Class 10 Maths Chapter 3 Exercise 3.4 Question 1

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Exercise 3.4 – Question 1: Reducing to Linear Form

📚 About Exercise 3.4

Focus: Solve pairs of equations by reducing them to linear form through substitution.

Key Strategy: When equations contain complex expressions (reciprocals, square roots, fractions), substitute them with simpler variables to create linear equations.

Types covered in this question:

  • Reciprocals: \( \frac{1}{x}, \frac{1}{y} \)
  • Square roots: \( \frac{1}{\sqrt{x}}, \frac{1}{\sqrt{y}} \)
  • Complex fractions: \( \frac{1}{x-1}, \frac{1}{y-2} \)
  • Expressions: \( \frac{1}{x+y}, \frac{1}{x-y} \)

📐 General Method

  1. Identify: Find the complex expressions in the equations
  2. Substitute: Let these expressions equal new variables (u, v, p, q, etc.)
  3. Simplify: Rewrite equations in terms of new variables (now linear)
  4. Solve: Use elimination or substitution to solve for new variables
  5. Back-substitute: Find original variables from new variables
  6. Verify: Check solution in original equations

📑 Question Parts

Part (i) – Reciprocal Equations

Equations:
\( \frac{1}{2x} + \frac{1}{3y} = 2 \) …(1)
\( \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \) …(2)
📝 Note: This is the same as Exercise 3.3 Question 5. We solved it there, so here’s a quick summary.

Step 1: Substitute

Let:

\( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)

Step 2: Rewrite Equations

Equation (1) becomes: \[ \frac{u}{2} + \frac{v}{3} = 2 \]

Multiply by 6: \( 3u + 2v = 12 \) …(3)

Equation (2) becomes: \[ \frac{u}{3} + \frac{v}{2} = \frac{13}{6} \]

Multiply by 6: \( 2u + 3v = 13 \) …(4)

Step 3: Solve

Using elimination:

Multiply (3) by 3: \( 9u + 6v = 36 \)

Multiply (4) by 2: \( 4u + 6v = 26 \)

Subtract: \( 5u = 10 \) → \( u = 2 \)

Substitute back: \( v = 3 \)

Step 4: Find x and y

Since \( u = \frac{1}{x} = 2 \): \[ x = \frac{1}{2} \] Since \( v = \frac{1}{y} = 3 \): \[ y = \frac{1}{3} \]
✅ Answer: \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \)

Part (ii) – Square Root Reciprocals

Equations:
\( \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \) …(1)
\( \frac{4}{\sqrt{x}} – \frac{9}{\sqrt{y}} = -1 \) …(2)

Step 1: Substitute

Let:

\( \frac{1}{\sqrt{x}} = u \) (so \( \sqrt{x} = \frac{1}{u} \) and \( x = \frac{1}{u^2} \))

\( \frac{1}{\sqrt{y}} = v \) (so \( \sqrt{y} = \frac{1}{v} \) and \( y = \frac{1}{v^2} \))

Step 2: Rewrite Equations

Equation (1) becomes: \[ 2u + 3v = 2 \quad \text{…(3)} \] Equation (2) becomes: \[ 4u – 9v = -1 \quad \text{…(4)} \]

Step 3: Solve Using Elimination

Multiply equation (3) by 3: \[ 6u + 9v = 6 \quad \text{…(5)} \] Add equations (4) and (5): \[ (4u – 9v) + (6u + 9v) = -1 + 6 \] \[ 10u = 5 \] \[ u = \frac{1}{2} \]
Substitute \( u = \frac{1}{2} \) in equation (3): \[ 2 \times \frac{1}{2} + 3v = 2 \] \[ 1 + 3v = 2 \] \[ 3v = 1 \] \[ v = \frac{1}{3} \]

Step 4: Find x and y

Since \( u = \frac{1}{\sqrt{x}} = \frac{1}{2} \): \[ \sqrt{x} = 2 \] \[ x = 4 \] Since \( v = \frac{1}{\sqrt{y}} = \frac{1}{3} \): \[ \sqrt{y} = 3 \] \[ y = 9 \]

Step 5: Verification

Check in equation (1): \( \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \)

LHS = \( \frac{2}{\sqrt{4}} + \frac{3}{\sqrt{9}} = \frac{2}{2} + \frac{3}{3} = 1 + 1 = 2 \) = RHS ✓

Check in equation (2): \( \frac{4}{\sqrt{x}} – \frac{9}{\sqrt{y}} = -1 \)

