Class 10 Maths Chapter 3.3 in English Medium – Solutions | SEBA/SCERT Assam 2026
Complete Guide for Class 10 Maths Chapter 3.3
Get complete Class 10 Maths Chapter 3.3 in English Medium solutions based on the latest SEBA/SCERT Assam 2026 syllabus. Learn Algebraic Methods of Solving a Pair of Linear Equations in Two Variables including Substitution Method, Elimination Method, and Cross-Multiplication Method with step-by-step explanations, solved examples, and free PDF notes from Digital Pipal Academy.
Are you searching for the best and most reliable Class 10 Maths Chapter 3.3 English Medium solutions? You’re in the right place!
Digital Pipal Academy brings you a complete and easy-to-understand guide for Algebraic Methods of solving Pair of Linear Equations in Two Variables, specially designed as per the latest SEBA/SCERT Assam 2026 syllabus.
This chapter is extremely important for your board exams because it teaches you how to solve linear equations using mathematical methods instead of graphs. Many students find algebraic methods tricky, but with our clear step-by-step explanations, you can master them easily.
1: Solve the following pair of linear equations by the elimination method and the substitution method:
(i) \(x + y = 5\) and \(2x - 3y = 4\)
(ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)
(iii) \(3x - 5y - 4 = 0\) and \(9x = 2y + 7\)
(iv) \(\frac{x}{2} + \frac{2y}{3} = -1\) and \(x - \frac{y}{3} = 3\)
(v) \(\frac{3y}{2} - \frac{5x}{3} = -2\) and \(\frac{y}{3} + \frac{x}{3} = \frac{13}{6}\)
(vi) \(x - y = 3\) and \(\frac{x}{3} + \frac{y}{2} = 6\)
(i) \(x + y = 5\) and \(2x - 3y = 4\)
By the method of elimination.
\(x + y = 5 \qquad (i)\)
\(2x - 3y = 4 \qquad (ii)\)
When the equation (i) is multiplied by 2, we get
\(2x + 2y = 10 \qquad (iii)\)
When equation (ii) is subtracted from (iii), we get
\(5y = 6\)
\(y = \frac{6}{5} \qquad (iv)\)
Substituting the value of \(y\) in eq. (i) we get
\(x = 5 - \frac{6}{5} = \frac{19}{5}\)
\(\therefore x = \frac{19}{5}, \; y = \frac{6}{5}\)
By the method of substitution.
From equation (i), we get
\(x = 5 - y \qquad (v)\)
Substituting in equation (ii), we get
\(2(5 - y) - 3y = 4\)
\(-5y = -6\)
\(y = \frac{6}{5}\)
Substituting in (v), we get
\(x = 5 - \frac{6}{5} = \frac{19}{5}\)
\(\therefore x = \frac{19}{5}, \; y = \frac{6}{5}\)
(ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)
By the method of elimination.
\(3x + 4y = 10 \qquad (i)\)
\(2x - 2y = 2 \qquad (ii)\)
Multiply equation (ii) by 2:
\(4x - 4y = 4 \qquad (iii)\)
Adding (i) and (iii):
\(7x = 14\)
\(x = 2 \qquad (iv)\)
Substituting in (i):
\(6 + 4y = 10\)
\(4y = 4\)
\(y = 1\)
Hence, \(x = 2\) and \(y = 1\)
By the method of substitution.
From (ii):
\(x = 1 + y \qquad (v)\)
Substitute in (i):
\(3(1 + y) + 4y = 10\)
\(7y = 7\)
\(y = 1\)
Substitute in (v):
\(x = 2\)
\(\therefore x = 2,\; y = 1\)
(iii) \(3x - 5y - 4 = 0\) and \(9x = 2y + 7\)
By the method of elimination.
\(3x - 5y - 4 = 0 \qquad (i)\)
\(9x - 2y - 7 = 0 \qquad (ii)\)
Multiply (i) by 3:
\(9x - 15y - 12 = 0 \qquad (iii)\)
Subtract (iii) from (ii):
\(13y = -5\)
\(y = -\frac{5}{13} \qquad (iv)\)
Substitute in (i):
\(3x + \frac{25}{13} - 4 = 0\)
\(3x = \frac{27}{13}\)
\(x = \frac{9}{13}\)
\(\therefore x = \frac{9}{13},\; y = -\frac{5}{13}\)
By the method of substitution.
