Find any two solutions for each of the following equations and draw their graphs:
- (i) 4x + 3y = 12
- (ii) 2x + 5y = 0
Step-by-step solution
Idea: Substitute x = 0 and y = 0 to get the points on the axes. When both give the origin (as in (ii), where there is no constant term), choose another value of x, preferably one that makes y a whole number.
(i) 4x + 3y = 12
- Put x = 0: 3y = 12, so y = 4. Solution (0, 4).½ mark
- Put y = 0: 4x = 12, so x = 3. Solution (3, 0).½ mark
- Plot B(0, 4) and A(3, 0) and draw the line through them: the graph of 4x + 3y = 12 (blue line in the diagram).1 mark
(ii) 2x + 5y = 0
- Put x = 0: 5y = 0, so y = 0. Solution (0, 0). Putting y = 0 gives the same point, so we need a different x.½ mark
- Put x = 1: 2 + 5y = 0, so y = −25. Put x = 5: 10 + 5y = 0, so y = −2. Solutions (1, −25) and (5, −2).½ mark
- Plot A(0, 0), B(1, −0.4), C(5, −2); they lie on one line through the origin: the graph of 2x + 5y = 0 (red line in the diagram).1 mark
Check: (i) 4(0) + 3(4) = 12 ✓, 4(3) + 0 = 12 ✓. (ii) 2(5) + 5(−2) = 0 ✓, 2(1) + 5(−25) = 2 − 2 = 0 ✓.
Answer to write in the exam
(i)
x = 0: 3y = 12 ⇒ y = 4 ⇒ (0, 4)
y = 0: 4x = 12 ⇒ x = 3 ⇒ (3, 0)
| x | 0 | 3 |
| y | 4 | 0 |
∴ Graph: the line through (0, 4) and (3, 0)
(ii)
x = 0: 5y = 0 ⇒ y = 0 ⇒ (0, 0)
x = 1: 5y = −2 ⇒ y = −25 ⇒ (1, −25)
x = 5: 5y = −10 ⇒ y = −2 ⇒ (5, −2)
| x | 0 | 1 | 5 |
| y | 0 | −25 | −2 |
∴ Graph: the line through (0, 0) and (5, −2)
Common mistakes that cost marks
- In (ii), using only (0, 0) twice and trying to draw a line through one point. A line needs two different points.
- Getting y = +25 in (ii): 5y = −2 gives a negative y.
- In (i), writing y = 3 when x = 0 by dividing 12 by the wrong coefficient.
How this can come in the exam
The graph of which equation passes through the origin?
- 3x + 4y = 1
- x − 2y = 5
- 7x − 3y = 0
- y = x + 2
Show answer
(C) 7x − 3y = 0
(0, 0) satisfies 7x − 3y = 0 since there is no constant term.
Find two solutions of 3x − 4y = 0 with whole-number coordinates.
Show answer
x = 0 ⇒ y = 0: (0, 0) (1 mark). x = 4 ⇒ 4y = 12 ⇒ y = 3: (4, 3) (1 mark).Try one yourself
Find two solutions of 5x + 2y = 10 and of x − 3y = 0.
Show answer
5x + 2y = 10: (0, 5), (2, 0). x − 3y = 0: (0, 0), (3, 1).
More questions like this
- In the equations shown below, a and b are unknown numbers.
3ax + 4y = −2
2x + by = 14.
If (−3, 4) is the solution of both equations, find the values of a and b. - Four friends — Ranju, Meena, Farhan, and Toshi — are solving problems.
Ranju: I noticed something! If c = 0 in the standard form of a line ax + by + c = 0, then the line must pass through the origin. Look, if I substitute x = 0 and y = 0, in equation ax + by = 0, the equation is satisfied. So, the origin lies on the line! We can also say that the line passes through the origin.
Farhan: Let us try for the equation 2x + 3y = 0. Here a = 2, b = 3 but c = 0. If we substitute x = 0, then we get 3y = 0 or y = 0. This means (0, 0) lies on the line. So yes, this line passes through the origin.
Toshi: Suppose our equation has b = c = 0, say, 5x = 0. This becomes x = 0 which is the equation of the y-axis. And the y-axis passes through the origin.
Meena: And if we take a = c = 0, say, 7y = 0, that means y = 0, which is the equation of the x-axis that also passes through the origin.
So, they conclude: Whenever c = 0, the line ax + by + c = 0 will always pass through the origin, irrespective of the values of a or b. Do you agree with them? - Verify if the ordered pair (4, 3) is a solution of 5x − 6y = 2. Explain your reasoning.
- Find any two solutions for each of the following equations:
- In the equations given below, m and n are unknown constants: 2mx + 3y = 7; 4x + ny = −10. If (2, −1) is the solution of both equations, find the values of m and n.
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