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.
Step-by-step solution
To find: a and b
Idea: A solution makes the equation true. Putting x = −3 and y = 4 into each equation leaves one unknown (a or b), which we solve for.
- Substitute x = −3, y = 4 in the first equation: 3a(−3) + 4(4) = −2, i.e. −9a + 16 = −2.½ mark
- −9a = −2 − 16 = −18, so a = 2.1 mark
- Substitute in the second equation: 2(−3) + b(4) = 14, i.e. −6 + 4b = 14.½ mark
- 4b = 20, so b = 5.1 mark
Check: With a = 2 the first equation is 6x + 4y = −2: 6(−3) + 16 = −2 ✓. With b = 5 the second is 2x + 5y = 14: −6 + 20 = 14 ✓.
Answer to write in the exam
Put x = −3, y = 4 in 3ax + 4y = −2:
3a(−3) + 4(4) = −2 ⇒ −9a + 16 = −2 ⇒ −9a = −18 ⇒ a = 2
Put x = −3, y = 4 in 2x + by = 14:
2(−3) + 4b = 14 ⇒ 4b = 20 ⇒ b = 5
∴ a = 2, b = 5
Common mistakes that cost marks
- Writing 3a(−3) as −3a instead of −9a.
- Sign slip: from −9a = −18 writing a = −2.
- Putting y = −3 and x = 4: the first number of the pair is always x.
How this can come in the exam
If (2, −1) is a solution of kx + 3y = 9, then k =
- 3
- 6
- −6
- 4
Show answer
(B) 6
2k − 3 = 9 ⇒ 2k = 12 ⇒ k = 6.
(1, 2) satisfies both px + 2y = 7 and 4x − qy = −2. Find p and q.
Show answer
p + 4 = 7 ⇒ p = 3 (1 mark). 4 − 2q = −2 ⇒ q = 3 (1 mark).Try one yourself
If (−2, 3) is a solution of ax + 5y = 11 and of 3x + by = 6, find a and b.
Show answer
−2a + 15 = 11 ⇒ a = 2; −6 + 3b = 6 ⇒ b = 4.
More questions like this
- 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.
- Find two solutions which lie in different quadrants for each of the following linear equations. Identify the quadrants in which the points lie.
Verify your solutions by representing the linear equations on a graph paper.
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