From (i):
\(x = \frac{5y + 4}{3} \qquad (v)\)
Substitute in (ii):
\(9\left(\frac{5y + 4}{3}\right) - 2y - 7 = 0\)
\(13y = -5\)
\(y = -\frac{5}{13}\)
Substitute in (v):
\(x = \frac{9}{13}\)
\(\therefore x = \frac{9}{13},\; y = -\frac{5}{13}\)
(iv) \(\frac{x}{2} + \frac{2y}{3} = -1\) and \(x - \frac{y}{3} = 3\)
By the method of elimination.
\(3x + 4y = -6 \qquad (i)\)
\(3x - y = 9 \qquad (ii)\)
Subtract (ii) from (i):
\(5y = -15\)
\(y = -3 \qquad (iii)\)
Substitute in (i):
\(3x - 12 = -6\)
\(3x = 6\)
\(x = 2\)
Hence, \(x = 2,\; y = -3\)
By the method of substitution.
From (ii):
\(x = \frac{y + 9}{3} \qquad (v)\)
Substitute in (i):
\(3\left(\frac{y + 9}{3}\right) + 4y = -6\)
\(5y = -15\)
\(y = -3\)
Substitute in (v):
\(x = 2\)
\(\therefore x = 2,\; y = -3\)
(v) \(\frac{3y}{2} - \frac{5x}{3} = -2\) and \(\frac{y}{3} + \frac{x}{3} = \frac{13}{6}\)
By the method of elimination.
\(9y - 10x = -12 \qquad (i)\)
\(2y + 2x = 13 \qquad (ii)\)
Multiply (ii) by 5:
\(10y + 10x = 65 \qquad (iii)\)
Add (i) and (iii):
\(19y = 53\)
\(y = \frac{53}{19}\)
Substitute in (ii):
\(2x = 13 - \frac{106}{19} = \frac{141}{19}\)
\(x = \frac{141}{38}\)
\(\therefore x = \frac{141}{38},\; y = \frac{53}{19}\)
By the method of substitution.
\(x = \frac{13 - 2y}{2}\)
Substitute:
\(9y - 10\left(\frac{13 - 2y}{2}\right) = -12\)
\(19y = 53\)
\(y = \frac{53}{19}\)
\(x = \frac{141}{38}\)
(vi) \(x - y = 3\) and \(\frac{x}{3} + \frac{y}{2} = 6\)
By the method of elimination.
\(x - y = 3 \qquad (i)\)
\(2x + 3y = 36 \qquad (ii)\)
Multiply (i) by 3:
\(3x - 3y = 9 \qquad (iii)\)
Add (ii) and (iii):
\(5x = 45\)
\(x = 9\)
Substitute in (i):
\(9 - y = 3\)
\(y = 6\)
\(\therefore x = 9,\; y = 6\)
By the method of substitution.
\(x = y + 3\)
Substitute:
\(2(y + 3) + 3y = 36\)
\(5y = 30\)
\(y = 6\)
\(x = 9\)
\(\therefore x = 9,\; y = 6\)
2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
Solution:
Let the fraction be \( \frac{a}{b} \).
According to the first condition,
\( \frac{a+1}{b-1} = 1 \)
\( \Rightarrow a + 1 = b - 1 \)
\( \Rightarrow a - b = -2 \qquad \dots (i) \)
According to the second condition,
\( \frac{a}{b+1} = \frac{1}{2} \)
\( \Rightarrow 2a = b + 1 \)
\( \Rightarrow 2a - b = 1 \qquad \dots (ii) \)
Subtracting equation (i) from (ii):
\( (2a - b) - (a - b) = 1 - (-2) \)
\( \Rightarrow a = 3 \)
Substituting \( a = 3 \) in equation (i):
\( 3 - b = -2 \)
\( \Rightarrow -b = -5 \)
\( \Rightarrow b = 5 \)
Hence, the required fraction is \( \frac{3}{5} \).
Solution:
Let the present age of Nuri be \( x \) years and the present age of Sonu be \( y \) years.
