Word Problems Solved Graphically
Solving word problems graphically involves converting a real-life situation into a pair of linear equations, plotting both lines on a graph, and reading the solution from their point of intersection.
This method provides a visual understanding of simultaneous equations. The graph shows whether the system has a unique solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (coincident lines).
Graphical methods are covered in Class 10 NCERT under the chapter "Pair of Linear Equations in Two Variables".
What is Word Problems Solved Graphically?
Definition: The graphical method of solving a pair of linear equations involves:
- Expressing each equation in the form y = mx + c (or any convenient form for plotting).
- Finding at least two solution points for each equation.
- Plotting both lines on the same coordinate plane.
- Reading the coordinates of the point of intersection — this is the solution.
Possible outcomes:
- Intersecting lines: Exactly one solution (consistent system).
- Parallel lines: No solution (inconsistent system).
- Coincident lines: Infinitely many solutions (dependent system).
Word Problems Solved Graphically Formula
General form of a linear equation:
ax + by + c = 0
Condition for consistency:
| Condition | Type of Lines | Solutions |
|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Intersecting | Unique solution |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel | No solution |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident | Infinite solutions |
Solved Examples
Example 1: Age Problem
Problem: The sum of the ages of a father and his son is 50. Five years ago, the father's age was 4 times the son's age. Find their present ages graphically.
Solution:
Let: Father's age = x, Son's age = y
Equations:
- x + y = 50 ... (i)
- (x − 5) = 4(y − 5) → x − 4y = −15 ... (ii)
Table for x + y = 50 (i.e., y = 50 − x):
| x | 20 | 30 | 40 |
| y | 30 | 20 | 10 |
Table for x − 4y = −15 (i.e., x = 4y − 15):
| y | 5 | 10 | 15 |
| x | 5 | 25 | 45 |
Plotting both lines, they intersect at (37, 13).
Answer: Father's age = 37 years, Son's age = 13 years.
Example 2: Cost Problem
Problem: 5 pencils and 7 pens cost Rs. 195. 7 pencils and 5 pens cost Rs. 177. Find the cost of each graphically.
Solution:
Let: Cost of pencil = x, Cost of pen = y
Equations:
- 5x + 7y = 195 → y = (195 − 5x)/7 ... (i)
- 7x + 5y = 177 → y = (177 − 7x)/5 ... (ii)
Table for equation (i):
| x | 0 | 7 | 14 |
| y | 27.86 | 22.86 | 17.86 |
Table for equation (ii):
| x | 1 | 6 | 11 |
| y | 34 | 27 | 20 |
Plotting both lines, they intersect at approximately (7, 22.86). Solving algebraically for exact values: x = 7, y = 195−35)/7 = 160/7 ≈ 22.86.
Since prices should be whole numbers, this indicates the graphical reading is approximate. Algebraic solution: multiply (i) by 5 and (ii) by 7, subtract → x = 3, y = (195−15)/7 = 180/7. Recheck: the exact answer is x = 3, y = 180/7, which is not an integer. The equations give non-integer pen cost.
Better re-read: Solving exactly: 25x + 35y = 975 and 49x + 35y = 1239. Subtract: 24x = 264, x = 11. Then y = (195−55)/7 = 140/7 = 20.
Answer: Pencil costs Rs. 11, Pen costs Rs. 20.
Example 3: Numbers Problem
Problem: The sum of two numbers is 15 and their difference is 3. Find the numbers graphically.
Solution:
Let: Numbers = x and y (x > y)
Equations:
- x + y = 15 → y = 15 − x ... (i)
- x − y = 3 → y = x − 3 ... (ii)
Table for y = 15 − x:
| x | 0 | 5 | 10 | 15 |
| y | 15 | 10 | 5 | 0 |
Table for y = x − 3:
| x | 3 | 6 | 9 | 12 |
| y | 0 | 3 | 6 | 9 |
The lines intersect at (9, 6).
Answer: The two numbers are 9 and 6.
Example 4: Speed-Distance Problem
Problem: 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:
Let: Speed of boat in still water = x km/h, Speed of stream = y km/h
Upstream speed = (x − y), Downstream speed = (x + y)
Let 1/(x−y) = u and 1/(x+y) = v (to make equations linear).
- 30u + 44v = 10 ... (i)
- 40u + 55v = 13 ... (ii)
From (i): u = (10 − 44v)/30. Substituting in (ii) and solving: v = 1/11 and u = 1/5.
So x − y = 5 and x + y = 11.
Plot y = x − 5 and y = 11 − x:
They intersect at (8, 3).
Answer: Speed of boat = 8 km/h, Speed of stream = 3 km/h.
Example 5: Fixed Charge Problem
Problem: A taxi charges a fixed amount plus a per-km rate. For 10 km the charge is Rs. 105 and for 15 km it is Rs. 155. Find the fixed charge and per-km rate graphically.
Solution:
Let: Fixed charge = x, Per-km rate = y
Equations:
- x + 10y = 105 → x = 105 − 10y ... (i)
- x + 15y = 155 → x = 155 − 15y ... (ii)
Table for (i): x = 105 − 10y:
| y | 0 | 5 | 10 |
| x | 105 | 55 | 5 |
Table for (ii): x = 155 − 15y:
| y | 0 | 5 | 10 |
| x | 155 | 80 | 5 |
The lines intersect at (5, 10) → x = 5, y = 10.
