Bunuel wrote:
Official Solution:
We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.
Given:
\(x^2 = y + 5\);
\(z = 2x\);
\(y = z - 2\), or which is the same \(y=(2x)-2\)
Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange: \(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)
So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;
The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.
(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.
(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.
Answer: D
Hello
Bunuel I am not sure what i missed, but in statement 2, y = 4, then x^2 = 9, then x = 3 or -3,
If x = 3 the equation results 49 which is divisible by 7
if x = -3 the equation results -5 which is not divisible by 7
So, i think statement 2 is not sufficient.
Can you please clarify my answer.