If x^2 = y + 5, y= z- 2 and z=2x, is x^3 + y^2 + z divisible by 7?
Please explain me how the equation got reduced to x^2 - 2*x - 3 = 0. Hence the question.
Please gimme the link, I could not find it. From 2, y=4, so I got z=6 and x = +-3, so if x>0 from 1, I had C.
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.
\(x^2 = y + 5\)
\(y = z - 2\) --> \(y=2x-2\).
\(z = 2x\)
\(x^2 =2x-2+ 5\) --> \(x^2-2x-3=0\) --> \(x=3\) 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.
Hope it helps.
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