Solved Examples

How to Use This Section

This is your practice powerhouse for Pair of Linear Equations. Below are 30+ fully worked problems spanning the whole chapter — checking solutions, graphical reasoning, substitution, elimination, cross-multiplication, consistency conditions, reducible equations, and word problems — arranged roughly easy to hard.

How to study: Try each problem with the solution covered, then check the steps. In the board exam, marks are awarded step by step — practise writing full solutions and stating word-problem answers in words.

Keep these handy:

• Ratios: unique $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$; infinite all equal; none $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$.
• Substitution, elimination, and cross-multiplication all give the same solution.
• Reducible: let $u = 1/x$, $v = 1/y$.

Solved Examples

Example 1: Solve by substitution

Solve: $x + y = 8$, $x - y = 2$.

Solution:

1. $x = 8 - y$; substitute: $(8 - y) - y = 2 \Rightarrow 8 - 2y = 2 \Rightarrow y = 3$.
2. $x = 5$.

Final Answer: $x = 5$, $y = 3$.

Takeaway: Isolate and substitute.

Example 2: Solve by elimination

Solve: $3x + 4y = 10$, $2x - 2y = 2$.

Solution:

1. Simplify the second: $x - y = 1 \Rightarrow x = y + 1$.
2. Substitute into first: $3(y+1) + 4y = 10 \Rightarrow 7y = 7 \Rightarrow y = 1$, $x = 2$.

Final Answer: $x = 2$, $y = 1$.

Takeaway: Simplify equations before solving.

Example 3: Classify by ratios

Classify $x + 2y = 5$ and $2x + 4y = 10$.

Solution:

1. $\dfrac{1}{2} = \dfrac{2}{4} = \dfrac{5}{10} = \dfrac{1}{2}$.
2. All ratios equal ⇒ coincident.

Final Answer: Infinitely many solutions.

Takeaway: All three ratios equal ⇒ infinitely many solutions.

Example 4: Cross-multiplication

Solve: $2x + y - 5 = 0$, $3x + 2y - 8 = 0$.

Solution:

1. $a_1b_2 - a_2b_1 = (2)(2) - (3)(1) = 1$.
2. $x = \dfrac{(1)(-8) - (2)(-5)}{1} = -8 + 10 = 2$.
3. $y = \dfrac{(-5)(3) - (-8)(2)}{1} = -15 + 16 = 1$.

Final Answer: $x = 2$, $y = 1$.

Takeaway: Cross-multiplication gives both at once.

Example 5: Two numbers

The sum of two numbers is 35 and their difference is 13. Find them.

Solution:

1. $x + y = 35$, $x - y = 13$.
2. Add: $2x = 48 \Rightarrow x = 24$; subtract: $2y = 22 \Rightarrow y = 11$.

Final Answer: 24 and 11.

Takeaway: Sum/difference problems solve instantly by add/subtract.

Example 6: Find $k$ — infinitely many

Find $k$: $2x + 3y = 2$ and $(k+1)x + 9y = 6$ has infinitely many solutions.

Solution:

1. Need $\dfrac{2}{k+1} = \dfrac{3}{9} = \dfrac{2}{6}$.
2. $\dfrac{3}{9} = \dfrac{1}{3}$, $\dfrac{2}{6} = \dfrac{1}{3}$ ✓.
3. $\dfrac{2}{k+1} = \dfrac{1}{3} \Rightarrow k + 1 = 6 \Rightarrow k = 5$.

Final Answer: $k = 5$.

Takeaway: Set all ratios equal; solve and verify.

Example 7: Reducible equations

Solve: $\dfrac{1}{2x} + \dfrac{1}{3y} = 2$ and $\dfrac{1}{3x} + \dfrac{1}{2y} = \dfrac{13}{6}$.

Solution:

1. Let $u = 1/x$, $v = 1/y$: $\dfrac{u}{2} + \dfrac{v}{3} = 2 \Rightarrow 3u + 2v = 12$; $\dfrac{u}{3} + \dfrac{v}{2} = \dfrac{13}{6} \Rightarrow 2u + 3v = 13$.
2. Solve: from Example earlier, $u = 2$, $v = 3$.
3. So $x = \dfrac{1}{2}$, $y = \dfrac{1}{3}$.

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

Takeaway: Clear fractions after substituting $u, v$.

Example 8: No solution by ratios

Classify $x + 2y = 3$ and $2x + 4y = 7$.

Solution:

1. $\dfrac{1}{2} = \dfrac{2}{4}$ but $\dfrac{3}{7} \neq \dfrac{1}{2}$.
2. So $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.

Final Answer: No solution (inconsistent).