Five years ago:
\( x - 5 = 3(y - 5) \)
\( \Rightarrow x - 3y = -10 \qquad \dots (1) \)
Ten years later:
\( x + 10 = 2(y + 10) \)
\( \Rightarrow x - 2y = 10 \qquad \dots (2) \)
Subtracting equation (1) from (2):
\( (x - 2y) - (x - 3y) = 10 - (-10) \)
\( \Rightarrow y = 20 \)
Substituting \( y = 20 \) in equation (1):
\( x - 3(20) = -10 \)
\( \Rightarrow x - 60 = -10 \)
\( \Rightarrow x = 50 \)
Therefore, Nuri is 50 years old and Sonu is 20 years old.
Solution:
Let the unit's digit be \( A \) and the ten's digit be \( B \).
Original number \( = 10B + A \).
Reversed number \( = 10A + B \).
According to the problem:
\( A + B = 9 \qquad \dots (i) \)
And,
\( 9(10B + A) = 2(10A + B) \)
\( \Rightarrow 90B + 9A = 20A + 2B \)
\( \Rightarrow 88B - 11A = 0 \)
\( \Rightarrow 8B - A = 0 \qquad \dots (ii) \)
Adding equations (i) and (ii):
\( 9B = 9 \Rightarrow B = 1 \)
Substituting \( B = 1 \) in equation (i):
\( A + 1 = 9 \Rightarrow A = 8 \)
Hence, the number is \( 10(1) + 8 = 18 \).
Solution:
Let the number of Rs. 50 notes be \( A \) and the number of Rs. 100 notes be \( B \).
According to the problem:
\( A + B = 25 \qquad \dots (i) \)
\( 50A + 100B = 2000 \)
\( \Rightarrow A + 2B = 40 \qquad \dots (ii) \) [Dividing by 50]
Subtracting equation (i) from (ii):
\( B = 15 \)
Substituting \( B = 15 \) in equation (i):
\( A + 15 = 25 \Rightarrow A = 10 \)
Hence, Meena received 10 notes of Rs. 50 and 15 notes of Rs. 100.
Solution:
Let the fixed charge for the first three days be Rs. \( A \) and the charge for each extra day be Rs. \( B \).
For Saritha (7 days = 3 fixed + 4 extra):
\( A + 4B = 27 \qquad \dots (i) \)
For Susy (5 days = 3 fixed + 2 extra):
\( A + 2B = 21 \qquad \dots (ii) \)
Subtracting equation (ii) from (i):
\( 2B = 6 \Rightarrow B = 3 \)
Substituting \( B = 3 \) in equation (ii):
\( A + 2(3) = 21 \)
\( \Rightarrow A + 6 = 21 \Rightarrow A = 15 \)
Hence, the fixed charge is Rs. 15 and the charge per extra day is Rs. 3.
Solution:
Let the fraction be \( \frac{a}{b} \).
According to the first condition,
\( \frac{a+1}{b-1} = 1 \)
\( \Rightarrow a + 1 = b - 1 \)
\( \Rightarrow a - b = -2 \qquad \dots (i) \)
According to the second condition,
\( \frac{a}{b+1} = \frac{1}{2} \)
\( \Rightarrow 2a = b + 1 \)
\( \Rightarrow 2a - b = 1 \qquad \dots (ii) \)
Subtracting equation (i) from (ii):
\( (2a - b) - (a - b) = 1 - (-2) \)
\( \Rightarrow a = 3 \)
Substituting \( a = 3 \) in equation (i):
\( 3 - b = -2 \)
\( \Rightarrow -b = -5 \)
\( \Rightarrow b = 5 \)
Hence, the required fraction is \( \frac{3}{5} \).
Solution:
Let the present age of Nuri be \( x \) years and the present age of Sonu be \( y \) years.
Five years ago:
\( x - 5 = 3(y - 5) \)
\( \Rightarrow x - 3y = -10 \qquad \dots (1) \)
Ten years later:
\( x + 10 = 2(y + 10) \)
\( \Rightarrow x - 2y = 10 \qquad \dots (2) \)
Subtracting equation (1) from (2):
\( (x - 2y) - (x - 3y) = 10 - (-10) \)
\( \Rightarrow y = 20 \)
Substituting \( y = 20 \) in equation (1):
\( x - 3(20) = -10 \)
\( \Rightarrow x - 60 = -10 \)
\( \Rightarrow x = 50 \)
Therefore, Nuri is 50 years old and Sonu is 20 years old.
