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In the following questions three equations numbered I, II and III are given.
You have to solve all the equations either together or separately, or two together and one
separately, or by any other method and Give answer
a: x< y=z
b: x ≤ y < z
c: x < y > z
d: x = y > z
e: x = y = z or if none of the above relationship is established
I. $$x=\sqrt{[(36)^{\frac{1}{2}}\times(1296)^{\frac{1}{4}}}]$$
II. $$2y + 3z = 33$$
III. $$6y + 5z = 71$$
From I : $$x=\sqrt{[(36)^{\frac{1}{2}}\times(1296)^{\frac{1}{4}}}]$$
=> $$x = \sqrt{6 \times 6} = \pm 6$$
Applying the operation : Eqn(II) $$\times 3$$ - Eqn(III)
=> $$(6y - 6y) + (9z - 5z) = (99 - 71)$$
=> $$4z = 28$$
=> $$z = \frac{28}{4} = 7$$
Putting it in Eqn(II), we get :
=> $$2y + 21 = 33$$
=> $$2y = 33 - 21 = 12$$
=> $$y = \frac{12}{2} = 6$$
$$\therefore x \leq y < z$$
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