Takeaway: Equal $a, b$ ratios but different $c$ ⇒ no solution.

Example 9: Age problem

Five years ago, A was thrice as old as B. Ten years later, A will be twice as old as B. Find their present ages.

Solution:

1. Let A $= x$, B $= y$. Five years ago: $x - 5 = 3(y - 5) \Rightarrow x - 3y = -10$.
2. Ten years later: $x + 10 = 2(y + 10) \Rightarrow x - 2y = 10$.
3. Subtract: $-y = -20 \Rightarrow y = 20$; then $x = 50$.

Final Answer: A is 50, B is 20.

Takeaway: Form one equation per time-condition.

Example 10: Cost of items

7 audio cassettes and 3 video cassettes cost ₹1110; 5 audio and 4 video cost ₹1350. Find each cost.

Solution:

1. $7x + 3y = 1110$, $5x + 4y = 1350$.
2. Multiply first by 4, second by 3: $28x + 12y = 4440$, $15x + 12y = 4050$. Subtract: $13x = 390 \Rightarrow x = 30$.
3. $7(30) + 3y = 1110 \Rightarrow 3y = 900 \Rightarrow y = 300$.

Final Answer: Audio ₹30, video ₹300.

Takeaway: Eliminate the variable with easy common multiple.

Example 11: Graphical reasoning

Without solving, state the number of solutions of $3x - y = 2$ and $6x - 2y = 4$.

Solution:

1. $\dfrac{3}{6} = \dfrac{-1}{-2} = \dfrac{2}{4} = \dfrac{1}{2}$.
2. All equal ⇒ coincident.

Final Answer: Infinitely many solutions.

Takeaway: Second equation is twice the first.

Example 12: Fraction problem

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

Solution:

1. Let fraction $= \dfrac{x}{y}$. $\dfrac{x+2}{y+2} = \dfrac{9}{11} \Rightarrow 11x - 9y = -4$.
2. $\dfrac{x+3}{y+3} = \dfrac{5}{6} \Rightarrow 6x - 5y = -3$.
3. Solve: multiply first by 5, second by 9: $55x - 45y = -20$, $54x - 45y = -27$. Subtract: $x = 7$; then $y = 9$.

Final Answer: The fraction is $\dfrac{7}{9}$.

Takeaway: Cross-multiply each condition, then solve.

Example 13: Speed of boat

A boat covers 24 km upstream in 6 hours and 36 km downstream in 6 hours. Find the speed of the boat in still water and the speed of the stream.

Solution:

1. Upstream speed $= 24/6 = 4$ km/h $= x - y$; downstream speed $= 36/6 = 6$ km/h $= x + y$.
2. Add: $2x = 10 \Rightarrow x = 5$; subtract: $2y = 2 \Rightarrow y = 1$.

Final Answer: Boat 5 km/h, stream 1 km/h.

Takeaway: Speed = distance/time gives the upstream/downstream speeds first.

Example 14: Find $k$ for unique solution

For what value of $k$ does $2x + 3y = 4$ and $(k+2)x + 6y = 3k + 2$ NOT have a unique solution?

Solution:

1. No unique solution when $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$: $\dfrac{2}{k+2} = \dfrac{3}{6} = \dfrac{1}{2}$.
2. $\dfrac{2}{k+2} = \dfrac{1}{2} \Rightarrow k + 2 = 4 \Rightarrow k = 2$.

Final Answer: $k = 2$.

Takeaway: 'Not unique' is the boundary where the first two ratios become equal.

Example 15: Elimination with scaling

Solve: $8x + 5y = 9$, $3x + 2y = 4$.

Solution:

1. Multiply first by 2, second by 5: $16x + 10y = 18$, $15x + 10y = 20$. Subtract: $x = -2$.
2. $3(-2) + 2y = 4 \Rightarrow 2y = 10 \Rightarrow y = 5$.

Final Answer: $x = -2$, $y = 5$.

Takeaway: Match the $y$-coefficients (LCM 10) and subtract.

Example 16: Geometry — angles

The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.

Solution:

1. Let the angles be $x$ and $y$ with $x > y$: $x + y = 180$ and $x - y = 18$.
2. Add: $2x = 198 \Rightarrow x = 99$; then $y = 81$.

Final Answer: 99° and 81°.

Takeaway: Supplementary ⇒ sum 180°; translate 'exceeds by' into a difference.

Example 17: Substitution with no solution

Solve: $3x + 2y = 4$, $9x + 6y = 7$.

Solution:

1. From the first: $y = \dfrac{4 - 3x}{2}$. Substitute: $9x + 6\cdot\dfrac{4 - 3x}{2} = 7 \Rightarrow 9x + 12 - 9x = 7 \Rightarrow 12 = 7$ (false).

