Reducing Equations to Linear Form

Class 10Pair of Linear Equations in Two Variables

Some pairs of equations are not linear in x and y, but can be converted to linear form by suitable substitution. The most common substitution is replacing 1/x with u and 1/y with v.

This topic is part of Chapter 3 — Pair of Linear Equations in Two Variables in the CBSE Class 10 syllabus. These problems appear as 4-5 mark questions in board exams.

Once the substitution converts the equations to linear form, standard methods (substitution, elimination, or cross-multiplication) are used to solve for the new variables. The final step is back-substitution to find the original unknowns x and y.

What is Reducing Equations to Linear Form?

Definition: Reducing to linear form is the process of transforming a pair of non-linear equations into a pair of linear equations by introducing new variables through substitution.

Common patterns that can be reduced:

Equations with 1/x and 1/y → substitute u = 1/x, v = 1/y.
Equations with 1/(x+a) and 1/(y+b) → substitute u = 1/(x+a), v = 1/(y+b).
Equations with √x and √y → substitute u = √x, v = √y.
Equations with x² and y² → substitute u = x², v = y² (if resulting equations are linear in u, v).

Reducing Equations to Linear Form Formula

Standard Substitution:

Let u = 1/x and v = 1/y

Process:

Step	Action
1	Identify the non-linear terms (1/x, 1/y, etc.)
2	Introduce substitution: u = 1/x, v = 1/y
3	Rewrite equations in terms of u and v (now linear)
4	Solve the linear system for u and v
5	Back-substitute: x = 1/u, y = 1/v

Conditions:

x ≠ 0 and y ≠ 0 (otherwise 1/x and 1/y are undefined).
After substitution, the resulting equations must be linear in u and v.

Derivation and Proof

Why does this method work?

An equation like a/x + b/y = c is not linear in x and y.
But if we write u = 1/x and v = 1/y, it becomes: au + bv = c — which IS linear in u and v.
Similarly, the second equation p/x + q/y = r becomes pu + qv = r.
Now we have a pair of linear equations in u and v, which can be solved by any standard method.
After finding u and v, we recover x = 1/u and y = 1/v.

Generalisation:

The same principle applies whenever a substitution converts non-linear expressions into linear ones.
For 1/(x+a) terms: let u = 1/(x+a). After solving, x = (1/u) − a.
The choice of substitution depends on the form of the given equations.

Types and Properties

Type 1: Equations with 1/x and 1/y

Example: 2/x + 3/y = 13, 5/x − 4/y = −2.
Substitution: u = 1/x, v = 1/y.

Type 2: Equations with 1/(x+a) and 1/(y+b)

Example: 1/(3x+y) + 1/(3x−y) = 3/4.
Substitution: u = 1/(3x+y), v = 1/(3x−y).

Type 3: Mixed equations (ax + b/y = c form)

Example: 4x + 6/y = 15, 6x − 8/y = 14.
Treat x as usual, substitute v = 1/y only.

Type 4: Word problems leading to 1/x, 1/y equations

Speed-distance-time problems where 1/speed appears naturally.
Work-rate problems where 1/(time to complete) is the rate.

Methods

Steps to solve by reducing to linear form:

Identify the non-linear terms in both equations.
Choose the substitution that makes both equations linear.
Rewrite both equations using the new variables.
Solve the linear system using elimination or substitution method.
Back-substitute to find x and y from u and v.
Verify by substituting x and y into the original equations.

Tips:

Write the substitution clearly at the start.
State that x ≠ 0 and y ≠ 0 (required for 1/x and 1/y to exist).
After finding u and v, check that they are non-zero (otherwise x or y would be undefined).
Always verify the solution in the original equations (not the reduced ones).

Solved Examples

Example 1: Basic 1/x and 1/y Substitution

Problem: Solve: 2/x + 3/y = 13 and 5/x − 4/y = −2, where x ≠ 0, y ≠ 0.

Solution:

Let u = 1/x, v = 1/y.

Equations become:

2u + 3v = 13 ... (i)
5u − 4v = −2 ... (ii)

Solving by elimination:

Multiply (i) by 4: 8u + 12v = 52
Multiply (ii) by 3: 15u − 12v = −6
Adding: 23u = 46 → u = 2
From (i): 2(2) + 3v = 13 → 3v = 9 → v = 3

Back-substitution:

x = 1/u = 1/2
y = 1/v = 1/3

Verification: 2/(1/2) + 3/(1/3) = 4 + 9 = 13. Correct.

Answer: x = 1/2, y = 1/3.