Solution:
Let the unit's digit be \( A \) and the ten's digit be \( B \).
Original number \( = 10B + A \).
Reversed number \( = 10A + B \).
According to the problem:
\( A + B = 9 \qquad \dots (i) \)
And,
\( 9(10B + A) = 2(10A + B) \)
\( \Rightarrow 90B + 9A = 20A + 2B \)
\( \Rightarrow 88B - 11A = 0 \)
\( \Rightarrow 8B - A = 0 \qquad \dots (ii) \)
Adding equations (i) and (ii):
\( 9B = 9 \Rightarrow B = 1 \)
Substituting \( B = 1 \) in equation (i):
\( A + 1 = 9 \Rightarrow A = 8 \)
Hence, the number is \( 10(1) + 8 = 18 \).
Solution:
Let the number of Rs. 50 notes be \( A \) and the number of Rs. 100 notes be \( B \).
According to the problem:
\( A + B = 25 \qquad \dots (i) \)
\( 50A + 100B = 2000 \)
\( \Rightarrow A + 2B = 40 \qquad \dots (ii) \) [Dividing by 50]
Subtracting equation (i) from (ii):
\( B = 15 \)
Substituting \( B = 15 \) in equation (i):
\( A + 15 = 25 \Rightarrow A = 10 \)
Hence, Meena received 10 notes of Rs. 50 and 15 notes of Rs. 100.
Solution:
Let the fixed charge for the first three days be Rs. \( A \) and the charge for each extra day be Rs. \( B \).
For Saritha (7 days = 3 fixed + 4 extra):
\( A + 4B = 27 \qquad \dots (i) \)
For Susy (5 days = 3 fixed + 2 extra):
\( A + 2B = 21 \qquad \dots (ii) \)
Subtracting equation (ii) from (i):
\( 2B = 6 \Rightarrow B = 3 \)
Substituting \( B = 3 \) in equation (ii):
\( A + 2(3) = 21 \)
\( \Rightarrow A + 6 = 21 \Rightarrow A = 15 \)
Hence, the fixed charge is Rs. 15 and the charge per extra day is Rs. 3.
3. The ratio of monthly incomes of A and B is \(9:7\) and the ratio of their monthly expenditures is \(4:3\). If each of them saves \(1600\) per month, find their monthly incomes.
Solution:
Let the monthly incomes of A and B be \(9x\) and \(7x\), respectively.
Let their monthly expenditures be \(4y\) and \(3y\), respectively.
We know that: Income - Expenditure = Savings
According to the problem, both save \(1600\) per month. Therefore, we can form the following equations:
\(9x - 4y = 1600 \qquad \dots (i)\)\(7x - 3y = 1600 \qquad \dots (ii)\)
To solve using the elimination method, multiply equation (i) by \(3\) and equation (ii) by \(4\):
\(27x - 12y = 4800 \qquad \dots (iii)\)\(28x - 12y = 6400 \qquad \dots (iv)\)
Subtracting equation (iii) from equation (iv):
\((28x - 27x) - (12y - 12y) = 6400 - 4800\)\(x = 1600\)
Now, we find their monthly incomes:
- Monthly income of A = \(9x = 9 \times 1600 = 14400\)
- Monthly income of B = \(7x = 7 \times 1600 = 11200\)
Hence, the monthly incomes of A and B are 14,400 and 11,200 respectively.
4. Students of a class are made to stand in rows. If \(3\) students are extra in a row, there would be \(1\) row less. If \(3\) students are less in a row, there would be \(2\) rows more. Find the number of students in the class.
Solution:
Let the number of rows be \(x\) and the number of students in each row be \(y\).
Therefore, the total number of students in the class = \(xy\).
According to the first condition, if \(3\) students are extra in a row, there would be \(1\) row less:
\(xy = (x - 1)(y + 3)\)\(xy = xy + 3x - y - 3\)
\(3x - y = 3 \qquad \dots (i)\)
According to the second condition, if \(3\) students are less in a row, there would be \(2\) rows more:
\(xy = (x + 2)(y - 3)\)\(xy = xy - 3x + 2y - 6\)
\(-3x + 2y = 6 \qquad \dots (ii)\)
Adding equation (i) and equation (ii):
\((3x - y) + (-3x + 2y) = 3 + 6\)\(y = 9\)
Substituting \(y = 9\) in equation (i):
\(3x - 9 = 3\)\(3x = 12\)
\(x = 4\)
Total number of students = \(x \times y = 4 \times 9 = 36\).
