You visited us 1 times! Enjoying our articles? Unlock Full Access!

Solving Inequalities

If 13 x+1<2 z...

Question

If 13x + 1 < 2z and z + 3 = 5 $y^{2}$ , then

x is necessarily less than y

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

x is necessarily greater than y

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

x is necessarily equal to y

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

None of the above is necessarily true

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

Open in App

Solution

The correct option is D None of the above is necessarily true
$13 x + 1 < 2 z$ and $z + 3 = 5 y^{2}$
$\Rightarrow 13 x + 1 < 2 (5 y^{2} - 3)$
$\Rightarrow 13 x + 7 < 10 y^{2} \Rightarrow 10 y^{2} > 13 x + 7$

Suggest Corrections

Similar questions

| z - 1 | < | z + 3 |

ω = 2 z + 3 - i

x^{2} + 5 y^{2} + z^{2} = 2 y (2 x + z)

x = 2 y

x = 2 z

2 x = z

z

| z - 1 | < | z + 3 |

ω = 2 z + 3 - i

i = \sqrt{- 1}

Examine the consistency of the system of equations.

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

Join BYJU'S Learning Program