Example 2: NCERT Exercise Type

Problem: Solve: 1/(2x) + 1/(3y) = 2 and 1/(3x) + 1/(2y) = 13/6, where x ≠ 0, y ≠ 0.

Solution:

Let u = 1/x, v = 1/y.

Equations become:

u/2 + v/3 = 2 ... (i)
u/3 + v/2 = 13/6 ... (ii)

Clear fractions:

(i) × 6: 3u + 2v = 12 ... (iii)
(ii) × 6: 2u + 3v = 13 ... (iv)

Solving:

(iii) × 3: 9u + 6v = 36
(iv) × 2: 4u + 6v = 26
Subtracting: 5u = 10 → u = 2
From (iii): 6 + 2v = 12 → v = 3

Back-substitution:

x = 1/u = 1/2
y = 1/v = 1/3

Answer: x = 1/2, y = 1/3.

Example 3: Substitution with (x+y) and (x−y)

Problem: Solve: 1/(x+y) + 2/(x−y) = 2 and 2/(x+y) + 1/(x−y) = 3, where x ≠ ±y.

Solution:

Let u = 1/(x+y), v = 1/(x−y).

Equations become:

u + 2v = 2 ... (i)
2u + v = 3 ... (ii)

Solving:

(i) × 2: 2u + 4v = 4
Subtracting (ii): 3v = 1 → v = 1/3
From (i): u + 2/3 = 2 → u = 4/3

Back-substitution:

x + y = 1/u = 3/4
x − y = 1/v = 3

Solving this simple system:

Adding: 2x = 3/4 + 3 = 15/4 → x = 15/8
Subtracting: 2y = 3/4 − 3 = −9/4 → y = −9/8

Answer: x = 15/8, y = −9/8.

Example 4: Speed-Distance Word Problem

Problem: A boat goes 30 km upstream and 44 km downstream in 10 hours. It can go 40 km upstream and 55 km downstream in 13 hours. Determine the speed of the stream and that of the boat in still water.

Solution:

Given:

Let speed of boat in still water = x km/h
Let speed of stream = y km/h
Upstream speed = (x − y) km/h
Downstream speed = (x + y) km/h

Setting up equations (time = distance/speed):

30/(x−y) + 44/(x+y) = 10 ... (i)
40/(x−y) + 55/(x+y) = 13 ... (ii)

Let u = 1/(x−y), v = 1/(x+y):

30u + 44v = 10 ... (iii)
40u + 55v = 13 ... (iv)

Solving:

(iii) × 4: 120u + 176v = 40
(iv) × 3: 120u + 165v = 39
Subtracting: 11v = 1 → v = 1/11
From (iii): 30u + 4 = 10 → 30u = 6 → u = 1/5

Back-substitution:

x − y = 1/u = 5
x + y = 1/v = 11

Solving:

2x = 16 → x = 8
y = 11 − 8 = 3

Answer: Speed of boat = 8 km/h, speed of stream = 3 km/h.

Example 5: Work Rate Problem

Problem: Ravi can finish a piece of work in x days and Mohan in y days. Together, they can finish it in 12 days. If Ravi worked for 6 days and Mohan for 9 days, half the work is done. Find x and y.

Solution:

Given:

Ravi's rate = 1/x per day, Mohan's rate = 1/y per day

Setting up equations:

1/x + 1/y = 1/12 ... (i)
6/x + 9/y = 1/2 ... (ii)

Let u = 1/x, v = 1/y:

u + v = 1/12 ... (iii)
6u + 9v = 1/2 ... (iv)

Solving:

(iii) × 6: 6u + 6v = 1/2
Subtracting from (iv): 3v = 0 → v = 0? Check: This gives impossible result.
Re-check (iv): 6u + 9v = 1/2
(iii) × 6: 6u + 6v = 6/12 = 1/2
Subtracting: 3v = 0 → v = 0

This means 1/y = 0, which is impossible. Let us reread: "half the work is done."

Correcting equation (ii): 6/x + 9/y = 1/2.

(iii) × 9: 9u + 9v = 9/12 = 3/4.

Subtracting (iv): 3u = 3/4 − 1/2 = 1/4 → u = 1/12.

From (iii): 1/12 + v = 1/12 → v = 0.

