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

Concept of Inequality

If x > y ; th...

Question

If x > y ; then which of the following statement holds true for the given inequation

x + z > y + z

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

x + z < y + z

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

x + z = y + z

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

x + z $\geq$ y + z

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

Open in App

Solution

The correct option is A
x + z > y + z

If x > y ; then when we add the same number on both the sides, then the inequality does not change,

x + z will be still greater than y + z

Suggest Corrections

Similar questions

x = \frac{4}{7}, y = \frac{1}{2}

z = (- \frac{3}{4})

x \div y \neq y \div x

(x \div y) \div z \neq x \div (y \div z)

x \div (y + z) \neq x \div y + x \div z

x \div (y - z) \neq x \div y - x \div z

∣ ∣ ∣ ∣ ∣ \begin{matrix} y z - x^{2} & z x - y^{2} & x y - z^{2} z x - y^{2} & x y - z^{2} & y z - x^{2} x y - z^{2} & y z - x^{2} & z x - y^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

(x + y + z)

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & y z y & y^{2} & z x z & z^{2} & x y \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (x y + y z + z x)

x + y + z = x y z

\frac{x + y}{1 - x y} + \frac{y + z}{1 - y z} + \frac{z + x}{1 - z x} = \frac{x + y}{1 - x y} \cdot \frac{y + z}{1 - y z} \cdot \frac{z + x}{1 - z x}

Join BYJU'S Learning Program