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.
- (i) 5x + 3y = 7
- (ii) 5x − 3y = 7
- (iii) −5x + 3y = 7
- (iv) −5x − 3y = 7
Step-by-step solution
Idea: Quadrant I: (+, +); II: (−, +); III: (−, −); IV: (+, −). Try small positive and negative values of x and keep those that give a whole-number y with the sign you need.
(i) 5x + 3y = 7
- x = 2: 10 + 3y = 7 ⇒ y = −1 → (2, −1), in Quadrant IV.½ mark
- x = −1: −5 + 3y = 7 ⇒ y = 4 → (−1, 4), in Quadrant II.½ mark
(ii) 5x − 3y = 7
- x = 2: 10 − 3y = 7 ⇒ y = 1 → (2, 1), in Quadrant I.½ mark
- x = −1: −5 − 3y = 7 ⇒ y = −4 → (−1, −4), in Quadrant III.½ mark
(iii) −5x + 3y = 7
- x = 1: −5 + 3y = 7 ⇒ y = 4 → (1, 4), in Quadrant I.½ mark
- x = −2: 10 + 3y = 7 ⇒ y = −1 → (−2, −1), in Quadrant III.½ mark
(iv) −5x − 3y = 7
- x = 1: −5 − 3y = 7 ⇒ y = −4 → (1, −4), in Quadrant IV.½ mark
- x = −2: 10 − 3y = 7 ⇒ y = 1 → (−2, 1), in Quadrant II. Plotting each pair of points and joining them gives the four lines in the diagram, which confirms the answers.½ mark
Check: (iii) −5(1) + 3(4) = 7 ✓, −5(−2) + 3(−1) = 7 ✓. (iv) −5(1) − 3(−4) = 7 ✓, −5(−2) − 3(1) = 7 ✓.
Answer to write in the exam
(i)
x = 2: 10 + 3y = 7 ⇒ y = −1 ⇒ (2, −1), Quadrant IV
x = −1: −5 + 3y = 7 ⇒ y = 4 ⇒ (−1, 4), Quadrant II
(ii)
x = 2: 10 − 3y = 7 ⇒ y = 1 ⇒ (2, 1), Quadrant I
x = −1: −5 − 3y = 7 ⇒ y = −4 ⇒ (−1, −4), Quadrant III
(iii)
x = 1: −5 + 3y = 7 ⇒ y = 4 ⇒ (1, 4), Quadrant I
x = −2: 10 + 3y = 7 ⇒ y = −1 ⇒ (−2, −1), Quadrant III
(iv)
x = 1: −5 − 3y = 7 ⇒ y = −4 ⇒ (1, −4), Quadrant IV
x = −2: 10 − 3y = 7 ⇒ y = 1 ⇒ (−2, 1), Quadrant II
Common mistakes that cost marks
- Mixing up quadrants II and IV. II has x negative and y positive; IV has x positive and y negative.
- Choosing a point on an axis, like (0, 73). Points on an axis are not in any quadrant.
- Losing a sign with −3y: in (iv), −3y = 12 gives y = −4, not 4.
How this can come in the exam
The point (−3, 2) lies on 2x + 5y = 4. In which quadrant is it?
- I
- II
- III
- IV
Show answer
(B) II
x < 0, y > 0 ⇒ Quadrant II. (Check: −6 + 10 = 4 ✓.)
Find a solution of x + y = 2 in Quadrant IV and one in Quadrant II.
Show answer
IV: (5, −3) since 5 − 3 = 2 (1 mark). II: (−1, 3) since −1 + 3 = 2 (1 mark).Try one yourself
Find two solutions of 3x − 2y = 5 lying in different quadrants.
Show answer
(3, 2) in Quadrant I and (1, −1) in Quadrant IV (also (−1, −4) in III).
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