Final Answer: No solution.

Takeaway: A contradiction confirms parallel lines.

Example 18: Coins problem

A piggy bank has ₹2 and ₹5 coins, 25 coins in all, worth ₹80. How many of each?

Solution:

1. Let ₹2 coins $= x$, ₹5 coins $= y$: $x + y = 25$, $2x + 5y = 80$.
2. From first $x = 25 - y$; substitute: $2(25 - y) + 5y = 80 \Rightarrow 50 + 3y = 80 \Rightarrow y = 10$, $x = 15$.

Final Answer: 15 two-rupee coins and 10 five-rupee coins.

Takeaway: One equation for count, one for value.

Example 19: Reducible — clear denominators first

Solve: $\dfrac{5}{x-1} + \dfrac{1}{y-2} = 2$ and $\dfrac{6}{x-1} - \dfrac{3}{y-2} = 1$.

Solution:

1. Let $u = \dfrac{1}{x-1}$, $v = \dfrac{1}{y-2}$: $5u + v = 2$, $6u - 3v = 1$.
2. From first $v = 2 - 5u$; substitute: $6u - 3(2 - 5u) = 1 \Rightarrow 21u = 7 \Rightarrow u = \dfrac{1}{3}$, $v = \dfrac{1}{3}$.
3. $x - 1 = 3 \Rightarrow x = 4$; $y - 2 = 3 \Rightarrow y = 5$.

Final Answer: $x = 4$, $y = 5$.

Takeaway: Substitute for the whole reciprocal expression, then convert back.

Example 20: Distance, two cars

Two cars start from places 100 km apart. Travelling toward each other they meet in 1 hour; in the same direction the faster overtakes in 5 hours. Find their speeds.

Solution:

1. Let speeds be $x > y$ km/h. Toward each other: $x + y = 100$. Same direction: $x - y = 20$ (since $100/5 = 20$).
2. Add: $2x = 120 \Rightarrow x = 60$; $y = 40$.

Final Answer: 60 km/h and 40 km/h.

Takeaway: Toward each other ⇒ sum of speeds; same direction ⇒ difference.

Example 21: Find both unknown constants

For what values of $a$ and $b$ does $2x + 3y = 7$ and $(a-b)x + (a+b)y = 3a + b - 2$ have infinitely many solutions?

Solution:

1. Need $\dfrac{2}{a-b} = \dfrac{3}{a+b} = \dfrac{7}{3a+b-2}$.
2. From first two: $2(a+b) = 3(a-b) \Rightarrow a = 5b$.
3. From $\dfrac{3}{a+b} = \dfrac{7}{3a+b-2}$ with $a = 5b$: $\dfrac{3}{6b} = \dfrac{7}{16b - 2} \Rightarrow 3(16b-2) = 42b \Rightarrow 48b - 6 = 42b \Rightarrow b = 1$, $a = 5$.

Final Answer: $a = 5$, $b = 1$.

Takeaway: Solve the ratio equations step by step for two unknowns.

Example 22: Elimination giving a fraction

Solve: $4x + 3y = 7$, $2x - y = 1$.

Solution:

1. From the second: $y = 2x - 1$. Substitute: $4x + 3(2x - 1) = 7 \Rightarrow 10x = 10 \Rightarrow x = 1$, $y = 1$.

Final Answer: $x = 1$, $y = 1$.

Takeaway: Substitution can be quicker when one coefficient is 1.

Example 23: Work/rate-style linear setup

The sum of a two-digit number and the number formed by reversing the digits is 66. If the digits differ by 2, find the number(s).

Solution:

1. $(10x + y) + (10y + x) = 66 \Rightarrow 11(x + y) = 66 \Rightarrow x + y = 6$.
2. $x - y = 2$ (or $y - x = 2$). With $x - y = 2$: $x = 4, y = 2$, number 42.
3. With $y - x = 2$: $x = 2, y = 4$, number 24.

Final Answer: 42 or 24.

Takeaway: A difference 'by 2' allows two cases; report both.

Example 24: Consistent and dependent

Show that $2x + 3y = 9$ and $4x + 6y = 18$ are consistent and find any two solutions.

Solution:

1. $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{9}{18} = \dfrac{1}{2}$ ⇒ coincident ⇒ consistent (dependent).
2. Solutions of $2x + 3y = 9$: $(0, 3)$ and $(3, 1)$.

Final Answer: Consistent; e.g. $(0,3)$ and $(3,1)$.

Takeaway: Dependent systems share every point of the common line.

Example 25: Geometry — perimeter

The length of a rectangle exceeds its breadth by 5 m. If the perimeter is 50 m, find the dimensions.

