Pair of Linear Equations in Two Variables • Topic 3 of 3

Applications in Daily Life

How are linear equations used in real life? Whenever a situation links two unknown quantities by two independent conditions, a pair of linear equations models it exactly. Translate carefully, solve by any algebraic method, and check the answer against the wording.

The five-step word-problem method.

StepAction
1Read twice; identify the two unknowns.
2Assign letters (often \(x\) and \(y\)) to the unknowns, with units.
3Turn each given condition into one equation — you need two.
4Solve by substitution, elimination, or cross-multiplication.
5Verify and state the answer with units; reject impossible values (e.g. negative ages).

Common problem families and how to set them up.

  • Age problems. If present ages are \(x\) and \(y\), then "\(n\) years ago" is \(x-n,\ y-n\) and "\(n\) years hence" is \(x+n,\ y+n\). Translate phrases like "was \(k\) times as old" into \(x-n=k(y-n)\).
  • Two-digit number / digit problems. If the tens digit is \(x\) and the units digit is \(y\), the number is \(10x+y\) and the reversed number is \(10y+x\). Their sum is \(11(x+y)\) and their difference is \(9(x-y)\).
  • Fixed + variable charges. Taxi fare, hostel charges, etc. take the form (fixed part) + (rate) × (units): e.g. \(\text{charge}=x+d\,y\) where \(x\) is the fixed amount and \(y\) the per-unit rate.
  • Fractions. Let the fraction be \(\dfrac{x}{y}\); each condition ("becomes \(\tfrac{a}{b}\) when…") gives one equation after cross-multiplying.
  • Geometry — cyclic quadrilateral. Opposite angles of a cyclic quadrilateral are supplementary: they add to \(180^{\circ}\).
  • Boat / stream and trains+cars. Downstream speed \(=x+y\), upstream speed \(=x-y\) (boat \(x\), current \(y\)). Using time \(=\dfrac{\text{distance}}{\text{speed}}\) often gives equations in \(\tfrac{1}{x+y},\tfrac{1}{x-y}\), which become linear after substitution.

Reducible equations in applications. Speed and work problems frequently produce terms like \(\dfrac{30}{x-y}\). Substitute \(u=\dfrac{1}{x-y}\) and \(v=\dfrac{1}{x+y}\) (or \(u=\tfrac1x,\ v=\tfrac1y\)) to linearise, solve for \(u,v\), then recover the original unknowns.

Common mistakes. (1) Writing the reversed two-digit number as \(yx\) instead of \(10y+x\). (2) Applying the "years ago/hence" shift to only one person. (3) Forgetting that opposite angles of a cyclic quadrilateral sum to \(180^{\circ}\), not \(360^{\circ}\). (4) Mixing up upstream (\(x-y\)) and downstream (\(x+y\)). (5) Accepting a negative or fractional answer where the context forbids it. (6) Dropping units in the final line.

Mixture and purchase problems. If \(x\) kg of one item and \(y\) kg of another are combined, one equation comes from the total quantity (\(x+y=\) total) and a second from the total cost (rate\(_1\,x+\) rate\(_2\,y=\) total cost). The same pattern models "\(a\) pencils and \(b\) pens cost ₹$M$" type questions: write one cost equation per purchase.

Sanity-checking the answer. Ages and counts must be positive whole numbers; speeds and prices must be positive; the upstream speed \(x-y\) must stay positive (the boat must actually make headway). If a value violates the context, recheck the equations rather than reporting an impossible result.

Four everyday word-problem setups modelled as a pair of linear equations Modelling everyday problems as two equations Ages f = father, s = son (present ages) f − 5 = 3(s − 5) f + s = 50 → s = 15, f = 35 Coins x = ₹1 coins, y = ₹2 coins x + y = 30 (count) x + 2y = 50 (value) → x = 10, y = 20 Speed · distance · time Car A Car B 2(x + y) = 400 10(x − y) = 400 → x = 120, y = 80 km/h Mixture a = kg Tea A, b = kg Tea B a + b = 10 (kg) 40a + 60b = 520 (₹) → a = 4, b = 6 kg
1
Worked Example
The sum of two numbers is \(25\). Twice the larger is \(5\) more than three times the smaller. Find the numbers.
Solution
  1. Step 1: Let the larger \(=x\), smaller \(=y\).
  2. Step 2: \(x+y=25\) and \(2x=3y+5\Rightarrow 2x-3y=5\).
  3. Step 3: Multiply the first by \(3\): \(3x+3y=75\); add to the second: \(5x=80\Rightarrow x=16\).
  4. Step 4: \(y=25-16=9\).

Answer: The numbers are \(16\) and \(9\).

2
Worked Example
Five years ago a father was three times as old as his son. Ten years hence he will be twice as old. Find their present ages.
Solution
  1. Step 1: Let father \(=x\) years, son \(=y\) years.
  2. Step 2: Five years ago: \(x-5=3(y-5)\Rightarrow x-3y=-10\).
  3. Step 3: Ten years hence: \(x+10=2(y+10)\Rightarrow x-2y=10\).
  4. Step 4: Subtract: \((x-3y)-(x-2y)=-10-10\Rightarrow -y=-20\Rightarrow y=20\); then \(x=2(20)+10=50\).
  5. Step 5: Verify: \(50-5=45=3(20-5)\); \(50+10=60=2(20+10)\).