LHS = \( \frac{4}{\sqrt{4}} – \frac{9}{\sqrt{9}} = \frac{4}{2} – \frac{9}{3} = 2 – 3 = -1 \) = RHS ✓

✅ Answer: \( x = 4 \) and \( y = 9 \)

Part (iii) – Mixed Reciprocal and Linear

Equations:
\( \frac{4}{x} + 3y = 14 \) …(1)
\( \frac{3}{x} – 4y = 23 \) …(2)

Step 1: Substitute

Let:

\( \frac{1}{x} = u \) (so \( x = \frac{1}{u} \))

Keep y as it is

Step 2: Rewrite Equations

Equation (1) becomes: \[ 4u + 3y = 14 \quad \text{…(3)} \] Equation (2) becomes: \[ 3u – 4y = 23 \quad \text{…(4)} \]

Step 3: Solve Using Elimination

Multiply equation (3) by 4 and equation (4) by 3:

Equation (3) × 4: \( 16u + 12y = 56 \) …(5)

Equation (4) × 3: \( 9u – 12y = 69 \) …(6)

Add equations (5) and (6): \[ 25u = 125 \] \[ u = 5 \]
Substitute \( u = 5 \) in equation (3): \[ 4(5) + 3y = 14 \] \[ 20 + 3y = 14 \] \[ 3y = -6 \] \[ y = -2 \]

Step 4: Find x

Since \( u = \frac{1}{x} = 5 \): \[ x = \frac{1}{5} \]

Step 5: Verification

Check in equation (1): \( \frac{4}{x} + 3y = 14 \)

LHS = \( \frac{4}{\frac{1}{5}} + 3(-2) = 4 \times 5 – 6 = 20 – 6 = 14 \) = RHS ✓

Check in equation (2): \( \frac{3}{x} – 4y = 23 \)

LHS = \( \frac{3}{\frac{1}{5}} – 4(-2) = 3 \times 5 + 8 = 15 + 8 = 23 \) = RHS ✓

✅ Answer: \( x = \frac{1}{5} \) and \( y = -2 \)

Part (iv) – Complex Fractions

Equations:
\( \frac{5}{x-1} + \frac{1}{y-2} = 2 \) …(1)
\( \frac{6}{x-1} – \frac{3}{y-2} = 1 \) …(2)

Step 1: Substitute

Let:

\( \frac{1}{x-1} = u \) (so \( x – 1 = \frac{1}{u} \) and \( x = \frac{1}{u} + 1 \))

\( \frac{1}{y-2} = v \) (so \( y – 2 = \frac{1}{v} \) and \( y = \frac{1}{v} + 2 \))

Step 2: Rewrite Equations

Equation (1) becomes: \[ 5u + v = 2 \quad \text{…(3)} \] Equation (2) becomes: \[ 6u – 3v = 1 \quad \text{…(4)} \]

Step 3: Solve Using Elimination

Multiply equation (3) by 3: \[ 15u + 3v = 6 \quad \text{…(5)} \] Add equations (4) and (5): \[ (6u – 3v) + (15u + 3v) = 1 + 6 \] \[ 21u = 7 \] \[ u = \frac{1}{3} \]
Substitute \( u = \frac{1}{3} \) in equation (3): \[ 5 \times \frac{1}{3} + v = 2 \] \[ \frac{5}{3} + v = 2 \] \[ v = 2 – \frac{5}{3} = \frac{6 – 5}{3} = \frac{1}{3} \]

Step 4: Find x and y

Since \( u = \frac{1}{x-1} = \frac{1}{3} \): \[ x – 1 = 3 \] \[ x = 4 \] Since \( v = \frac{1}{y-2} = \frac{1}{3} \): \[ y – 2 = 3 \] \[ y = 5 \]

Step 5: Verification

Check in equation (1): \( \frac{5}{x-1} + \frac{1}{y-2} = 2 \)

LHS = \( \frac{5}{4-1} + \frac{1}{5-2} = \frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2 \) = RHS ✓

Check in equation (2): \( \frac{6}{x-1} – \frac{3}{y-2} = 1 \)

LHS = \( \frac{6}{3} – \frac{3}{3} = 2 – 1 = 1 \) = RHS ✓

✅ Answer: \( x = 4 \) and \( y = 5 \)

Part (v) – Algebraic Fractions

Equations:
\( \frac{7x – 2y}{xy} = 5 \) …(1)
\( \frac{8x + 7y}{xy} = 15 \) …(2)

