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# If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z

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If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]  29 Apr 2012, 01:29
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If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

(1) x > 0
(2) y = 4
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Jul 2013, 01:29, edited 2 times in total.
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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]  29 Apr 2012, 04:24
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keiraria wrote:
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

(1) x>0
(2) y = 4

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:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]  22 Sep 2012, 09:28
Bunuel wrote:
keiraria wrote:
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

(1) x>0
(2) y = 4

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:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

Your explanation is good but i want to know where am i making a mistake.
z=2x
y= 2x-2
x^2 = 2x+3
$$x^3 + y^2 +z = x(2x+3) + (2x-2)^2 + 2x = 6x^2 -3x +4$$
Why cant we use this equation to solve the question.
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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]  05 Feb 2015, 16:10
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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z   [#permalink] 05 Feb 2015, 16:10
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