \Rightarrow x = 45 \)
Putting \( x = 45 \) in (i): \( 45 + y = 75 \Rightarrow y = 30 \)
\( \therefore \) Numbers are 45, 30.

Question. Find the value of \( \alpha \) and \( \beta \) for which the following pair of linear equations has infinite number of solutions:
\( 2x + 3y = 7; \alpha x + (\alpha + \beta)y = 28 \)
Answer: Sol. For infinite solutions: \( \frac{2}{\alpha} = \frac{3}{\alpha + \beta} = \frac{7}{28} = \frac{1}{4} \)
\( \frac{2}{\alpha} = \frac{1}{4} \Rightarrow \alpha = 8 \)
\( \frac{3}{\alpha + \beta} = \frac{1}{4} \Rightarrow \alpha + \beta = 12 \)
\( 8 + \beta = 12 \Rightarrow \beta = 4 \)
\( \therefore \alpha = 8, \beta = 4 \)

Question. Solve the following pair of linear equations by the cross multiplication method:
\( x + 2y = 2; x - 3y = 7 \)
Answer: Sol. \( x + 2y - 2 = 0 \) and \( x - 3y - 7 = 0 \)
\( \frac{x}{2(-7) - (-3)(-2)} = \frac{y}{(-2)(1) - (-7)(1)} = \frac{1}{1(-3) - (1)(2)} \)
\( \frac{x}{-14 - 6} = \frac{y}{-2 + 7} = \frac{1}{-3 - 2} \)
\( \frac{x}{-20} = \frac{y}{5} = \frac{1}{-5} \)
\( x = \frac{-20}{-5} = 4 \) and \( y = \frac{5}{-5} = -1 \)
\( \therefore x = 4, y = -1 \)

Question. A man earns Rs 600 per month more than his wife. One-tenth of the man’s salary and 1/6th of the wife’s salary amount to Rs 1,500, which is saved every month. Find their incomes.
Answer: Sol. Let wife’s monthly income \( = \text{Rs } x \)
Then man’s monthly income \( = \text{Rs }(x + 600) \)
According to question: \( \frac{1}{10}(x + 600) + \frac{1}{6}(x) = 1500 \)
\( \frac{3(x + 600) + 5x}{30} = 1500 \)
\( 3x + 1800 + 5x = 45000 \Rightarrow 8x = 43200 \)
\( x = \frac{43200}{8} = 5400 \)
Wife’s income \( = \text{Rs } 5,400 \)
Man’s income \( = \text{Rs } (5400 + 600) = \text{Rs } 6,000 \)

Long Answer 

Question. The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.
Answer: Sol. Let unit and tens digit be \( x \) and \( y \).
Original number \( = x + 10y \)
Reversed number \( = 10x + y \)
According to question: \( x + y = 8 \Rightarrow y = 8 - x \) ...(ii)
Also, \( (1x + 10y) - (10x + y) = 18 \Rightarrow 9y - 9x = 18 \Rightarrow y - x = 2 \)
Substituting \( y \): \( (8 - x) - x = 2 \Rightarrow 8 - 2 = 2x \Rightarrow 2x = 6 \therefore x = 3 \)
\( y = 8 - 3 = 5 \)
Original number \( = 3 + 10(5) = 53 \)

Question. The age of the father is twice the sum of the ages of his 2 children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Answer: Sol. Let present ages of children be \( x \) and \( y \). Present age of father \( = 2(x + y) \) ...(i)
After 20 years: Father's age \( = 2(x + y) + 20 \). Children's ages sum \( = (x + 20) + (y + 20) \)
According to Question: \( 2(x + y) + 20 = x + y + 40 \)
\( x + y = 20 \)
Present age of father \( = 2(20) = 40 \text{ years} \)

Question. A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.
Answer: Sol. Let unit’s digit be \( x \) and ten’s digit \( y \). Original number \( = x + 10y \)
\( x + 10y = 7(x + y) \Rightarrow 3y = 6x \Rightarrow y = 2x \) ...(i)
\( (x + 10y) - (10x + y) = 18 \Rightarrow 9y - 9x = 18 \Rightarrow y - x = 2 \)
Substituting \( y \): \( 2x - x = 2 \Rightarrow x = 2 \). Then \( y = 4 \).
Original number \( = 2 + 10(4) = 42 \)

