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1) we know that sum of all angles equal to \(180\) so we can write such equation: \(x^2+6x+243-22x=180\) \(x^2-16x+63=0\) \(x\) can be equal to \(7\) or \(9\) Insufficient

2) we know that sum of all angles equal to \(180\) so we can write such equation: \(x^2+6x+261-24x=180\) \(x^2-18x+81=0\) \(x\) can be equal only to \(9\) Sufficient

The angles of any triangle add up to 180°, so we can write x^2+6x+y=180. However, we cannot figure out angle B, which measures 6x°, until we turn to the additional premises.

Statement (1) gives y=243–22x. Substituting this equation into our original equation gives x^2+6x+243–22x=180. Standardizing and simplifying, this reduces to x^2–16x+63=0. This quadratic gives two possible values of x. It isn’t really necessary to proceed further, but we’ll point out that it factors as (x–7)(x–9)=0 and x=7 or 9. Angle B could measure either 42° or 54°, so statement (1) standing alone is insufficient.

Statement (2) gives y=261–24x. Substituting this equation into our original equation gives x^2+6x+261–24x=180. Standardizing and simplifying, this reduces to x^2–18x+81=0. Unlike the quadratic from statement (1), however, this quadratic is a common algebraic equation – a perfect square. It factors as (x–9)^2=0, and it has only one solution: x=9. In turn, angle B measures 54°. So statement (2) standing alone is sufficient.

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