CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 28

Elimination Method

Question 1. Solve the following pair of linear equations by the elimination method and the substitution method: \( x + y = 5,\ 2x - 3y = 4 \).
Answer:
By the elimination method.
\( x + y = 5 \) ...(i)    \( 2x - 3y = 4 \) ...(ii)
Multiplying (i) by 3: \( 3x + 3y = 15 \) ...(iii)
Adding (iii) and (ii): \( 5x = 19 \), so \( x = \frac{19}{5} \).
Putting \( x = \frac{19}{5} \) in (i): \( \frac{19}{5} + y = 5 \), so \( y = \frac{6}{5} \).
Hence, \( x = \frac{19}{5},\ y = \frac{6}{5} \).
By the substitution method.
\( x + y = 5 \) ...(i)    \( 2x - 3y = 4 \) ...(ii)
From (i): \( x = 5 - y \) ...(iii)
Substituting in (ii): \( -5y = -6 \), so \( y = \frac{6}{5} \).
Putting this in (iii): \( x = \frac{19}{5} \).
Hence, \( x = \frac{19}{5},\ y = \frac{6}{5} \).
In simple words: Solve it twice, once with each method; both give the same point.

Teacher's Note:
1. Make the coefficients of one variable equal, then add or subtract.
2. Both methods must give the same answer, so use one to verify the other.

Question 2. Solve the following pair of linear equations by the elimination method and the substitution method: \( 3x + 4y = 10,\ 2x - 2y = 2 \).
Answer:
By the elimination method.
\( 3x + 4y = 10 \) ...(i)    \( 2x - 2y = 2 \) ...(ii)
Multiplying (ii) by 2: \( 4x - 4y = 4 \) ...(iii)
Adding (i) and (iii): \( 7x = 14 \), so \( x = 2 \).
Putting \( x = 2 \) in (i): \( 6 + 4y = 10 \), so \( y = 1 \).
Hence, \( x = 2,\ y = 1 \).
By the substitution method.
\( 3x + 4y = 10 \) ...(i)    \( 2x - 2y = 2 \) ...(ii)
From (ii): \( x = 1 + y \) ...(iii)
Substituting in (i): \( 7y = 7 \), so \( y = 1 \).
Putting this in (iii): \( x = 2 \).
Hence, \( x = 2,\ y = 1 \).
In simple words: Solve it twice, once with each method; both give the same point.

Teacher's Note:
1. Make the coefficients of one variable equal, then add or subtract.
2. Both methods must give the same answer, so use one to verify the other.

Question 3. Solve the following pair of linear equations by the elimination method and the substitution method: \( 3x - 5y - 4 = 0,\ 9x = 2y + 7 \).
Answer:
Rewriting: \( 3x - 5y = 4 \) and \( 9x - 2y = 7 \).
By the elimination method.
\( 3x - 5y = 4 \) ...(i)    \( 9x - 2y = 7 \) ...(ii)
Multiplying (i) by 2: \( 6x - 10y = 8 \) ...(iii)
Multiplying (ii) by 5: \( 45x - 10y = 35 \) ...(iv)
Subtracting (iv) from (iii): \( -39x = -27 \), so \( x = \frac{9}{13} \).
Putting \( x = \frac{9}{13} \) in (i): \( \frac{27}{13} - 5y = 4 \), so \( y = -\frac{5}{13} \).
Hence, \( x = \frac{9}{13},\ y = -\frac{5}{13} \).
By the substitution method.
\( 3x - 5y = 4 \) ...(i)    \( 9x - 2y = 7 \) ...(ii)
From (i): \( y = -\frac{4}{5} + \frac{3}{5}x \) ...(iii)
Substituting in (ii): \( \frac{39}{5}x = \frac{27}{5} \), so \( x = \frac{9}{13} \).
Putting this in (iii): \( y = -\frac{5}{13} \).
Hence, \( x = \frac{9}{13},\ y = -\frac{5}{13} \).
In simple words: Rearrange to standard form first, then use either method.

