Equations Reducible to Linear Form | Class 10 Maths

Quick answer: Some equations are not linear but can be reduced to linear form by substitution. If the variables appear as $\frac{1}{x}$ and $\frac{1}{y}$, put $u = \frac{1}{x}$ and $v = \frac{1}{y}$; if they appear as $\frac{1}{x + y}$ and $\frac{1}{x - y}$, put $u = \frac{1}{x + y}$ and $v = \frac{1}{x - y}$. Solve the new linear equations for $u$ and $v$, then substitute back to find $x$ and $y$.

Some pairs of equations have the variables in the denominators, for example:

$\frac{2}{x} + \frac{3}{y} = 13 \qquad \frac{5}{x} - \frac{4}{y} = -2$

These are not linear equations in $x$ and $y$. But if we replace $\frac{1}{x}$ by $u$ and $\frac{1}{y}$ by $v$, they become a pair of linear equations in $u$ and $v$, which we already know how to solve.

Syllabus note: The section on equations reducible to a pair of linear equations has been removed from the current rationalised NCERT Class 10 textbook (Chapter 3, Pair of Linear Equations in Two Variables). These problems still appear in older papers, NCERT Exemplar and competitive exams. Check your current syllabus before your exam.

What You Will Learn

• How substitution turns these equations into linear equations
• Type 1: $u = \frac{1}{x}$, $v = \frac{1}{y}$ (and the square-root version)
• Type 2: $u = \frac{1}{x + y}$, $v = \frac{1}{x - y}$
• Three fully worked examples
• Why you must substitute back at the end

1. The Method

Step 1: Substitute. Replace the repeated expressions in the denominators with new variables $u$ and $v$.

Step 2: Solve for $u$ and $v$. Use elimination or substitution on the new linear equations.

Step 3: Substitute back. Use the definitions of $u$ and $v$ to find $x$ and $y$.

2. Example 1: Reciprocals of x and y

Question

Solve: $\frac{2}{x} + \frac{3}{y} = 13$ and $\frac{5}{x} - \frac{4}{y} = -2$.

Solution

Step 1: Substitute $u = \frac{1}{x}$ and $v = \frac{1}{y}$:

$2u + 3v = 13 \qquad \text{...(1)}$

$5u - 4v = -2 \qquad \text{...(2)}$

Step 2: Eliminate $v$. Multiply (1) by $4$ and (2) by $3$, then add:

$8u + 12v = 52 \qquad 15u - 12v = -6$

$23u = 46 \implies u = 2$

From (1): $4 + 3v = 13 \implies v = 3$.

Step 3: Substitute back.

$x = \frac{1}{u} = \frac{1}{2} \qquad y = \frac{1}{v} = \frac{1}{3}$

Answer: $x = \frac{1}{2}$, $y = \frac{1}{3}$.

Check: $\frac{2}{1/2} + \frac{3}{1/3} = 4 + 9 = 13$. ✓

3. Example 2: Square Roots in the Denominators

Question

Solve: $\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2$ and $\frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1$.

Solution

Step 1: Substitute $u = \frac{1}{\sqrt{x}}$ and $v = \frac{1}{\sqrt{y}}$:

$2u + 3v = 2 \qquad \text{...(1)}$

$4u - 9v = -1 \qquad \text{...(2)}$

Step 2: Eliminate $v$. Multiply (1) by $3$ and add to (2):

$6u + 9v = 6$

$10u = 5 \implies u = \frac{1}{2}$

From (1): $1 + 3v = 2 \implies v = \frac{1}{3}$.

Step 3: Substitute back.

$\sqrt{x} = 2 \implies x = 4 \qquad \sqrt{y} = 3 \implies y = 9$

Answer: $x = 4$, $y = 9$.

Check: $\frac{2}{2} + \frac{3}{3} = 2$ and $\frac{4}{2} - \frac{9}{3} = -1$. ✓

4. Example 3: Denominators x + y and x − y

Question

Solve: $\frac{10}{x + y} + \frac{2}{x - y} = 4$ and $\frac{15}{x + y} - \frac{5}{x - y} = -2$.

Solution

Step 1: Substitute $u = \frac{1}{x + y}$ and $v = \frac{1}{x - y}$:

$10u + 2v = 4 \implies 5u + v = 2 \qquad \text{...(1)}$

$15u - 5v = -2 \qquad \text{...(2)}$

Step 2: Solve for $u$ and $v$. From (1), $v = 2 - 5u$. Substituting in (2):

$15u - 5(2 - 5u) = -2 \implies 40u = 8 \implies u = \frac{1}{5}$

$v = 2 - 5 \times \frac{1}{5} = 1$

Step 3: Substitute back. This gives a second pair of linear equations:

$x + y = 5 \qquad \text{...(3)}$

$x - y = 1 \qquad \text{...(4)}$

Step 4: Solve the second pair. Adding (3) and (4): $2x = 6 \implies x = 3$, and then $y = 2$.

Answer: $x = 3$, $y = 2$.

Check: $\frac{10}{5} + \frac{2}{1} = 4$ and $\frac{15}{5} - \frac{5}{1} = -2$. ✓

5. Summary Table

6. Exam Tips and Common Mistakes

Exam Tip: Finding $u$ and $v$ is only part of the answer. Always substitute back to find $x$ and $y$.

Exam Tip: In Type 2 problems, substituting back gives a second pair of linear equations ($x + y = \dots$ and $x - y = \dots$). Solve it to finish.

Common Mistake: Clearing all the denominators by cross-multiplying. This creates terms like $x^2 - y^2$ and makes the problem much harder. Use substitution instead.

Common Mistake: Forgetting to square in the square-root type. If $\sqrt{x} = 2$, then $x = 4$, not $2$.

Frequently Asked Questions

What are equations reducible to a pair of linear equations?

They are equations that are not linear in x and y, but become linear after a substitution such as u = 1/x and v = 1/y.

Which substitution is used for 1/(x + y) and 1/(x − y)?

Put u = 1/(x + y) and v = 1/(x − y). After finding u and v, solve x + y = 1/u and x − y = 1/v.

Why do we substitute back at the end?

The question asks for x and y, not for u and v, so we use the definitions of u and v to find x and y.

Is this topic in the current Class 10 syllabus?

It has been removed from the rationalised NCERT Class 10 textbook, but it is still useful for older papers, NCERT Exemplar and competitive exams.