Question. Sita Devi wants to make a rectangular pond... The area of the pond will be decreased by 3 square feet if its length is decreased by 2 ft. and breadth is increased by 1 ft. Its area will be increased by 4 square feet if the length is increased by 1 ft. and breadth remains same. Find the dimensions of the pond.
Answer: Sol. Let length \( = x \), breadth \( = y \). Area \( = xy \)
According to Question: \( (x - 2)(y + 1) = xy - 3 \Rightarrow xy + x - 2y - 2 = xy - 3 \Rightarrow x - 2y = -1 \) ...(i)
Also, \( (x + 1)y = xy + 4 \Rightarrow xy + y = xy + 4 \therefore y = 4 \)
Putting \( y = 4 \) in (i): \( x - 2(4) = -1 \Rightarrow x = 7 \)
Length \( = 7 \text{ ft} \), Breadth \( = 4 \text{ ft} \)

Question. On reversing the digits of a two digit number, number obtained is 9 less than three times the original number. If difference of these two numbers is 45, find the original number.
Answer: Sol. Let unit digit be \( x \), tens digit \( y \). Original number \( = x + 10y \)
\( 10x + y = 3(x + 10y) - 9 \Rightarrow 7x - 29y = -9 \) ...(i)
\( (10x + y) - (x + 10y) = 45 \Rightarrow 9x - 9y = 45 \Rightarrow x - y = 5 \Rightarrow x = 5 + y \) ...(ii)
Solving (i) and (ii): \( 7(5 + y) - 29y = -9 \Rightarrow 35 + 7y - 29y = -9 \Rightarrow -22y = -44 \therefore y = 2 \)
Then \( x = 5 + 2 = 7 \). Original number \( = 27 \)

Question. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Answer: Sol. Let speed of stream \( = x \text{ km/hr} \). Speed upstream \( = 15 - x \), downstream \( = 15 + x \)
According to question: \( \frac{30}{15 - x} + \frac{30}{15 + x} = 4.5 \text{ hrs} \)
\( 30 \left[ \frac{15 + x + 15 - x}{225 - x^2} \right] = \frac{9}{2} \Rightarrow \frac{30 \times 30}{225 - x^2} = \frac{9}{2} \)
\( 1800 = 9(225 - x^2) \Rightarrow 200 = 225 - x^2 \Rightarrow x^2 = 25 \therefore x = 5 \text{ km/hr} \)

Question. Taxi charges in city comprises of fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and for journey of 20 km, the charge paid is Rs 145. What will a person have to pay for travelling a distance of 30 km?
Answer: Sol. Let fixed charges \( = \text{Rs } x \), charge per km \( = \text{Rs } y \)
\( x + 12y = 89 \) ...(i)
\( x + 20y = 145 \) ...(ii)
Subtracting (i) from (ii): \( 8y = 56 \Rightarrow y = 7 \)
\( x + 12(7) = 89 \Rightarrow x + 84 = 89 \therefore x = 5 \)
Total fare for 30 km \( = x + 30y = 5 + 30(7) = 5 + 210 = \text{Rs } 215 \)

Question. A boat takes 4 hours to go 44 km downstream and it can go 20 km upstream in the same time. Find the speed of the stream and that of the boat in still water.
Answer: Sol. Let boat speed \( = x \), stream speed \( = y \)
\( \frac{44}{x + y} = 4 \Rightarrow x + y = 11 \) ...(i)
\( \frac{20}{x - y} = 4 \Rightarrow x - y = 5 \)
Solving: \( 2x = 16 \Rightarrow x = 8 \text{ km/hr} \). Then \( y = 3 \text{ km/hr} \).
Boat speed \( = 8 \text{ km/hr} \), Stream speed \( = 3 \text{ km/hr} \)

Question. A man travels 300 km partly by train and partly by car. He takes 4 hours if he travels 60 km by train and the rest by car. If he travels 100 km by train and the remaining by car, he takes 10 minutes longer. Find the speeds of the train and the car separately.
Answer: Sol. Let train speed \( = x \text{ km/hr} \), car speed \( = y \text{ km/hr} \)
\( \frac{60}{x} + \frac{240}{y} = 4 \) ...(i)
\( \frac{100}{x} + \frac{200}{y} = 4 \text{ hr } 10 \text{ min} = \frac{25}{6} \) ...(ii)
Solving: \( x = 60 \text{ km/hr}, y = 80 \text{ km/hr} \)