Question

please answer fast.............​

Answer 1

Given that,



x = $\frac{7}{12}$

y = $\frac{0}{12}$

z = $\frac{-6}{12}$

To find,

Whether, x > y, y = z or z > x

Let's try the first condition,

x > y

As we can see,

x = $\frac{7}{12}\:and\:y \:= \frac{0}{12}$

As the denominators are same we don't have to make both the fractions equivalent.

If the denominators were different, then we'd to equivalent to make the denominators equal and then we've to compare it.

Here, the denominators are equal so, we'll directly compare both the ratios.

But, y = $\frac{0}{12}$

Which can be also expressed as 0.

And we know that,

Any positive integer is always greater than the zero.

Therefore, we can say that,

$\frac{7}{12}$ > 0

$\boxed{\bold{x > y}}$

Now, let's try out the second condition which is,

y = z

As we know,

y = 0 and z = $\frac{-6}{12}$

But z when converted in simplest ratio it can also be expressed as,

$\frac{-6}{12} = \frac{\cancel{-6}}{\cancel{12}} = \frac{-1}{2}$

Now, by comparing the values of y and z, we find that,

0 < $\frac{-1}{2}$

Therefore, we cannot say that,

y = z

Therefore, y ≠ z.

Now, let's try out the third one.

z > x

We know that,

z = $\frac{-1}{2}$ and y = $\frac{7}{12}$

As the denominators are equal, we don't wanna to equivalent them (as said earlier).

Now by comparing z and y, we find that,

z < y

Therefore, the above given condition is wrong.

Therefore, $\boxed{\bold{z < y}}$

The correct condition is,

$\boxed{\bold{x > y}}$