CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 27

Substitution Method

Question 1. Solve the following pair of linear equations by the substitution method: \( x + y = 14,\ x - y = 4 \).
Answer:
\( x + y = 14 \) ...(i)    \( x - y = 4 \) ...(ii)
From (i): \( x = 14 - y \) ...(iii)
Substituting in (ii): \( -2y = -10 \), so \( y = 5 \).
Putting this in (iii): \( x = 9 \).
Hence, \( x = 9,\ y = 5 \).
In simple words: Write x in terms of y from one equation and use it in the other.

Teacher's Note:
1. Solve one equation for one variable, substitute it in the other, then back-substitute.
2. Check the answer in both original equations.

Question 2. Solve the following pair of linear equations by the substitution method: \( s - t = 3,\ \frac{s}{3} + \frac{t}{2} = 6 \).
Answer:
Multiplying the second equation by 6: \( 2s + 3t = 36 \).
\( s - t = 3 \) ...(i)    \( 2s + 3t = 36 \) ...(ii)
From (i): \( s = 3 + t \) ...(iii)
Substituting in (ii): \( 5t = 30 \), so \( t = 6 \).
Putting this in (iii): \( s = 9 \).
Hence, \( s = 9,\ t = 6 \).
In simple words: Clear the fractions first, then substitute.

Teacher's Note:
1. Multiply by the LCM of the denominators to remove fractions.
2. Use letters s and t exactly as given in the question.

Question 3. Solve the following pair of linear equations by the substitution method: \( 3x - y = 3,\ 9x - 3y = 9 \).
Answer:
From the first equation, \( y = 3x - 3 \).
Substituting in the second: \( 9x - 3(3x - 3) = 9 \Rightarrow 9x - 9x + 9 = 9 \Rightarrow 9 = 9 \), which is always true.
So the two equations represent the same line (\( \frac{3}{9} = \frac{-1}{-3} = \frac{3}{9} \)). The pair has infinitely many solutions: every pair \( (x,\ 3x - 3) \) with x any real number.
In simple words: The second equation is just three times the first, so every point of the line is a solution.

Teacher's Note:
1. An identity such as 9 = 9 means infinitely many solutions.
2. A false statement such as 0 = 5 would mean no solution.

Question 4. Solve the following pair of linear equations by the substitution method: \( 0.2x + 0.3y = 1.3,\ 0.4x + 0.5y = 2.3 \).
Answer:
Multiplying both equations by 10: \( 2x + 3y = 13 \) and \( 4x + 5y = 23 \).
\( 2x + 3y = 13 \) ...(i)    \( 4x + 5y = 23 \) ...(ii)
From (i): \( y = \frac{13}{3} - \frac{2}{3}x \) ...(iii)
Substituting in (ii): \( \frac{2}{3}x = \frac{4}{3} \), so \( x = 2 \).
Putting this in (iii): \( y = 3 \).
Hence, \( x = 2,\ y = 3 \).
In simple words: Remove the decimals by multiplying by 10 and then substitute.

Teacher's Note:
1. Multiplying by 10 turns decimals into integers.
2. Verify: \( 0.2(2) + 0.3(3) = 1.3 \) and \( 0.4(2) + 0.5(3) = 2.3 \).

Question 5. Solve the following pair of linear equations by the substitution method: \( \sqrt{2}x + \sqrt{3}y = 0,\ \sqrt{3}x - \sqrt{8}y = 0 \).
Answer:
From the first equation, \( y = -\frac{\sqrt{2}}{\sqrt{3}}x \).
Substituting in the second: \( \sqrt{3}x + \sqrt{8}\cdot\frac{\sqrt{2}}{\sqrt{3}}x = 0 \Rightarrow \sqrt{3}x + \frac{4}{\sqrt{3}}x = 0 \Rightarrow 3x + 4x = 0 \Rightarrow x = 0\).
Then \( y = 0 \). Hence \( x = 0,\ y = 0 \).
In simple words: Both equations pass through the origin, so the only common point is (0, 0).

Teacher's Note:
1. Since \( \sqrt{8}\sqrt{2} = 4 \), the second equation simplifies neatly.
2. Homogeneous equations (constant term 0) with non-parallel lines meet at the origin.

Question 6. Solve the following pair of linear equations by the substitution method: \( \frac{3x}{2} - \frac{5y}{3} = -2,\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \).
Answer:
Multiplying the first equation by 6: \( 9x - 10y = -12 \). Multiplying the second equation by 6: \( 2x + 3y = 13 \).
\( 9x - 10y = -12 \) ...(i)    \( 2x + 3y = 13 \) ...(ii)
From (i): \( y = \frac{6}{5} + \frac{9}{10}x \) ...(iii)
Substituting in (ii): \( \frac{47}{10}x = \frac{47}{5} \), so \( x = 2 \).
Putting this in (iii): \( y = 3 \).
Hence, \( x = 2,\ y = 3 \).
In simple words: Clear the fractions in each equation, then substitute.

Teacher's Note:
1. Use the LCM of the denominators of each equation separately.
2. Check: \( \frac{3(2)}{2} - \frac{5(3)}{3} = 3 - 5 = -2 \).