Answer: Father \(=50\) years, son \(=20\) years.

3
Worked Example
A two-digit number is four times the sum of its digits and twice the product of its digits. Find the number.
Solution
  1. Step 1: Tens digit \(=x\), units digit \(=y\); number \(=10x+y\).
  2. Step 2: \(10x+y=4(x+y)\Rightarrow 6x-3y=0\Rightarrow y=2x\).
  3. Step 3: \(10x+y=2xy\). Put \(y=2x\): \(10x+2x=2x(2x)\Rightarrow 12x=4x^{2}\Rightarrow x=3\) (taking \(x\ne0\)).
  4. Step 4: \(y=2(3)=6\); number \(=36\).
  5. Step 5: Check: \(4(3+6)=36\) and \(2(3\times6)=36\).

Answer: The number is \(36\).

4
Worked Example
The sum of a two-digit number and the number obtained by reversing the digits is \(66\). If the digits differ by \(2\), find the number.
Solution
  1. Step 1: Tens digit \(=x\), units digit \(=y\). Number \(=10x+y\), reversed \(=10y+x\).
  2. Step 2: Sum: \((10x+y)+(10y+x)=66\Rightarrow 11(x+y)=66\Rightarrow x+y=6\).
  3. Step 3: Digits differ by \(2\): \(x-y=2\) or \(y-x=2\).
  4. Step 4: With \(x-y=2\): adding gives \(2x=8\Rightarrow x=4,\ y=2\Rightarrow 42\). With \(y-x=2\): \(24\).

Answer: The number is \(42\) or \(24\).

5
Worked Example
A taxi charges a fixed amount plus a rate per km. For \(10\) km the charge is ₹105 and for \(15\) km it is ₹155. Find the fixed charge and the rate per km, then the fare for \(25\) km.
Solution
  1. Step 1: Fixed charge \(=x\), per-km rate \(=y\).
  2. Step 2: \(x+10y=105\) and \(x+15y=155\).
  3. Step 3: Subtract: \(5y=50\Rightarrow y=10\); then \(x=105-100=5\).
  4. Step 4: Fare for \(25\) km \(=x+25y=5+250=255\).

Answer: Fixed = ₹5, rate = ₹10/km; fare for \(25\) km = ₹255.

6
Worked Example
In a hostel the monthly charge is a fixed amount plus a cost per day of food. Staying \(20\) days costs ₹1000; staying \(26\) days costs ₹1180. Find the fixed charge and daily food cost.
Solution
  1. Step 1: Fixed charge \(=x\), daily food cost \(=y\).
  2. Step 2: \(x+20y=1000\) and \(x+26y=1180\).
  3. Step 3: Subtract: \(6y=180\Rightarrow y=30\); then \(x=1000-600=400\).

Answer: Fixed charge = ₹400, food cost = ₹30/day.

7
Worked Example
A fraction becomes \(\dfrac{1}{3}\) when \(1\) is subtracted from the numerator, and \(\dfrac{1}{4}\) when \(8\) is added to the denominator. Find the fraction.
Solution
  1. Step 1: Let the fraction be \(\dfrac{x}{y}\).
  2. Step 2: \(\dfrac{x-1}{y}=\dfrac13\Rightarrow 3(x-1)=y\Rightarrow 3x-y=3\).
  3. Step 3: \(\dfrac{x}{y+8}=\dfrac14\Rightarrow 4x=y+8\Rightarrow 4x-y=8\).
  4. Step 4: Subtract: \((4x-y)-(3x-y)=8-3\Rightarrow x=5\); then \(y=3x-3=12\).

Answer: The fraction is \(\dfrac{5}{12}\).

8
Worked Example
The difference of two numbers is \(26\) and one number is three times the other. Find them.
Solution
  1. Step 1: Let the numbers be \(x\) (larger) and \(y\).
  2. Step 2: \(x-y=26\) and \(x=3y\).
  3. Step 3: Substitute: \(3y-y=26\Rightarrow 2y=26\Rightarrow y=13\).
  4. Step 4: \(x=3(13)=39\).

Answer: The numbers are \(39\) and \(13\).

9
Worked Example
The angles of a cyclic quadrilateral \(ABCD\) are \(\angle A=(4y+20)^{\circ}\), \(\angle B=(3y-5)^{\circ}\), \(\angle C=(4x)^{\circ}\), \(\angle D=(7x+5)^{\circ}\). Find the angles.
Solution
  1. Step 1: Opposite angles of a cyclic quadrilateral are supplementary: \(\angle A+\angle C=180\) and \(\angle B+\angle D=180\).
  2. Step 2: \((4y+20)+(4x)=180\Rightarrow 4x+4y=160\Rightarrow x+y=40\).
  3. Step 3: \((3y-5)+(7x+5)=180\Rightarrow 7x+3y=180\).
  4. Step 4: From \(x+y=40,\ y=40-x\); substitute: \(7x+3(40-x)=180\Rightarrow 4x+120=180\Rightarrow x=15,\ y=25\).
  5. Step 5: \(\angle A=120^{\circ},\ \angle B=70^{\circ},\ \angle C=60^{\circ},\ \angle D=110^{\circ}\).