Answer: Fixed charge = Rs. 5, Per-km rate = Rs. 10.
Example 6: Parallel Lines — No Solution
Problem: Is the following system consistent? 2x + 3y = 12 and 4x + 6y = 30. Verify graphically.
Solution:
Check ratios: a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/30 = 2/5.
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
Plotting:
- Line 1: y = (12−2x)/3 → points (0, 4), (3, 2), (6, 0)
- Line 2: y = (30−4x)/6 → points (0, 5), (3, 3), (6, 1)
The lines are parallel and never meet.
Answer: The system has no solution (inconsistent).
Example 7: Ticket Pricing Problem
Problem: In a class test, the total marks obtained by two students together is 40. If first student got 4 more marks than the second, find their individual marks graphically.
Solution:
Let: First student = x, Second student = y
Equations:
- x + y = 40 → y = 40 − x
- x − y = 4 → y = x − 4
Tables and plot: The two lines intersect at (22, 18).
Verification: 22 + 18 = 40 ✓, 22 − 18 = 4 ✓
Answer: First student got 22 marks, second got 18 marks.
Example 8: Fraction Problem
Problem: The numerator of a fraction is 1 less than its denominator. If 1 is added to both numerator and denominator, the fraction becomes 1/2. Find the fraction.
Solution:
Let: Numerator = x, Denominator = y
Equations:
- x = y − 1 → x − y = −1 ... (i)
- (x+1)/(y+1) = 1/2 → 2x + 2 = y + 1 → 2x − y = −1 ... (ii)
From (i): y = x + 1. From (ii): y = 2x + 1.
Plot y = x + 1 and y = 2x + 1. They intersect at (0, 1).
Answer: The fraction is 0/1 = 0.
Recheck: If x = 0, y = 1. Fraction = 0/1 = 0. Adding 1: 1/2. ✓
Real-World Applications
The graphical method of solving linear equations applies in:
- Business: Finding break-even points where revenue equals cost.
- Physics: Finding where two objects moving at different speeds meet.
- Economics: Supply and demand equilibrium is the intersection of two linear graphs.
- Engineering: Determining the conditions under which two constraints are simultaneously satisfied.
- Everyday Life: Comparing two pricing plans (fixed + variable) to find the crossover point.
Key Points to Remember
- Always form two equations with two unknowns from the word problem.
- Find at least 3 points for each line (the third point serves as a check).
- Choose a suitable scale on graph paper for accurate plotting.
- The point of intersection gives the unique solution.
- If lines are parallel, the system has no solution.
- If lines coincide, the system has infinitely many solutions.
- Graphical solutions are approximate — algebraic verification is recommended.
- Speed-distance problems often need substitution (u = 1/(x−y)) to become linear.
- Label axes clearly and write the coordinates of the intersection point.
- In board exams, show the table of values and the graph clearly.
Practice Problems
- The sum of two numbers is 20 and their difference is 4. Find the numbers graphically.
- 3 chairs and 2 tables cost Rs. 1200. 5 chairs and 3 tables cost Rs. 1900. Find the cost of each graphically.
- A father is 3 times as old as his son. After 12 years, the father will be twice as old as the son. Find their ages graphically.
- The perimeter of a rectangular garden is 40 m and its area is 96 m². Find its length and breadth graphically.
- Show graphically that 3x + 2y = 6 and 6x + 4y = 18 have no solution.
- A student scores a total of 180 marks in two subjects. If Maths marks are 20 more than Science marks, find the marks in each subject graphically.
Frequently Asked Questions
Q1. Why is the graphical method less accurate than algebraic methods?
The graphical method depends on the precision of plotting and reading coordinates from graph paper. Small errors in plotting lead to approximate answers. Algebraic methods give exact answers.
Q2. How many points are needed to draw a straight line?
Two points are sufficient to draw a straight line. However, plotting a third point is recommended to verify correctness. If the third point does not lie on the line, there is an error in calculation.
Q3. What does it mean if two lines are parallel on the graph?
Parallel lines never intersect, which means the system of equations has no solution. The two conditions described in the word problem are contradictory.
Q4. Can every word problem be solved graphically?
Only problems that reduce to a pair of linear equations in two variables can be solved graphically in this chapter. Non-linear problems (quadratic, exponential) need different methods.
Q5. How do you choose the scale for graph paper?
Choose a scale that fits the range of values on the paper. If values go up to 50, use 1 cm = 5 units. The scale should allow the intersection point to be clearly visible.
Q6. Is graphical method asked in board exams?
Yes. CBSE often asks 4-mark or 5-mark questions requiring students to solve a word problem graphically, including the table of values and the graph.
Related Topics
- Graphical Method for Linear Equations
- Pair of Linear Equations in Two Variables
- Word Problems on Pair of Linear Equations
- Consistency of Linear Equations
- Substitution Method
- Elimination Method
- Cross-Multiplication Method
- Reducing Equations to Linear Form
- Conditions for Solvability
- Linear Equations in Real Life
