Linear Equations in Two Variables: Word Problems Mastery Guide Class 10

In CBSE Class 10 Mathematics, Chapter 3 (Pair of Linear Equations in Two Variables) features heavily in Section C and Section D of the board examination. While solving a pair of standard equations like $2x + 3y = 7$ using substitution or elimination is mechanical, word problems require converting English sentences into a system of two simultaneous linear equations: $a_1 x + b_1 y + c_1 = 0 \quad \text{and} \quad a_2 x + b_2 y + c_2 = 0$

Every linear equation word problem on the board exam belongs to one of four classical archetypes: Age Problems, Fraction Problems, Fixed and Running Charge Systems, and Reciprocal Upstream-Downstream Systems.

In this guide, we break down the translation templates for each archetype and work through fully solved board examination problems.

What You Will Learn

• The 4 foundational linear algebraic word problem blueprints
• Archetype 1: Age Problems (Past vs. Future relations)
• Archetype 2: Fraction Manipulation Problems (Numerator and Denominator equations)
• Archetype 3: Fixed Charges and Per-Day / Per-Kilometre Rates (Hostel mess & Taxi fares)
• Archetype 4: Reciprocal Systems (Solving $\frac{10}{x+y} + \frac{2}{x-y} = 4$ using substitution)
• Elimination method shortcuts and presentation formats

1. Archetype 1: Age Problems (Past and Future Relations)

In age problems, always assign variables to PRESENT AGES:

• Let the father's present age be $x\text{ years}$.
• Let the son's present age be $y\text{ years}$.

    Time Frame               Father's Age           Son's Age
    -------------------------------------------------------------------
    Present Age              x years                y years
    5 Years Ago (Past)       (x - 5) years          (y - 5) years
    5 Years Hence (Future)   (x + 5) years          (y + 5) years
    -------------------------------------------------------------------

Solved Example: The Jacob and Son Problem (NCERT Classic)

Problem: Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

Solution:

1. Let the present age of Jacob be $x\text{ years}$, and the present age of his son be $y\text{ years}$.
2. Condition 1 (Five years hence):
   • Jacob's age $= x + 5$; Son's age $= y + 5$. $(x + 5) = 3(y + 5)$ $x + 5 = 3y + 15 \implies \mathbf{x - 3y = 10} \quad \text{--- (Equation 1)}$
3. Condition 2 (Five years ago):
   • Jacob's age $= x - 5$; Son's age $= y - 5$. $(x - 5) = 7(y - 5)$ $x - 5 = 7y - 35 \implies \mathbf{x - 7y = -30} \quad \text{--- (Equation 2)}$
4. Solve by Elimination (Subtract Eq. 2 from Eq. 1): $(x - 3y) - (x - 7y) = 10 - (-30)$ $-3y + 7y = 10 + 30$ $4y = 40 \implies \mathbf{y = 10\text{ years}}$
5. Substitute $y = 10$ into Equation 1: $x - 3(10) = 10 \implies x - 30 = 10 \implies \mathbf{x = 40\text{ years}}$
6. Therefore, <u>the present age of Jacob is $40 ext{ years}$ and the present age of his son is $10 ext{ years}$</u>.

2. Archetype 2: Fraction Manipulation Problems

In fraction problems, define the numerator as $x$ and denominator as $y$: $\text{Original Fraction} = \mathbf{\frac{x}{y}}$

Solved Example: Fraction Problem (NCERT Classic)

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

Solution:

1. Let the fraction be $\frac{x}{y}$ (where $x$ is numerator, $y$ is denominator).
2. Condition 1: Adding 2 to both: $\frac{x + 2}{y + 2} = \frac{9}{11}$ $11(x + 2) = 9(y + 2) \implies 11x + 22 = 9y + 18$ $\mathbf{11x - 9y = -4} \quad \text{--- (Equation 1)}$
3. Condition 2: Adding 3 to both: $\frac{x + 3}{y + 3} = \frac{5}{6}$ $6(x + 3) = 5(y + 3) \implies 6x + 18 = 5y + 15$ $\mathbf{6x - 5y = -3} \quad \text{--- (Equation 2)}$
4. Solve by Elimination: Multiply Eq. 1 by 5 and Eq. 2 by 9: $55x - 45y = -20$ $54x - 45y = -27$ Subtracting the two equations: $(55x - 54x) = -20 - (-27) \implies \mathbf{x = 7}$
5. Substitute $x = 7$ into Eq. 2: $6(7) - 5y = -3 \implies 42 - 5y = -3 \implies 5y = 45 \implies \mathbf{y = 9}$
6. Therefore, <u>the required fraction is rac{7}{9}</u>.

3. Archetype 3: Fixed and Variable Running Charges

Many commercial services charge a base fee plus a rate per unit: $\mathbf{\text{Total Cost} = \text{Fixed Charge } (x) + [\text{Number of Units } (n) \times \text{Unit Charge } (y)]}$

Solved Example: The Lending Library

Problem: A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹$27$ for a book kept for seven days, while Susy paid ₹$21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

1. Let the fixed charge for the first 3 days be ₹$x$.
2. Let the additional charge per extra day be ₹$y$.
3. Saritha's Case (7 Days Total):
   • First 3 days covered by fixed charge $x$.
   • Extra days $= 7 - 3 = 4\text{ days}$. $\mathbf{x + 4y = 27} \quad \text{--- (Equation 1)}$
4. Susy's Case (5 Days Total):
   • First 3 days covered by fixed charge $x$.
   • Extra days $= 5 - 3 = 2\text{ days}$. $\mathbf{x + 2y = 21} \quad \text{--- (Equation 2)}$
5. Subtract Eq. 2 from Eq. 1: $(x + 4y) - (x + 2y) = 27 - 21$ $2y = 6 \implies \mathbf{y = ₹\,3\text{ per extra day}}$
6. Substitute $y = 3$ into Eq. 2: $x + 2(3) = 21 \implies x + 6 = 21 \implies \mathbf{x = ₹\,15}$
7. Therefore, <u>the fixed charge for the first three days is ₹$15$ and the charge for each extra day is ₹$3$</u>.

4. Archetype 4: Reciprocal Upstream-Downstream Systems

When variables appear in denominators, substitute dummy variables: Let $\frac{1}{x+y} = u$ and $\frac{1}{x-y} = v$.

$\frac{10}{x+y} + \frac{2}{x-y} = 4 \implies 10u + 2v = 4$ $\frac{15}{x+y} - \frac{5}{x-y} = -2 \implies 15u - 5v = -2$ Solving yields $u = \frac{1}{5} \implies x + y = 5$, and $v = 1 \implies x - y = 1$, giving $x = 3\text{ km/h}$ (boat) and $y = 2\text{ km/h}$ (stream).

5. Summary and Examination Tips

Exam Tip: In fixed charge problems (like library books or taxi fares), remember to subtract the base period! If the fixed charge covers the first 3 days, a 7-day rental has $7 - 3 = 4$ extra days, NOT 7!

Common Mistake: Forgetting to state the final answer in terms of the original question. Solving for $x = 7, y = 9$ is not enough; you must explicitly write: "The fraction is $\frac{7}{9}$"!