Teacher's Note:
1. Make the coefficients of one variable equal, then add or subtract.
2. Both methods must give the same answer, so use one to verify the other.

Question 4. Solve the following pair of linear equations by the elimination method and the substitution method: \( \frac{x}{2} + \frac{2y}{3} = -1,\ x - \frac{y}{3} = 3 \).
Answer:
Multiplying the first equation by 6: \( 3x + 4y = -6 \). Multiplying the second equation by 3: \( 3x - y = 9 \).
By the elimination method.
\( 3x + 4y = -6 \) ...(i)    \( 3x - y = 9 \) ...(ii)
Multiplying (ii) by 4: \( 12x - 4y = 36 \) ...(iii)
Adding (i) and (iii): \( 15x = 30 \), so \( x = 2 \).
Putting \( x = 2 \) in (i): \( 6 + 4y = -6 \), so \( y = -3 \).
Hence, \( x = 2,\ y = -3 \).
By the substitution method.
\( 3x + 4y = -6 \) ...(i)    \( 3x - y = 9 \) ...(ii)
From (ii): \( y = -9 + 3x \) ...(iii)
Substituting in (i): \( 15x = 30 \), so \( x = 2 \).
Putting this in (iii): \( y = -3 \).
Hence, \( x = 2,\ y = -3 \).
In simple words: Clear the fractions, then solve with both methods.

Teacher's Note:
1. Make the coefficients of one variable equal, then add or subtract.
2. Both methods must give the same answer, so use one to verify the other.

Question 5. Form the pair of linear equations in the following problem and find its solution (if it exists) by the elimination method: If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \( \frac{1}{2} \) if we only add 1 to the denominator. What is the fraction?
Answer:
Let the fraction be \( \frac{x}{y} \). Then \( \frac{x + 1}{y - 1} = 1 \Rightarrow x - y = -2 \) ...(i) and \( \frac{x}{y + 1} = \frac{1}{2} \Rightarrow 2x - y = 1 \) ...(ii).
Subtracting (i) from (ii): \( x = 3 \). Putting in (i): \( 3 - y = -2 \Rightarrow y = 5 \).
Hence the fraction is \( \frac{3}{5} \).
In simple words: Two conditions on the same fraction give two equations in its numerator and denominator.

Teacher's Note:
1. Cross-multiply each condition before eliminating.
2. Check: \( \frac{3+1}{5-1} = 1 \) and \( \frac{3}{5+1} = \frac{1}{2} \).

Question 6. Form the pair of linear equations in the following problem and find its solution (if it exists) by the elimination method: Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Answer:
Let Nuri's present age be x years and Sonu's be y years. Five years ago: \( x - 5 = 3(y - 5) \Rightarrow x - 3y = -10 \) ...(i). Ten years later: \( x + 10 = 2(y + 10) \Rightarrow x - 2y = 10 \) ...(ii).
Subtracting (i) from (ii): \( y = 20 \). Putting in (ii): \( x = 50 \).
Hence Nuri is 50 years and Sonu is 20 years old.
In simple words: Write one age equation for the past and one for the future.

Teacher's Note:
1. Subtract years from both ages for the past and add for the future.
2. Check: 45 = 3 × 15 and 60 = 2 × 30.

Question 7. Form the pair of linear equations in the following problem and find its solution (if it exists) by the elimination method: The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Answer:
Let the units digit be x and the tens digit be y, so the number is \( 10y + x \). Then \( x + y = 9 \) ...(i) and \( 9(10y + x) = 2(10x + y) \Rightarrow 88y = 11x \Rightarrow x = 8y \) ...(ii).
Putting (ii) in (i): \( 9y = 9 \Rightarrow y = 1 \), so \( x = 8 \).
Hence the number is 18 (check: 9 × 18 = 162 = 2 × 81).
In simple words: Represent the number as 10 times the tens digit plus the units digit.

Teacher's Note:
1. The reversed number is \( 10x + y \).
2. Always check the result against both conditions.