Solution:

1. Let length $= x$, breadth $= y$: $x - y = 5$ and $2(x + y) = 50 \Rightarrow x + y = 25$.
2. Add: $2x = 30 \Rightarrow x = 15$; $y = 10$.

Final Answer: Length 15 m, breadth 10 m.

Takeaway: Perimeter $= 2(l + b)$ gives one equation.

Example 26: Reducible with a constant term

Solve: $\dfrac{4}{x} + 3y = 14$ and $\dfrac{3}{x} - 4y = 23$.

Solution:

1. Let $u = 1/x$: $4u + 3y = 14$, $3u - 4y = 23$.
2. Multiply first by 4, second by 3: $16u + 12y = 56$, $9u - 12y = 69$. Add: $25u = 125 \Rightarrow u = 5$.
3. $4(5) + 3y = 14 \Rightarrow 3y = -6 \Rightarrow y = -2$. So $x = \dfrac{1}{5}$.

Final Answer: $x = \dfrac{1}{5}$, $y = -2$.

Takeaway: Only the $1/x$ part needs substitution; $y$ stays as is.

Example 27: Solve and verify (clean integers)

Solve: $x - y = 3$, $\dfrac{x}{3} + \dfrac{y}{2} = 6$.

Solution:

1. Clear fractions in the second: multiply by 6: $2x + 3y = 36$.
2. From the first $x = y + 3$; substitute: $2(y+3) + 3y = 36 \Rightarrow 5y = 30 \Rightarrow y = 6$, $x = 9$.

Final Answer: $x = 9$, $y = 6$.

Takeaway: Clear fractions first by multiplying through by the LCM of denominators.

Example 28: Cross-multiplication with negatives

Solve by cross-multiplication: $3x - 5y - 11 = 0$, $5x + 2y - 7 = 0$.

Solution:

1. $a_1b_2 - a_2b_1 = (3)(2) - (5)(-5) = 6 + 25 = 31$.
2. $x = \dfrac{(-5)(-7) - (2)(-11)}{31} = \dfrac{35 + 22}{31} = \dfrac{57}{31}$.
3. $y = \dfrac{(-11)(5) - (-7)(3)}{31} = \dfrac{-55 + 21}{31} = \dfrac{-34}{31}$.

Final Answer: $x = \dfrac{57}{31}$, $y = -\dfrac{34}{31}$.

Takeaway: Track signs carefully; non-integer answers are fine.

Example 29: Time and work-style mixture

2 men and 5 women finish a piece of work in 4 days, while 3 men and 6 women finish it in 3 days. If a man's one-day work is $x$ and a woman's is $y$, then $4(2x + 5y) = 1$ and $3(3x + 6y) = 1$. Find $x$ and $y$.

Solution:

1. $8x + 20y = 1$ and $9x + 18y = 1$.
2. Multiply first by 9, second by 8: $72x + 180y = 9$, $72x + 144y = 8$. Subtract: $36y = 1 \Rightarrow y = \dfrac{1}{36}$.
3. $8x + 20\cdot\dfrac{1}{36} = 1 \Rightarrow 8x = 1 - \dfrac{5}{9} = \dfrac{4}{9} \Rightarrow x = \dfrac{1}{18}$.

Final Answer: Man $\dfrac{1}{18}$/day, woman $\dfrac{1}{36}$/day.

Takeaway: Rate problems become linear in the per-day work variables.

Example 30: Determine consistency for three cases

For $x + 2y = 1$, decide the second equation that makes the pair (a) inconsistent, (b) dependent.

Solution:

1. (a) Inconsistent: keep $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$ but change $c$: e.g. $2x + 4y = 5$ (since $\dfrac{1}{2} = \dfrac{2}{4} \neq \dfrac{1}{5}$).
2. (b) Dependent: all ratios equal: e.g. $2x + 4y = 2$.

Final Answer: (a) $2x + 4y = 5$; (b) $2x + 4y = 2$.

Takeaway: Control the $c$-ratio to switch between no-solution and infinitely-many.

Example 31: Mixed multi-step

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes $\dfrac{1}{2}$. Find the fraction.

Solution:

1. Let fraction $= \dfrac{x}{y}$: $x + y = 12$ and $\dfrac{x}{y+3} = \dfrac{1}{2} \Rightarrow 2x = y + 3$.
2. From the first $y = 12 - x$; substitute: $2x = (12 - x) + 3 \Rightarrow 3x = 15 \Rightarrow x = 5$, $y = 7$.

Final Answer: The fraction is $\dfrac{5}{7}$.

Takeaway: Translate both conditions, then solve by substitution.