Graphical Solution Class 9 ICSE — Maths Chapter 27
Graphical Solution for Class 9 ICSE is Chapter 27 in your Class 9 Maths book. It teaches you to draw graphs of linear equations in two variables. You will learn the five-step graphing method, read missing values from a plotted line, and solve simultaneous equations graphically. The chapter also covers the breakeven cost problem. The CISCE website carries the official ICSE syllabus for Class 9 Mathematics.
ICSE Class 9 Mathematics Chapter 27 Graphical Solution PDF
What is inside Chapter 27
6 pages · 1 sections · 6 worked examples · 13 exercise questions
| Section | Title |
|---|---|
| 27.1 | GRAPHS OF LINEAR EQUATIONS IN TWO VARIABLES |
- EXERCISE 27(A): 13 questions
27.1 GRAPHS OF LINEAR EQUATIONS IN TWO VARIABLES
An equation of the form ax + by + c = 0 is called a linear equation in two variables. Here x and y are the variables, and a, b, c are the constants. Every such equation produces a straight line when plotted. This is why the chapter is named Graphical Solution.
The five step method to draw a graph
The chapter gives five steps for drawing the graph of any linear equation ax + by + c = 0.
- Make x or y the subject of the equation.
- Give at least three suitable values to the variable on the right-hand side and find the corresponding values of the variable on the left-hand side.
- Construct a table for the different pairs of values of x and y.
- Plot at least three ordered pairs from the table on graph paper.
- Draw a straight line passing through the plotted points.
You need at least three points, never two. Three points let you check the line is straight and catch arithmetic errors in your values.
Here is the pattern with original numbers. Draw the graph of 2x + y - 6 = 0. Make y the subject: y = -2x + 6. Give x three values.
- x = 0: y = -2(0) + 6 = 6
- x = 1: y = -2(1) + 6 = 4
- x = 2: y = -2(2) + 6 = 2
The table of pairs is:
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 6 | 4 | 2 |
Plot the points (0, 6), (1, 4) and (2, 2) on graph paper. Draw a straight line through them. That line is the graph of 2x + y - 6 = 0.
Choose which variable to make the subject by checking which gives simpler integer pairs. The chapter’s first example makes y the subject because y has coefficient 1. The second example makes x the subject from 2x - 3y + 12 = 0, giving x = \frac{3y - 12}{2}.
| When to make y the subject | When to make x the subject |
|---|---|
| y has coefficient 1, as in 3x + y - 4 = 0 | x has coefficient 1, as in x - 5y + 4 = 0 |
| Gives y = -3x + 4, clean pairs for simple x | Gives x = 5y - 4, clean pairs for simple y |
| Chapter’s first worked example uses this | Chapter’s second worked example makes x the subject |
Some exercise equations contain fractions, such as \frac{1}{3}x + \frac{1}{5}y = 1. Multiply through by a common multiple to clear the fractions before making a variable the subject.
Reading missing values from a drawn graph
Exercise questions often give a table with unknown letters such as a, b, m and n. The graph provides their values. To find a y-value, draw a vertical line through the given x until it meets the graph. Then draw a horizontal line from that meeting point to the y-axis. The number you reach is the missing value.
To find an x-value, reverse the order. Draw a horizontal line through the given y until it meets the graph. Then draw a vertical line to the x-axis. The number you reach is the missing value.
Here is the pattern with original numbers. A table has x = 1, 3, 5 and y = -2, 2, 6. Plot the points (1, -2), (3, 2) and (5, 6) and draw the line. To find the missing y at x = 4, draw a vertical line through x = 4 to the graph, then a horizontal line to the y-axis. The value is 4.
To find the missing x at y = 8, draw a horizontal line through y = 8 to the graph, then a vertical line to the x-axis. The value is 6.
This pattern runs through the exercise. Questions 6 and 7 of EXERCISE 27(A) ask for a, b and c from tables in this way.
Finding the linear relation y = mx + c
When two points on a line are known, you can find its equation. Let the relation be y = mx + c. Substitute each known point into the equation to form two equations. Solve for c first, then m.
When one known point has x = 0, c is read directly. The y-coordinate of that point is the value of c.
Here is the pattern with original numbers. The points (0, 3) and (2, 7) lie on a line. Let the relation be y = mx + c. From (0, 3): 3 = m(0) + c, so c = 3. From (2, 7): 7 = m(2) + 3, so 2m = 4 and m = 2. The required relation is y = 2x + 3.
