Graphical Solution
ICSE Class 9 · Mathematics · Chapter 24
What this chapter covers
- Why a graph can solve a pair of equations
- Linear equations in two variables and their graphs
- Special lines: parallel to an axis or through the origin
- Choosing a scale and drawing accurately
- The method of graphical solution
- Solutions with negative or fractional values
- Where the lines meet the axes, and the triangles they form
- When the lines do not meet in one point
- Word problems solved graphically
- Common mistakes and how to avoid them
Key points
- Every linear equation in two variables has a straight line as its graph, and every point on that line satisfies the equation.
- A pair of simultaneous linear equations is solved graphically by drawing both lines on the same axes and reading the coordinates of their point of intersection.
- Make a table of at least three points for each line, choosing values of x that give whole-number values of y.
- The line x = a is parallel to the y-axis, the line y = b is parallel to the x-axis, and a line y = mx passes through the origin.
- Choose the scale after making the tables, state it on the graph, and extend both lines far enough to meet.
- A line meets the x-axis where y = 0 and the y-axis where x = 0.
- The triangle formed by two lines and an axis has its base on that axis and its height equal to the distance of the meeting point from that axis.
- Intersecting lines give exactly one solution, parallel lines give no solution, and coincident lines give infinitely many solutions.
- Always check the pair read from the graph by substituting it in both original equations.
Questions answered
Define the graphical solution of a pair of simultaneous linear equations.
A graphical solution of a pair of simultaneous linear equations is the ordered pair given by the coordinates of the point where the graphs of the two equations intersect. That point lies on both lines, so it satisfies both equations.
Is (2, -1) a solution of 3x + 2y = 4? Does the point (2, -1) lie on the graph of this equation?
3(2) + 2(-1) = 6 - 2 = 4, so (2, -1) is a solution, and the point lies on the line 3x + 2y = 4.
Complete a table of values for y = 3x - 2 using x = -1, 0, 2.
For x = -1, y = -5; for x = 0, y = -2; for x = 2, y = 4. The points are (-1, -5), (0, -2) and (2, 4).
Why should you plot at least three points for each line, when two points are enough to fix a straight line?
Two points always give a straight line, even if one of them is wrong. If three correctly worked points do not lie in a straight line, there is an error in the table or the plotting, so the third point acts as a check.
For 5x - 2y = 8, which values of x give whole-number values of y: odd or even? Give three such points.
y = (5x - 8)/(2). Even values of x make 5x - 8 even, so y is a whole number. For x = 0, 2, 4 the points are (0, -4), (2, 1) and (4, 6).
Write down the point of intersection of the lines x = -4 and y = 5. Describe each line.
(-4, 5). The line x = -4 is parallel to the y-axis, 4 units to its left; the line y = 5 is parallel to the x-axis, 5 units above it.
Solve graphically: x + y = 5 and 2x - y = 4.
Points for x + y = 5: (0, 5), (2, 3), (5, 0). Points for 2x - y = 4, that is y = 2x - 4: (0, -4), (2, 0), (3, 2). The lines meet at (3, 2), so x = 3, y = 2. Check: 3 + 2 = 5 and 6 - 2 = 4.
Solve graphically: x + 2y = 1 and x - y = 4.
Points for x + 2y = 1: (-1, 1), (1, 0), (3, -1). Points for x - y = 4, that is y = x - 4: (0, -4), (3, -1), (4, 0). The lines meet at (3, -1) in the fourth quadrant, so x = 3, y = -1. Check: 3 - 2 = 1 and 3 + 1 = 4.
Past board exam questions
ICSE 2019 sample paper · 4 marks
Solve graphically the following system of linear equations (use graph sheet): x-3y=3 2x+3y=6 Also, find the area of the triangle formed by these two lines and the y-axis.
Show answer
x - 3y = 3: points (0, −1), (3, 0), (6, 1).
2x + 3y = 6: points (0, 2), (3, 0), (−3, 4).
Plot both lines on the graph sheet; they intersect at (3, 0). Solution: x = 3, y = 0.
The lines meet the y-axis at (0, −1) and (0, 2). The triangle has vertices (3, 0), (0, −1), (0, 2): base on the y-axis = 3 units, height = 3 units.
Area = (1)/(2) × 3 × 3 = 4.5 square units.