Hence, the total number of students in the class is 36.
5. A and B each have a certain number of mangoes. A said to B, "If you give me 30 of yours, the number of mangoes I have will be equal to twice the number of mangoes remaining with you." B replied, "If you give me 10 of your mangoes, the number of mangoes I have will be three times the number of mangoes remaining with you." Find the number of mangoes each of them has.
Solution:
Let the number of mangoes with A be \(x\) and the number of mangoes with B be \(y\).
According to the first condition, if B gives 30 mangoes to A:
\(x + 30 = 2(y - 30)\)\(x + 30 = 2y - 60\)
\(x - 2y = -90 \qquad \dots (i)\)
According to the second condition, if A gives 10 mangoes to B:
\(y + 10 = 3(x - 10)\)\(y + 10 = 3x - 30\)
\(3x - y = 40 \qquad \dots (ii)\)
To solve these equations, multiply equation (ii) by 2:
\(6x - 2y = 80 \qquad \dots (iii)\)Subtracting equation (i) from equation (iii):
\((6x - 2y) - (x - 2y) = 80 - (-90)\)\(5x = 170\)
\(x = 34\)
Substituting \(x = 34\) in equation (ii):
\(3(34) - y = 40\)\(102 - y = 40\)
\(y = 62\)
Hence, A has 34 mangoes and B has 62 mangoes.
6. If the pair of equations \(x + ay = b\) and \(ax + y = 1\) has infinitely many solutions, then choose the correct option from the alternatives mentioned below.
(i) \(a = 1,\; b = 1\)
(ii) \(a = -1,\; b = -1\)
(iii) \(a = 1,\; b = -1\)
(iv) \(a = -1,\; b = 1\)
(a) Both (i) and (iii) are true.
(b) Both (ii) and (iv) are true.
(c) Both (i) and (ii) are true.
(d) Both (iii) and (iv) are true.
Solution:
The given pair of linear equations is:
\(1x + ay = b \qquad \dots (1)\)\(ax + 1y = 1 \qquad \dots (2)\)
For a pair of linear equations \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\) to have infinitely many solutions, the condition is:
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
Substituting the values from our equations:
\(\frac{1}{a} = \frac{a}{1} = \frac{b}{1}\)Step 1: Solve for \(a\)
\(\frac{1}{a} = \frac{a}{1}\)\(a^2 = 1\)
\(a = 1 \quad \text{or} \quad a = -1\)
Step 2: Solve for \(b\)
- If \(a = 1\):
From the ratio \(\frac{a}{1} = \frac{b}{1}\), we get \(1 = b\).
So, \(a = 1, b = 1\) is a valid solution. (Option i) - If \(a = -1\):
From the ratio \(\frac{a}{1} = \frac{b}{1}\), we get \(-1 = b\).
So, \(a = -1, b = -1\) is also a valid solution. (Option ii)
Since both (i) and (ii) result in infinitely many solutions, the correct alternative is (c).
Correct Option: (c) Both (i) and (ii) are true.
7. For what value of \(k\) will the following pair of linear equations have infinitely many solutions?
\(kx + 3y - (k - 3) = 0\)
\(12x + ky - k = 0\)
(a) \(k = -6\)
(b) \(k = 4\)
(c) \(k = -4\)
(d) \(k = 6\)
Solution:
The given pair of linear equations is:
\(kx + 3y - (k - 3) = 0 \qquad \dots (1)\)\(12x + ky - k = 0 \qquad \dots (2)\)
For a pair of linear equations \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) to have infinitely many solutions, the condition is:
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
Comparing the given equations with the standard form, we have:
\(a_1 = k, \quad b_1 = 3, \quad c_1 = -(k - 3)\)\(a_2 = 12, \quad b_2 = k, \quad c_2 = -k\)
Applying the condition:
\(\frac{k}{12} = \frac{3}{k} = \frac{-(k - 3)}{-k}\)\(\frac{k}{12} = \frac{3}{k}\)
\(\Rightarrow k^2 = 36\)
\(\Rightarrow k = \pm 6\)
\(\frac{3}{k} = \frac{k - 3}{k}\)
\(\Rightarrow 3k = k(k - 3)\)
\(\Rightarrow 3k = k^2 - 3k\)
\(\Rightarrow k^2 - 6k = 0\)
\(\Rightarrow k(k - 6) = 0\)
\(\Rightarrow k = 0 \quad \text{or} \quad k = 6\)
Since the value \(k = 6\) satisfies both conditions, the required value is \(k = 6\).