Question 8. Form the pair of linear equations in the following problem and find its solution (if it exists) by the elimination method: Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.
Answer:
Let the number of Rs. 50 notes be x and the number of Rs. 100 notes be y. Then \( x + y = 25 \) ...(i) and \( 50x + 100y = 2000 \Rightarrow x + 2y = 40 \) ...(ii).
Subtracting (i) from (ii): \( y = 15 \). Then \( x = 10 \).
Meena received 10 notes of Rs. 50 and 15 notes of Rs. 100.
In simple words: Count of notes and total value give the two equations.

Teacher's Note:
1. One equation counts notes, the other adds up the money.
2. Check: 10 × 50 + 15 × 100 = 2000.

Question 9. Form the pair of linear equations in the following problem and find its solution (if it exists) by the elimination method: A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Answer:
Let the fixed charge be Rs. x and the charge for each extra day be Rs. y. Seven days means 4 extra days: \( x + 4y = 27 \) ...(i). Five days means 2 extra days: \( x + 2y = 21 \) ...(ii).
Subtracting (ii) from (i): \( 2y = 6 \Rightarrow y = 3 \). Then \( x = 21 - 6 = 15 \).
The fixed charge is Rs. 15 and the charge for each extra day is Rs. 3.
In simple words: The first three days are covered by the fixed charge, so only the remaining days are charged extra.

Teacher's Note:
1. Subtract the 3 days covered by the fixed charge.
2. Check: 15 + 4(3) = 27 and 15 + 2(3) = 21.

Conditions for Solutions of a Pair of Linear Equations

Question 10. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it (by the elimination method): \( x - 3y - 3 = 0 \text{ and } 3x - 9y - 2 = 0 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{1}{3},\ \frac{b_1}{b_2} = \frac{1}{3},\ \frac{c_1}{c_2} = \frac{3}{2} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel: the pair is inconsistent and has no solution.
In simple words: Compare the ratios of coefficients to classify the pair; solve only if the lines meet in one point.

Teacher's Note:
1. Unique solution: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \); no solution: parallel; infinite: coincident.
2. Do not solve the pair if the ratios already show no solution.

Question 11. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it (by the elimination method): \( 2x + y = 5 \text{ and } 3x + 2y = 8 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{2}{3},\ \frac{b_1}{b_2} = \frac{1}{2},\ \frac{c_1}{c_2} = \frac{5}{8} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at exactly one point: the pair is consistent with a unique solution.
Solving by elimination:
\( 2x + y = 5 \) ...(i)    \( 3x + 2y = 8 \) ...(ii)
Multiplying (i) by 2: \( 4x + 2y = 10 \) ...(iii)
Subtracting (ii) from (iii): \( x = 2 \), so \( x = 2 \).
Putting \( x = 2 \) in (i): \( 4 + y = 5 \), so \( y = 1 \).
Hence, \( x = 2,\ y = 1 \).
In simple words: Compare the ratios of coefficients to classify the pair; solve only if the lines meet in one point.

Teacher's Note:
1. Unique solution: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \); no solution: parallel; infinite: coincident.
2. Do not solve the pair if the ratios already show no solution.

Question 12. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it (by the elimination method): \( 3x - 5y = 20 \text{ and } 6x - 10y = 40 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{1}{2},\ \frac{b_1}{b_2} = \frac{1}{2},\ \frac{c_1}{c_2} = \frac{1}{2} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident: the pair is consistent with infinitely many solutions.
In simple words: Compare the ratios of coefficients to classify the pair; solve only if the lines meet in one point.

Teacher's Note:
1. Unique solution: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \); no solution: parallel; infinite: coincident.
2. Do not solve the pair if the ratios already show no solution.

