Ordered pair(s) that satisfy the inequation $x + y + 1 < 0$ is _____.

(0, 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 2, 4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(2, - 4)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Both (B) and (C)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $(2, - 4)$
Lets check each option one by one,
$(0, 1)$ so put $x = 0, y = 1$ in inequation, we get $x + y + 1 = 0 + 1 + 1 = 2$ which
is greater than zero
$(- 2, 4)$ so put $x = - 2, y = 4$ in inequation, we get $x + y + 1 = - 2 + 4 + 1 = 3$ which
is greater than zero
$(2, - 4)$ so put $x = 2, y = - 4$ in inequation, we get $x + y + z = 2 - 4 + 1 = - 1$ which
is less than zero
Final answer: only option (C) satisfies the inequality

Suggest Corrections

Similar questions

(x, y)

0 \leq x \leq 2 π

2^{{sec}^{2} x} \sqrt{y^{2} - 2 y + 2} \leq 2

(x, y)

2 \sqrt{{sin}^{2} x - 2 sin x + 5} . \frac{1}{4^{{sin}^{2} y}} \leq 1

x + y + 1 < 0

(x, y)

2^{\sqrt{{sin}^{2} x - 2 sin x + 5}} \cdot \frac{1}{4^{{sin}^{2} y}} \leq 1

y \leq 3 x + 1

x - y > 1

Join BYJU'S Learning Program