Fraction Word Problems in Linear Equations | Class 10 Maths

Quick answer: Let the fraction be $\frac{x}{y}$, turn each condition into an equation, cross-multiply to get two linear equations, and solve them. For example, if adding 2 to both parts gives $\frac{9}{11}$ and adding 3 gives $\frac{5}{6}$, the equations are $11x - 9y = -4$ and $6x - 5y = -3$, so the fraction is 7/9.

How to set up a fraction problem

Step 1: Let the numerator be $x$ and the denominator be $y$, so the fraction is $\frac{x}{y}$.

Step 2: Write each condition as an equation.

Step 3: Cross-multiply to remove fractions and write each equation in the form $ax + by = c$.

Step 4: Solve the pair by elimination or substitution, then write the fraction.

Problem 1: the fraction becomes 9/11 and 5/6

A fraction becomes $\frac{9}{11}$ if 2 is added to both the numerator and the denominator. If 3 is added to both, it becomes $\frac{5}{6}$. Find the fraction.

Step 1: First condition.

$\frac{x + 2}{y + 2} = \frac{9}{11}$

$11x + 22 = 9y + 18$

$11x - 9y = -4 \qquad \text{(1)}$

Step 2: Second condition.

$\frac{x + 3}{y + 3} = \frac{5}{6}$

$6x + 18 = 5y + 15$

$6x - 5y = -3 \qquad \text{(2)}$

Step 3: Eliminate $y$. Multiply (1) by 5 and (2) by 9:

$55x - 45y = -20$

$54x - 45y = -27$

Subtract:

$x = 7$

Step 4: Find $y$. From (2): $42 - 5y = -3$, so $y = 9$.

Answer: The fraction is $\frac{7}{9}$. Check: $\frac{9}{11}$ ✓ and $\frac{10}{12} = \frac{5}{6}$ ✓.

Problem 2: the fraction becomes 1/3 and 1/4

A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator, and $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.

Step 1: First condition.

$\frac{x - 1}{y} = \frac{1}{3} \quad \Rightarrow \quad 3x - y = 3 \qquad \text{(1)}$

Step 2: Second condition.

$\frac{x}{y + 8} = \frac{1}{4} \quad \Rightarrow \quad 4x - y = 8 \qquad \text{(2)}$

Step 3: Subtract (1) from (2).

$x = 5$

Step 4: Find $y$. From (1): $y = 3 \times 5 - 3 = 12$.

Answer: The fraction is $\frac{5}{12}$. Check: $\frac{4}{12} = \frac{1}{3}$ ✓ and $\frac{5}{20} = \frac{1}{4}$ ✓.

Summary table

Exam tips

• Start with "Let the fraction be $\frac{x}{y}$".
• Read each condition carefully: is the change made to the numerator, the denominator, or both?
• Choose elimination when the coefficients can be matched easily; use substitution when one variable has coefficient 1.
• Finish by writing the fraction, not just the values of $x$ and $y$, and check it against both conditions.

Common mistakes

• Adding the number only to the numerator when the question says "both".
• Cross-multiplying wrongly, for example writing $9(x + 2) = 11(y + 2)$ instead of $11(x + 2) = 9(y + 2)$.
• Sign errors when moving terms, such as $11x - 9y = 4$ instead of $-4$.
• Simplifying the answer when the question expects the original fraction (both 7/9 and 5/12 are already in lowest terms here).

Frequently Asked Questions

How do you solve fraction word problems with linear equations?

Let the fraction be x/y, write each condition as an equation, cross-multiply to get two linear equations, solve them, and write the fraction.

What is the fraction that becomes 9/11 when 2 is added to both parts and 5/6 when 3 is added?

The fraction is 7/9. The conditions give 11x − 9y = −4 and 6x − 5y = −3, which solve to x = 7 and y = 9.

What is the fraction that becomes 1/3 when 1 is subtracted from the numerator and 1/4 when 8 is added to the denominator?

The fraction is 5/12, from the equations 3x − y = 3 and 4x − y = 8.

Should I use elimination or substitution?

Either works. Elimination is quicker when coefficients can be matched easily; substitution is quicker when one variable has a coefficient of 1.

How do I check my answer?

Substitute the numerator and denominator back into both conditions in the question and confirm that you get the given fractions.