Question 7. Solve \( 2x + 3y = 11 \) and \( 2x - 4y = -24 \) and hence find the value of m for which \( y = mx + 3 \).
Answer:
\( 2x + 3y = 11 \) ...(i)    \( 2x - 4y = -24 \) ...(ii)
From (ii): \( x = -12 + 2y \) ...(iii)
Substituting in (i): \( 7y = 35 \), so \( y = 5 \).
Putting this in (iii): \( x = -2 \).
Hence, \( x = -2,\ y = 5 \).
Now \( y = mx + 3 \) must hold for this solution: \( 5 = m(-2) + 3 \Rightarrow -2m = 2 \Rightarrow m = -1 \).
Hence m = −1.
In simple words: Find the common solution first and then put it in the third equation to get m.

Teacher's Note:
1. The third equation must be satisfied by the same point (x, y).
2. Substitute carefully with the negative value of x.

Question 8. The difference between two numbers is 26 and one number is three times the other. Find them.
Answer:
Let the numbers be x and y with \( y > x \). Then \( y = 3x \) ...(i) and \( y - x = 26 \) ...(ii).
Putting (i) in (ii): \( 3x - x = 26 \Rightarrow x = 13 \). Then \( y = 39 \).
Hence the numbers are 13 and 39.
In simple words: One number is 3 times the other and they differ by 26, so the smaller one is 13.

Teacher's Note:
1. Take the larger number as y so that \( y - x = 26 \).
2. Check: 39 − 13 = 26 and 39 = 3 × 13.

Question 9. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer:
Let the angles be x° (larger) and y° (smaller). Supplementary: \( x + y = 180 \) ...(i). Difference: \( x - y = 18 \) ...(ii).
From (ii), \( x = y + 18 \). Putting in (i): \( 2y + 18 = 180 \Rightarrow y = 81 \), so \( x = 99 \).
Hence the angles are 99° and 81°.
In simple words: Supplementary angles add to 180 degrees; combine this with the given difference.

Teacher's Note:
1. Supplementary means the sum is 180°, complementary means 90°.
2. Check: 99 + 81 = 180 and 99 − 81 = 18.

Question 10. The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Answer:
Let a bat cost Rs. x and a ball cost Rs. y. Then \( 7x + 6y = 3800 \) ...(i) and \( 3x + 5y = 1750 \) ...(ii).
From (i): \( y = \frac{1900}{3} - \frac{7}{6}x \) ...(iii)
Substituting in (ii): \( -\frac{17}{6}x = -\frac{4250}{3} \), so \( x = 500 \).
Putting this in (iii): \( y = 50 \).
Hence, \( x = 500,\ y = 50 \).
Hence a bat costs Rs. 500 and a ball costs Rs. 50.
In simple words: Write one equation for each purchase and solve the pair.

Teacher's Note:
1. Check: 7(500) + 6(50) = 3800 and 3(500) + 5(50) = 1750.
2. State the answer with units (Rs.).

Question 11. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Answer:
Let the fixed charge be Rs. x and the charge per km be Rs. y. Then \( x + 10y = 105 \) ...(i) and \( x + 15y = 155 \) ...(ii).
From (i), \( x = 105 - 10y \). Putting in (ii): \( 105 - 10y + 15y = 155 \Rightarrow 5y = 50 \Rightarrow y = 10 \), so \( x = 5 \).
Fixed charge = Rs. 5, charge per km = Rs. 10. Fare for 25 km = \( 5 + 25 \times 10 = \) Rs. 255.
In simple words: Total fare equals the fixed part plus rate times distance.

Teacher's Note:
1. Fare = fixed charge + (rate per km × distance).
2. Use the two given journeys to find the two unknowns, then apply them to 25 km.

Question 12. A fraction becomes \( \frac{9}{11} \) if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes \( \frac{5}{6} \). Find the fraction.
Answer:
Let the fraction be \( \frac{x}{y} \). Then \( \frac{x + 2}{y + 2} = \frac{9}{11} \Rightarrow 11x - 9y = -4 \) ...(i) and \( \frac{x + 3}{y + 3} = \frac{5}{6} \Rightarrow 6x - 5y = -3 \) ...(ii).
From (i): \( y = \frac{4}{9} + \frac{11}{9}x \) ...(iii)
Substituting in (ii): \( -\frac{1}{9}x = -\frac{7}{9} \), so \( x = 7 \).
Putting this in (iii): \( y = 9 \).
Hence, \( x = 7,\ y = 9 \).
Hence the fraction is \( \frac{7}{9} \).
In simple words: Cross-multiply each condition to get two linear equations in the numerator and denominator.

Teacher's Note:
1. Cross-multiplying \( \frac{x+2}{y+2} = \frac{9}{11} \) gives \( 11(x + 2) = 9(y + 2) \).
2. Check: \( \frac{9}{11} \) from \( \frac{7+2}{9+2} \) and \( \frac{5}{6} \) from \( \frac{7+3}{9+3} \).

Question 13. Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?
Answer:
Let Jacob's present age be x years and his son's present age be y years. After five years: \( x + 5 = 3(y + 5) \Rightarrow x - 3y = 10 \) ...(i). Five years ago: \( x - 5 = 7(y - 5) \Rightarrow x - 7y = -30 \) ...(ii).
From (i), \( x = 3y + 10 \). Putting in (ii): \( 3y + 10 - 7y = -30 \Rightarrow -4y = -40 \Rightarrow y = 10 \), so \( x = 40 \).
Hence Jacob is 40 years old and his son is 10 years old.
In simple words: Use the present ages to write the condition for five years ahead and five years back.

Teacher's Note:
1. Adding or subtracting the years must be done on both persons' ages.
2. Check: 45 = 3 × 15 and 35 = 7 × 5.