The system is inconsistent with the given constraints (Mohan's contribution is zero, contradicting the premise). This indicates the problem conditions need adjustment. Let us modify: suppose together they finish in 15 days.

u + v = 1/15 ... (iii')
6u + 9v = 1/2 ... (iv)

(iii') × 6: 6u + 6v = 6/15 = 2/5
Subtracting from (iv): 3v = 1/2 − 2/5 = 1/10 → v = 1/30
u = 1/15 − 1/30 = 1/30

Back-substitution: x = 30, y = 30.

Answer: Ravi takes 30 days, Mohan takes 30 days.

Example 6: Mixed Type: ax + b/y = c

Problem: Solve: 4x + 6/y = 15 and 6x − 8/y = 14, where y ≠ 0.

Solution:

Let v = 1/y. (x remains as x.)

Equations become:

4x + 6v = 15 ... (i)
6x − 8v = 14 ... (ii)

Solving by elimination:

(i) × 4: 16x + 24v = 60
(ii) × 3: 18x − 24v = 42
Adding: 34x = 102 → x = 3
From (i): 12 + 6v = 15 → 6v = 3 → v = 1/2

Back-substitution:

y = 1/v = 2

Verification: 4(3) + 6/2 = 12 + 3 = 15. 6(3) − 8/2 = 18 − 4 = 14. Both correct.

Answer: x = 3, y = 2.

Example 7: NCERT Textbook Example

Problem: Solve: 6x + 3y = 6xy and 2x + 4y = 5xy, where x ≠ 0, y ≠ 0.

Solution:

Divide both equations by xy:

6x/(xy) + 3y/(xy) = 6 → 6/y + 3/x = 6
2x/(xy) + 4y/(xy) = 5 → 2/y + 4/x = 5

Let u = 1/x, v = 1/y:

3u + 6v = 6 → u + 2v = 2 ... (i)
4u + 2v = 5 ... (ii)

Solving:

Subtracting (i) from (ii): 3u = 3 → u = 1
From (i): 1 + 2v = 2 → v = 1/2

Back-substitution:

x = 1/u = 1
y = 1/v = 2

Verification: 6(1) + 3(2) = 6(1)(2) → 12 = 12. Correct.

Answer: x = 1, y = 2.

Example 8: Train Problem

Problem: A train covered a distance of 300 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 2 hours less. Find the speed of the train.

Solution:

Given:

Let speed = s km/h, time = t hours
s × t = 300 → t = 300/s
(s + 5)(t − 2) = 300

Steps:

(s + 5)(300/s − 2) = 300
300 − 2s + 1500/s − 10 = 300
−2s + 1500/s − 10 = 0
Multiply by s: −2s² + 1500 − 10s = 0
2s² + 10s − 1500 = 0
s² + 5s − 750 = 0
s = (−5 ± √(25 + 3000))/2 = (−5 ± √3025)/2 = (−5 ± 55)/2
s = 25 or s = −30 (rejected, speed is positive)

Answer: The speed of the train is 25 km/h.

Example 9: Substitution with (ax+b) Terms

Problem: Solve: 5/(x−1) + 1/(y−2) = 2 and 6/(x−1) − 3/(y−2) = 1, where x ≠ 1, y ≠ 2.

Solution:

Let u = 1/(x−1), v = 1/(y−2):

5u + v = 2 ... (i)
6u − 3v = 1 ... (ii)

Solving:

(i) × 3: 15u + 3v = 6
Adding (ii): 21u = 7 → u = 1/3
From (i): 5/3 + v = 2 → v = 1/3

Back-substitution:

x − 1 = 1/u = 3 → x = 4
y − 2 = 1/v = 3 → y = 5

Verification: 5/(4−1) + 1/(5−2) = 5/3 + 1/3 = 2. Correct.

Answer: x = 4, y = 5.

Example 10: Fraction Problem

Problem: Solve: 10/(x+y) + 2/(x−y) = 4 and 15/(x+y) − 5/(x−y) = −2.

Solution:

Let u = 1/(x+y), v = 1/(x−y):

10u + 2v = 4 → 5u + v = 2 ... (i)
15u − 5v = −2 ... (ii)

Solving:

(i) × 5: 25u + 5v = 10
Adding (ii): 40u = 8 → u = 1/5
From (i): 1 + v = 2 → v = 1

Back-substitution:

x + y = 5
x − y = 1

Solving:

2x = 6 → x = 3, y = 2

Answer: x = 3, y = 2.

Real-World Applications

Speed-Distance-Time Problems:

Upstream and downstream problems naturally produce equations with 1/(x+y) and 1/(x−y).

Work-Rate Problems:

If A completes work in x days, A's rate is 1/x. Equations with 1/x and 1/y arise naturally.