Question 13. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it (by the elimination method): \( x - 3y - 7 = 0 \text{ and } 3x - 3y - 15 = 0 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{1}{3},\ \frac{b_1}{b_2} = 1,\ \frac{c_1}{c_2} = \frac{7}{15} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at exactly one point: the pair is consistent with a unique solution.
Solving by elimination:
\( x - 3y = 7 \) ...(i)    \( 3x - 3y = 15 \) ...(ii)
Subtracting (ii) from (i): \( -2x = -8 \), so \( x = 4 \).
Putting \( x = 4 \) in (i): \( 4 - 3y = 7 \), so \( y = -1 \).
Hence, \( x = 4,\ y = -1 \).
In simple words: Compare the ratios of coefficients to classify the pair; solve only if the lines meet in one point.

Teacher's Note:
1. Unique solution: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \); no solution: parallel; infinite: coincident.
2. Do not solve the pair if the ratios already show no solution.

Question 14. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? \( 2x + 3y = 7 \) and \( (a - b)x + (a + b)y = 3a + b - 2 \).
Answer:
For infinitely many solutions \( \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3a + b - 2} \).
From \( \frac{2}{a - b} = \frac{3}{a + b} \): \( 2a + 2b = 3a - 3b \Rightarrow a = 5b \) ...(i).
From \( \frac{2}{a - b} = \frac{7}{3a + b - 2} \): \( 6a + 2b - 4 = 7a - 7b \Rightarrow a - 9b = -4 \) ...(ii).
Putting \( a = 5b \) in (ii): \( -4b = -4 \Rightarrow b = 1 \), so \( a = 5 \).
Hence a = 5 and b = 1.
In simple words: Equate all three ratios and solve the two resulting equations for a and b.

Teacher's Note:
1. Infinitely many solutions means the three ratios are equal.
2. Check: with a = 5, b = 1 the second equation is \( 4x + 6y = 14 \), twice the first.

Question 15. (ii) For which value of k will the following pair of linear equations have no solution? \( 3x + y = 1 \) and \( (2k - 1)x + (k - 1)y = 2k + 1 \).
Answer:
For no solution \( \frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{1}{2k + 1} \).
\( 3(k - 1) = 2k - 1 \Rightarrow 3k - 3 = 2k - 1 \Rightarrow k = 2 \).
For k = 2 the third ratio is \( \frac{1}{5} \), and \( \frac{3}{3} = 1 \neq \frac{1}{5} \), so the condition holds.
Hence k = 2.
In simple words: Make the x and y ratios equal and then confirm that the constant ratio is different.

Teacher's Note:
1. Always verify that the third ratio is not equal, otherwise the lines would coincide.
2. Cross-multiplying the first equality gives a linear equation in k.

Question 16. Solve the following pair of linear equations by the substitution method and by the elimination method: \( 8x + 5y = 9 \), \( 3x + 2y = 4 \).
Answer:
Substitution method.
\( 8x + 5y = 9 \) ...(i)    \( 3x + 2y = 4 \) ...(ii)
From (i): \( y = \frac{9}{5} - \frac{8}{5}x \) ...(iii)
Substituting in (ii): \( -\frac{1}{5}x = \frac{2}{5} \), so \( x = -2 \).
Putting this in (iii): \( y = 5 \).
Hence, \( x = -2,\ y = 5 \).
Elimination method.
\( 8x + 5y = 9 \) ...(i)    \( 3x + 2y = 4 \) ...(ii)
Multiplying (i) by 2: \( 16x + 10y = 18 \) ...(iii)
Multiplying (ii) by 5: \( 15x + 10y = 20 \) ...(iv)
Subtracting (iv) from (iii): \( x = -2 \), so \( x = -2 \).
Putting \( x = -2 \) in (i): \( -16 + 5y = 9 \), so \( y = 5 \).
Hence, \( x = -2,\ y = 5 \).
In simple words: Solve it twice, once with each method; the answers agree.

Teacher's Note:
1. The book asks for the cross-multiplication method as well; the elimination method gives the same result.
2. Verify in both equations: \( 8(-2) + 5(5) = 9 \) and \( 3(-2) + 2(5) = 4 \).