Correct Option: (d) \(k = 6\)
8. The steps for solving a word problem using linear equations in two variables are mentioned below. Choose the correct sequence from the options provided.
(i) Represent the unknowns using variables.
(ii) Solve the equations.
(iii) Form the required equations.
(iv) Interpret the results.
(a) (i) → (ii) → (iii) → (iv)
(b) (i) → (iii) → (ii) → (iv)
(c) (iv) → (iii) → (ii) → (i)
(d) (ii) → (i) → (iii) → (iv)
Solution:
To solve a word problem using linear equations, we follow a logical progression of steps:
- Represent the unknowns: First, identify what needs to be found and assign variables (like x and y) to those unknown quantities.
- Form the equations: Translate the conditions given in the word problem into mathematical statements (equations) using those variables.
- Solve the equations: Use algebraic methods (like substitution or elimination) to find the values of the variables.
- Interpret the results: Finally, relate the numerical values back to the original problem to answer the question asked.
Based on these steps, the correct sequence is (i) → (iii) → (ii) → (iv).
Correct Option: (b) (i) → (iii) → (ii) → (iv)
9. Solve the following pairs of linear equations:
(i)
\(\frac{2}{x} + \frac{3}{y} = 2\)
\(\frac{4}{x} - \frac{9}{y} = -1\)
(ii)
\(\frac{1}{2x} + \frac{1}{3y} = 2\)
\(\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}\)
(iii)
\(\frac{5}{x - 1} + \frac{1}{y - 2} = 2\)
\(\frac{6}{x - 1} - \frac{3}{y - 2} = 1\)
Solution:
(i)
Let \(\frac{1}{x} = u\) and \(\frac{1}{y} = v\). The equations become:\(2u + 3v = 2 \qquad \dots (1)\)
\(4u - 9v = -1 \qquad \dots (2)\)
Multiplying equation (1) by 3:
\(6u + 9v = 6 \qquad \dots (3)\)
Adding equations (2) and (3):
\((4u - 9v) + (6u + 9v) = -1 + 6\)
\(10u = 5 \Rightarrow u = \frac{1}{2}\)
Substituting \(u = \frac{1}{2}\) in equation (1):
\(2(\frac{1}{2}) + 3v = 2 \Rightarrow 1 + 3v = 2 \Rightarrow 3v = 1 \Rightarrow v = \frac{1}{3}\)
Now, \(\frac{1}{x} = \frac{1}{2} \Rightarrow x = 2\) and \(\frac{1}{y} = \frac{1}{3} \Rightarrow y = 3\).
Result: \(x = 2, y = 3\)
(ii)
Let \(\frac{1}{x} = u\) and \(\frac{1}{y} = v\). The equations become:\(\frac{u}{2} + \frac{v}{3} = 2 \Rightarrow 3u + 2v = 12 \qquad \dots (1)\)
\(\frac{u}{3} + \frac{v}{2} = \frac{13}{6} \Rightarrow 2u + 3v = 13 \qquad \dots (2)\)
To use the elimination method, multiply equation (1) by 3 and equation (2) by 2:
\(9u + 6v = 36 \qquad \dots (3)\)
\(4u + 6v = 26 \qquad \dots (4)\)
Subtracting equation (4) from equation (3):
\(5u = 10 \Rightarrow u = 2\)
Substituting \(u = 2\) in equation (1):
\(3(2) + 2v = 12 \Rightarrow 6 + 2v = 12 \Rightarrow 2v = 6 \Rightarrow v = 3\)
Now, \(\frac{1}{x} = 2 \Rightarrow x = \frac{1}{2}\) and \(\frac{1}{y} = 3 \Rightarrow y = \frac{1}{3}\).