Answer: \(\angle A=120^{\circ},\ \angle B=70^{\circ},\ \angle C=60^{\circ},\ \angle D=110^{\circ}\).

10
Worked Example
A boat goes \(30\) km upstream and \(44\) km downstream in \(10\) hours. It goes \(40\) km upstream and \(55\) km downstream in \(13\) hours. Find the speed of the boat in still water and the speed of the stream.
Solution
  1. Step 1: Let boat speed \(=x\) km/h, stream speed \(=y\) km/h; upstream \(=x-y\), downstream \(=x+y\).
  2. Step 2: Using time \(=\dfrac{\text{distance}}{\text{speed}}\): \(\dfrac{30}{x-y}+\dfrac{44}{x+y}=10\) and \(\dfrac{40}{x-y}+\dfrac{55}{x+y}=13\).
  3. Step 3: Let \(u=\dfrac{1}{x-y},\ v=\dfrac{1}{x+y}\): \(30u+44v=10\) and \(40u+55v=13\).
  4. Step 4: Multiply first by \(4\), second by \(3\): \(120u+176v=40\) and \(120u+165v=39\); subtract: \(11v=1\Rightarrow v=\dfrac{1}{11}\).
  5. Step 5: Then \(30u+4=10\Rightarrow u=\dfrac15\). So \(x-y=5\) and \(x+y=11\).
  6. Step 6: Add: \(2x=16\Rightarrow x=8\); \(y=3\).

Answer: Boat \(=8\) km/h, stream \(=3\) km/h.

11
Worked Example
A man travels \(600\) km partly by train and partly by car. Covering \(400\) km by train and the rest by car takes \(6.5\) hours; covering \(200\) km by train and the rest by car takes \(7\) hours. Find the speeds of the train and the car.
Solution
  1. Step 1: Train speed \(=t\), car speed \(=c\). Case 1: \(\dfrac{400}{t}+\dfrac{200}{c}=6.5\). Case 2: \(\dfrac{200}{t}+\dfrac{400}{c}=7\).
  2. Step 2: Let \(u=\dfrac1t,\ v=\dfrac1c\): \(400u+200v=6.5\) and \(200u+400v=7\).
  3. Step 3: Multiply the first by \(2\): \(800u+400v=13\); subtract the second: \(600u=6\Rightarrow u=\dfrac{1}{100}\Rightarrow t=100\).
  4. Step 4: Substitute \(u=\dfrac{1}{100}\) into \(400u+200v=6.5\): \(4+200v=6.5\Rightarrow v=\dfrac{2.5}{200}=\dfrac{1}{80}\Rightarrow c=80\).
  5. Step 5: Verify Case 2: \(\dfrac{200}{100}+\dfrac{400}{80}=2+5=7\).

Answer: Train \(=100\) km/h, car \(=80\) km/h.

12
Worked Example
\(5\) pencils and \(7\) pens cost ₹115; \(7\) pencils and \(5\) pens cost ₹125. Find the cost of one pencil and one pen.
Solution
  1. Step 1: Pencil \(=p\), pen \(=q\) (in rupees). \(5p+7q=115\) and \(7p+5q=125\).
  2. Step 2: Add: \(12p+12q=240\Rightarrow p+q=20\).
  3. Step 3: Subtract first from second: \(2p-2q=10\Rightarrow p-q=5\).
  4. Step 4: Add the two reduced equations: \(2p=25\Rightarrow p=12.5\); then \(q=20-12.5=7.5\).

Answer: Pencil = ₹12.50, pen = ₹7.50.

Key Points

  • Identify two unknowns and assign variables clearly
  • Translate word statements into mathematical equations
  • Use the method (substitution/elimination) that seems easiest
  • Check if your answer makes logical sense in the original problem
  • Common applications: age, money, speed-distance, mixtures, numbers
  • Always include units (₹, km/h, years, etc.) in your final answer
  • Two-digit number \(=10x+y\); reversed \(=10y+x\); their sum \(=11(x+y)\), difference \(=9(x-y)\)
  • Opposite angles of a cyclic quadrilateral are supplementary (sum \(=180^{\circ}\))
  • Boat: downstream speed \(=x+y\), upstream \(=x-y\); use \(u=\tfrac{1}{x-y},\ v=\tfrac{1}{x+y}\)
  • For "years ago/hence" subtract/add the same number from BOTH ages
Tap an option to check your answer0 / 4
Q1.Word problems are solved by first forming:
Explanation: Translate words into equations.
Q2.Two numbers have sum $20$ and difference $4$. They are:
Explanation: $x+y=20,\ x-y=4$.
Q3.If $x+y=12$ and $x=2y$, then $y=$
Explanation: $3y=12\Rightarrow y=4$.
Q4.If the cost of $2$ pens is the variable, the model uses:
Explanation: Typically two variables.