Is −3 ≤ x ≤ 3 ? (1) x^2 + y^2 = 9 (2) x^2 + y ≤ 9

Statement 1-

\(x^2+y^2=9\)
\(x^2=9-y^2\)

As \(y^2≥0\)
\(0≤x^2≤9\)

\(-3≤x≤3\)

Sufficient

Statement 2-

\(x^2+y≤9\)
\(y ≤ 9-x^2\)
x∈ [-∞, ∞)

Insufficient

Can someone please explain the approach to solve this question? please

Niha01

Statement 1: x^2 + y^2 = 9 => y^2 = \sqrt{9 - x^2}

Since root of negative number is not possible, sufficient to answer −3 ≤ x ≤ 3

Statement 2: A range of values are possible as this is an inequality and we do not know exact values of x or y. i.e. no unique value.

Nothing much in the question stem except a range for x. But let us simplify it.
\(−3 ≤ x ≤ 3\) = \(| x |≤ 3\) or \(x^2\leq{9}\)

(1) \(x^2 + y^2 = 9\)
\(y^2\geq{0}\), as all squares are non-negative.
\(x^2=9-y^2\) That is, the value of x^2 decreases from 9 to 0 as value of y^2 increases from 0 to 9
So \({x^2}\leq 9\) always irrespective of value of y.
Sufficient

(2) \(x^2 + y ≤ 9\)
\(x^2\leq 9-y\)
So \(x^2\) will be greater than 9, when y is negative, but will be lesser than 9 when y is positive.
Insufficient

A

Bunuel chetan2u

For statement 1, can't we just square both sides of the equation?

By doing so, we'd be left with... x + y = 3

In which case...

0+3= 3
1+2 = 3
2+1 =3
3+0 = 3

All satisfy the question, therefore the statement is sufficient.

Please let me know if this approach is acceptable - thanks!

I think you mean taking the square root rather than squaring. The problem with that is that \(\sqrt{x^2 + y^2}\) generally does not equal to x + y (it does when on of the unknowns is 0 and another is nonzero). Fore example, \(\sqrt{3^2 + 4^2}=5\), while 3 + 4 = 7.

Moreover, even if (1) were x + y = 3, it would be insufficient. Any value of x will satisfy it when y = 3 - x. For example, if x = -100, then y = 103.

I believe you intended to say take the square root instead of square . However, this presents a problem as \(\sqrt{x^2 + y^2}\) does not typically equal x + y (with the exception of one variable being zero and the other nonnegative). For instance, \(\sqrt{3^2 + 4^2}=5\), while 3 + 4 = 7.

Furthermore, even if we assume that (1) is x + y = 3, it would still be insufficient. Any value of x could satisfy this equation as long as y = 3 - x. For instance, if x = -100, then y would equal 103.

Hope it's clear.

APPRACH 1:

Is \(−3 ≤ x ≤ 3\)?

(1) \(x^2 + y^2 = 9\)

If x were greater than 3 or less than -3, then \(x^2\) would be greater than 9. Since \(y^2 \geq 0\), then \(x^2 + y^2\) would be greater than 9, and not equal to it. Therefore, x cannot be greater than 3 or less than -3, which means that \(-3 \leq x \leq 3\).

Sufficient.

(2) \(x^2 + y \leq 9\)

Rearranging the inequality, we get \(x^2 ≤ 9 - y\). However, since there is no constraint on the value of y, the statement is just saying that \(x^2\) is less than or equal to some unknown number (the statement might as well be x^2 ≤ z, where we know nothing about z), which is clearly insufficient.

Not sufficient.


APPRACH 2:

Is \(−3 ≤ x ≤ 3\)?

(1) \(x^2 + y^2 = 9\)

The equation \(x^2 + y^2 = 9\) represents a circle centered at the origin with a radius of \(3\). Therefore, the x-coordinate of any point on the circle must be between \(-3\) and \(3\). Hence, the answer is \(-3 \leq x \leq 3\).


Sufficient.

(2) \(x^2 + y \leq 9\)

The inequality \(y \leq -x^2 + 9\) represents the region below a downward-facing parabola that opens towards the negative y-axis. The vertex of this parabola is at the origin and its arms extend indefinitely. Therefore, x can take any value, so this statement is also insufficient.


Not sufficient.

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