Step 1: Simplify

Equation (1): \[ \frac{7x – 2y}{xy} = 5 \] \[ \frac{7x}{xy} – \frac{2y}{xy} = 5 \] \[ \frac{7}{y} – \frac{2}{x} = 5 \quad \text{…(3)} \] Equation (2): \[ \frac{8x + 7y}{xy} = 15 \] \[ \frac{8x}{xy} + \frac{7y}{xy} = 15 \] \[ \frac{8}{y} + \frac{7}{x} = 15 \quad \text{…(4)} \]

Step 2: Substitute

Let:

\( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)

Equation (3) becomes: \[ 7v – 2u = 5 \quad \text{…(5)} \] Equation (4) becomes: \[ 8v + 7u = 15 \quad \text{…(6)} \]

Step 3: Solve Using Elimination

Multiply equation (5) by 7 and equation (6) by 2:

Equation (5) × 7: \( 49v – 14u = 35 \) …(7)

Equation (6) × 2: \( 16v + 14u = 30 \) …(8)

Add equations (7) and (8): \[ 65v = 65 \] \[ v = 1 \]
Substitute \( v = 1 \) in equation (5): \[ 7(1) – 2u = 5 \] \[ 7 – 2u = 5 \] \[ -2u = -2 \] \[ u = 1 \]

Step 4: Find x and y

Since \( u = \frac{1}{x} = 1 \): \[ x = 1 \] Since \( v = \frac{1}{y} = 1 \): \[ y = 1 \]

Step 5: Verification

Check in equation (1): \( \frac{7x – 2y}{xy} = 5 \)

LHS = \( \frac{7(1) – 2(1)}{1 \times 1} = \frac{7 – 2}{1} = 5 \) = RHS ✓

Check in equation (2): \( \frac{8x + 7y}{xy} = 15 \)

LHS = \( \frac{8(1) + 7(1)}{1 \times 1} = \frac{8 + 7}{1} = 15 \) = RHS ✓

✅ Answer: \( x = 1 \) and \( y = 1 \)

Part (vi) – Product Form

Equations:
\( 6x + 3y = 6xy \) …(1)
\( 2x + 4y = 5xy \) …(2)

Step 1: Divide by xy

Divide equation (1) by xy: \[ \frac{6x + 3y}{xy} = \frac{6xy}{xy} \] \[ \frac{6x}{xy} + \frac{3y}{xy} = 6 \] \[ \frac{6}{y} + \frac{3}{x} = 6 \quad \text{…(3)} \] Divide equation (2) by xy: \[ \frac{2x + 4y}{xy} = \frac{5xy}{xy} \] \[ \frac{2x}{xy} + \frac{4y}{xy} = 5 \] \[ \frac{2}{y} + \frac{4}{x} = 5 \quad \text{…(4)} \]

Step 2: Substitute

Let:

\( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)

Equation (3) becomes: \[ 6v + 3u = 6 \] \[ 3u + 6v = 6 \]

Divide by 3: \( u + 2v = 2 \) …(5)

Equation (4) becomes: \[ 2v + 4u = 5 \] \[ 4u + 2v = 5 \quad \text{…(6)} \]

Step 3: Solve Using Elimination

Multiply equation (5) by 4: \[ 4u + 8v = 8 \quad \text{…(7)} \] Subtract equation (6) from equation (7): \[ (4u + 8v) – (4u + 2v) = 8 – 5 \] \[ 6v = 3 \] \[ v = \frac{1}{2} \]
Substitute \( v = \frac{1}{2} \) in equation (5): \[ u + 2 \times \frac{1}{2} = 2 \] \[ u + 1 = 2 \] \[ u = 1 \]

Step 4: Find x and y

Since \( u = \frac{1}{x} = 1 \): \[ x = 1 \] Since \( v = \frac{1}{y} = \frac{1}{2} \): \[ y = 2 \]

Step 5: Verification

Check in equation (1): \( 6x + 3y = 6xy \)

LHS = \( 6(1) + 3(2) = 6 + 6 = 12 \)

RHS = \( 6(1)(2) = 12 \)

LHS = RHS ✓

Check in equation (2): \( 2x + 4y = 5xy \)

LHS = \( 2(1) + 4(2) = 2 + 8 = 10 \)

RHS = \( 5(1)(2) = 10 \)