The chapter’s worked example uses points (-2, -3) and (0, 1) to find c = 1 and m = 2, giving the relation y = 2x + 1.
Lines parallel to the axes
Four facts about lines parallel to the axes appear in the Remember box.
- The x-axis has equation y = 0.
- The y-axis has equation x = 0.
- The graph of x = a is a vertical line parallel to the y-axis, at distance a units from it.
- The graph of y = b is a horizontal line parallel to the x-axis, at distance b units from it.
The constant a or b can be positive or negative. A negative value places the line on the other side of the origin. Exercise question 1 asks you to draw graphs like x = 5, x + 5 = 0, y = 7 and y + 7 = 0.
Note that x + 5 = 0 simplifies to x = -5, and y + 7 = 0 simplifies to y = -7.
Solving simultaneous equations graphically
To solve two linear equations graphically, use three steps. First, draw a graph for each equation on the same graph paper. Second, find the coordinates of the point where the two lines meet. Third, read that point as the solution.
The intersection point must satisfy both equations. You can verify by substituting the x and y values into each original equation.
Here is the pattern with original numbers. Solve x + y = 5 and x - y = 1 graphically. For x + y = 5, make y the subject: y = 5 - x. Points: (0, 5), (2, 3), (5, 0). For x - y = 1, make y the subject: y = x - 1. Points: (0, -1), (2, 1), (5, 4).
Both lines meet at (3, 2).
Check: 3 + 2 = 5 and 3 - 2 = 1. Both equations hold. So the solution is x = 3, y = 2. The chapter’s worked example solves 3x - 2y = 4 and 5x - 2y = 0, which meet at (-2, -5). This graphical method builds directly on the plotting skills from Chapter 26 Co-ordinate Geometry.
Breakeven problems in the exercise
Exercise 27(A) includes a word problem about manufacturing cost and selling price. The breakeven point is where cost price equals selling price. On the graph, that is where the two lines meet.
The cost of manufacturing x articles is a fixed amount plus a per-article cost. The selling price is a per-article amount times x. Draw both lines on the same axes. The intersection gives the number of articles needed to break even.
Here is the pattern with original numbers. Let cost be \text{₹}(30 + 3x) and selling price be \text{₹}4.5x. Table for cost: x = 0 gives 30, x = 10 gives 60, x = 20 gives 90. Table for selling price: x = 0 gives 0, x = 10 gives 45, x = 20 gives 90. The lines meet at x = 20.
So 20 articles must be made and sold to break even.
At x = 10, cost is ₹60 and selling price is ₹45, a loss of ₹15. At x = 30, cost is ₹120 and selling price is ₹135, a profit of ₹15. The chapter’s example uses cost \text{₹}(20 + 2x) and selling price \text{₹}2.5x, with breakeven at 40 articles.
Three mistakes appear often in this chapter. Plotting only two points instead of three. Two points define a line but cannot reveal an arithmetic error. Misreading the axes when drawing the guide lines. Confusing the equations of the axes: the x-axis is y = 0 and the y-axis is x = 0.
What to practise from this chapter
The chapter contains one exercise block. Explore more resources in the ICSE Class 9 books section.
- EXERCISE 27(A): 13 questions
The questions ask you to draw graphs, find where a line meets the axes, compute the area of a triangle formed by a line and the axes, read unknown letters from a table, and solve the breakeven cost problem.
Frequently asked questions
How many points do I need to plot to draw the graph of a linear equation?
You need at least three points. Three points let you check that the line is straight and catch arithmetic errors. Two points define a line but cannot reveal a mistake in your values.
What does it mean to make x or y the subject of the equation?
Making a variable the subject means rewriting the equation so that variable stands alone on one side. For example, from 2x + y - 6 = 0, making y the subject gives y = -2x + 6. This lets you choose values for x and compute y easily.
How do I find the values of a and b from a drawn graph?
To find a y-value like a, draw a vertical line through the given x to the graph, then a horizontal line to the y-axis. To find an x-value like b, reverse the steps: horizontal line through the given y to the graph, then vertical line to the x-axis.
How do I solve simultaneous linear equations graphically?
Draw both lines on the same graph paper. Find the point where they meet. The coordinates of that point are the solution. Check the point satisfies both original equations.
What is the breakeven point in the cost and selling price problem?
The breakeven point is where cost price equals selling price. On the graph, the cost line and the selling price line meet there. The x-coordinate of that meeting point tells you how many articles must be made and sold for no profit and no loss.
Reference: ICSE Class 9 Mathematics textbook material, chapter 27: Graphical Solution — CISCE-aligned.