Result: \(x = \frac{1}{2}, y = \frac{1}{3}\)
(iii)
Let \(\frac{1}{x - 1} = u\) and \(\frac{1}{y - 2} = v\). The equations become:\(5u + v = 2 \qquad \dots (1)\)
\(6u - 3v = 1 \qquad \dots (2)\)
Multiplying equation (1) by 3:
\(15u + 3v = 6 \qquad \dots (3)\)
Adding equations (2) and (3):
\(21u = 7 \Rightarrow u = \frac{7}{21} = \frac{1}{3}\)
Substituting \(u = \frac{1}{3}\) in equation (1):
\(5(\frac{1}{3}) + v = 2 \Rightarrow v = 2 - \frac{5}{3} \Rightarrow v = \frac{1}{3}\)
Now, \(\frac{1}{x - 1} = \frac{1}{3} \Rightarrow x - 1 = 3 \Rightarrow x = 4\)
And \(\frac{1}{y - 2} = \frac{1}{3} \Rightarrow y - 2 = 3 \Rightarrow y = 5\).
Result: \(x = 4, y = 5\)
10. The sum of the digits of a two-digit number is 9. If the positions of the digits are interchanged, the number increases by 9. Find the number.
Solution:
Let the digit at the tens place be \(x\) and the digit at the units place be \(y\).
The original number is \(10x + y\).
According to the first condition (sum of digits):
\(x + y = 9 \qquad \dots (1)\)
According to the second condition (digits interchanged):
The new number is \(10y + x\).
\((10y + x) - (10x + y) = 9\)
\(9y - 9x = 9\)
\(y - x = 1 \quad \text{or} \quad -x + y = 1 \qquad \dots (2)\)
Adding equations (1) and (2):
\((x + y) + (-x + y) = 9 + 1\)
\(2y = 10 \Rightarrow y = 5\)
Substituting \(y = 5\) in equation (1):
\(x + 5 = 9 \Rightarrow x = 4\)
The number is: \(10(4) + 5 = 45\).
11. The sum of the digits of a two-digit number is 15. If the positions of the digits are interchanged, the number decreases by 27. Find the number.
Solution:
Let the tens digit be \(x\) and the units digit be \(y\).
The original number is \(10x + y\).
According to the first condition:
\(x + y = 15 \qquad \dots (1)\)
According to the second condition:
\((10x + y) - (10y + x) = 27\)
\(9x - 9y = 27\)
\(x - y = 3 \qquad \dots (2)\)
Adding equations (1) and (2):
\(2x = 18 \Rightarrow x = 9\)
Substituting \(x = 9\) in equation (1):
\(9 + y = 15 \Rightarrow y = 6\)
The number is: \(10(9) + 6 = 96\).
12. The difference between two numbers is 4. If three times the larger number is added to twice the smaller number, the sum is 82. Find the two numbers.
Solution:
Let the larger number be \(x\) and the smaller number be \(y\).
According to the first condition:
\(x - y = 4 \Rightarrow x = y + 4 \qquad \dots (1)\)
According to the second condition:
\(3x + 2y = 82 \qquad \dots (2)\)
Substituting the value of \(x\) from (1) into (2):
\(3(y + 4) + 2y = 82\)
\(3y + 12 + 2y = 82\)
\(5y = 70 \Rightarrow y = 14\)
Now, finding \(x\):
\(x = 14 + 4 = 18\)
The two numbers are 18 and 14.
📚 What You Will Learn in Chapter 3.3
Elimination Method
Cross-Multiplication Method
Solving real-life problems using linear equations
Step-by-step exam-oriented solutions
📚 Focus Keywords
Class 10 Maths 3.3 English Medium
SEBA Class 10 Maths Chapter 3.3
Algebraic Methods Class 10 Maths
Pair of Linear Equations English Medium
SCERT Assam Maths Solutions 2026
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✔️ Simple Assamese + English Explanation
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❓ Frequently Asked Questions (FAQs)
Q1. What is Chapter 3.3 in Class 10 Maths Assamese Medium?
Chapter 3.3 focuses on Algebraic Methods of solving a pair of linear equations in two variables.
Q2. Which methods are included in Exercise 3.3?
Substitution Method, Elimination Method, and Cross-Multiplication Method.
Q3. Is Chapter 3.3 important for SEBA exams?
Yes, it is very important and carries significant marks in board exams.
Q4. Which method is easiest to solve linear equations?
The Substitution Method is often considered easiest for beginners, but it depends on the question.
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