LHS = RHS ✓

✅ Answer: \( x = 1 \) and \( y = 2 \)

Part (vii) – Sum and Difference

Equations:
\( \frac{10}{x+y} + \frac{2}{x-y} = 4 \) …(1)
\( \frac{15}{x+y} – \frac{5}{x-y} = -2 \) …(2)

Step 1: Substitute

Let:

\( \frac{1}{x+y} = u \) (so \( x + y = \frac{1}{u} \))

\( \frac{1}{x-y} = v \) (so \( x – y = \frac{1}{v} \))

Step 2: Rewrite Equations

Equation (1) becomes: \[ 10u + 2v = 4 \]

Divide by 2: \( 5u + v = 2 \) …(3)

Equation (2) becomes: \[ 15u – 5v = -2 \quad \text{…(4)} \]

Step 3: Solve Using Elimination

Multiply equation (3) by 5: \[ 25u + 5v = 10 \quad \text{…(5)} \] Add equations (4) and (5): \[ (15u – 5v) + (25u + 5v) = -2 + 10 \] \[ 40u = 8 \] \[ u = \frac{1}{5} \]
Substitute \( u = \frac{1}{5} \) in equation (3): \[ 5 \times \frac{1}{5} + v = 2 \] \[ 1 + v = 2 \] \[ v = 1 \]

Step 4: Find x and y

Since \( u = \frac{1}{x+y} = \frac{1}{5} \): \[ x + y = 5 \quad \text{…(6)} \] Since \( v = \frac{1}{x-y} = 1 \): \[ x – y = 1 \quad \text{…(7)} \]
Add equations (6) and (7): \[ 2x = 6 \] \[ x = 3 \] Substitute in equation (6): \[ 3 + y = 5 \] \[ y = 2 \]

Step 5: Visual Representation

📊 Finding x and y from Sum and Difference

Visual representation of solving x + y = 5 and x – y = 1

Step 6: Verification

Check in equation (1): \( \frac{10}{x+y} + \frac{2}{x-y} = 4 \)

LHS = \( \frac{10}{3+2} + \frac{2}{3-2} = \frac{10}{5} + \frac{2}{1} = 2 + 2 = 4 \) = RHS ✓

Check in equation (2): \( \frac{15}{x+y} - \frac{5}{x-y} = -2 \)

LHS = \( \frac{15}{5} - \frac{5}{1} = 3 - 5 = -2 \) = RHS ✓

✅ Answer: \( x = 3 \) and \( y = 2 \)

📊 Summary of All Solutions

PartTypeSolution
(i)Reciprocals\( x = \frac{1}{2}, y = \frac{1}{3} \)
(ii)Square root reciprocals\( x = 4, y = 9 \)
(iii)Mixed\( x = \frac{1}{5}, y = -2 \)
(iv)Complex fractions\( x = 4, y = 5 \)
(v)Algebraic fractions\( x = 1, y = 1 \)
(vi)Product form\( x = 1, y = 2 \)
(vii)Sum and difference\( x = 3, y = 2 \)

⚠️ Common Mistakes to Avoid

❌ Mistake 1: Forgetting to back-substitute to find original variables.

✅ Correct: Always convert u, v back to x, y at the end.
❌ Mistake 2: Incorrect simplification of complex fractions.

✅ Correct: \( \frac{7x - 2y}{xy} = \frac{7}{y} - \frac{2}{x} \), not \( \frac{7}{x} - \frac{2}{y} \)
❌ Mistake 3: Sign errors when dealing with square roots.

✅ Correct: If \( \frac{1}{\sqrt{x}} = \frac{1}{2} \), then \( \sqrt{x} = 2 \), so \( x = 4 \) (not -4)
❌ Mistake 4: Not verifying the solution in original equations.

✅ Correct: Always substitute back into the original (not simplified) equations.

💡 Key Points to Remember

  • Substitution Strategy:
    • Identify complex expressions in equations
    • Replace with simple variables (u, v, p, q)
    • Solve the linear system
    • Convert back to original variables
  • Common Substitutions:
    • \( \frac{1}{x} = u \) → \( x = \frac{1}{u} \)
    • \( \frac{1}{\sqrt{x}} = u \) → \( \sqrt{x} = \frac{1}{u} \) → \( x = \frac{1}{u^2} \)
    • \( \frac{1}{x-a} = u \) → \( x - a = \frac{1}{u} \) → \( x = \frac{1}{u} + a \)
    • \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \)
  • Simplification Tips:

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