Electrical Circuits:

Resistors in parallel: 1/R_total = 1/R₁ + 1/R₂. Similar substitution techniques apply.

Mixture Problems:

Concentration calculations involving reciprocals of volumes or weights.

Key Points to Remember

Equations with 1/x and 1/y can be reduced to linear form by substituting u = 1/x and v = 1/y.
After substitution, solve the resulting linear system by elimination or substitution.
Back-substitute to find x and y from u and v.
Always state that x ≠ 0 and y ≠ 0 (or the relevant conditions for the denominators).
Equations with 1/(x+a) and 1/(y+b) use u = 1/(x+a), v = 1/(y+b).
Verify the solution in the original equations, not the reduced ones.
Speed-distance problems produce 1/(speed ± current) terms — use this method.
Dividing by xy can convert equations like ax + by = cxy into linear form.
The method extends to any substitution that linearises the equations.
This is a frequently tested topic in CBSE board exams (4-5 marks).

Practice Problems

Solve: 1/x + 1/y = 7 and 2/x + 3/y = 17, where x ≠ 0, y ≠ 0.
Solve: 2/(3x+2y) + 3/(3x−2y) = 17/5 and 5/(3x+2y) + 1/(3x−2y) = 2.
A boat goes 24 km upstream and 28 km downstream in 6 hours. It can go 30 km upstream and 21 km downstream in 6 hours 30 minutes. Find the speed of the boat and the stream.
Solve: 3/x − 1/y = −9 and 2/x + 3/y = 5, where x ≠ 0, y ≠ 0.
Solve: 2x + 3/y = 9 and 3x + 2/y = 7, where y ≠ 0.
Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Solve: 4/(x+y) + 3/(x−y) = 14 and 3/(x+y) − 4/(x−y) = 23.
A and B together can complete a task in 24 days. After A works alone for 10 days and B for 12 days, 5/8 of the work is completed. Find the number of days each takes alone.

Frequently Asked Questions

Q1. What does 'reducing to linear form' mean?

It means using a substitution (like u = 1/x, v = 1/y) to convert a pair of non-linear equations into a pair of linear equations, which can then be solved by standard methods.

Q2. What substitution is most commonly used?

The most common substitution is u = 1/x and v = 1/y. For equations with (x+a) or (x−b) in the denominator, use u = 1/(x+a) or u = 1/(x−b).

Q3. Why must x and y be non-zero?

Because the substitution involves 1/x and 1/y. Division by zero is undefined, so x = 0 and y = 0 must be excluded.

Q4. How do you solve boat speed problems?

Let boat speed = x, stream speed = y. Upstream speed = (x−y), downstream speed = (x+y). Time equations give terms like distance/(x−y) and distance/(x+y). Substitute u = 1/(x−y), v = 1/(x+y) and solve.

Q5. What if the problem gives equations like ax + by = cxy?

Divide both sides by xy to get: a/y + b/x = c. Then substitute u = 1/x, v = 1/y to get: bu + av = c, which is linear.

Q6. Can this method be used for more than two variables?

In Class 10, this method is applied only to two variables. Extensions to three or more variables follow the same principle but require more equations.

Q7. How many marks do these problems carry in CBSE exams?

These are typically 4-5 mark questions. Full marks require showing the substitution, solving the linear system, back-substituting, and verifying.

Q8. What is the difference between this and the substitution method?

The substitution method solves a linear system by expressing one variable in terms of the other. Reducing to linear form is a pre-processing step that converts a non-linear system into a linear one. After that, any solving method can be used.

Reducing Equations to Linear Form

What is Reducing Equations to Linear Form?

Reducing Equations to Linear Form Formula

Derivation and Proof

Types and Properties

Methods

Solved Examples

Example 1: Basic 1/x and 1/y Substitution

Example 2: NCERT Exercise Type

Example 3: Substitution with (x+y) and (x−y)

Example 4: Speed-Distance Word Problem

Example 5: Work Rate Problem

Example 6: Mixed Type: ax + b/y = c

Example 7: NCERT Textbook Example

Example 8: Train Problem

Example 9: Substitution with (ax+b) Terms

Example 10: Fraction Problem

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. What does 'reducing to linear form' mean?

Q2. What substitution is most commonly used?

Q3. Why must x and y be non-zero?

Q4. How do you solve boat speed problems?

Q5. What if the problem gives equations like ax + by = cxy?

Q6. Can this method be used for more than two variables?

Q7. How many marks do these problems carry in CBSE exams?

Q8. What is the difference between this and the substitution method?

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