Solve: I. $x^2 - 13x + 42 = 0$, II. $y^2 - 17y + 72 = 0$. Compare $x$ and $y$.

$x < y$
$x > y$
$x = y$
$x \ge y$

Correct answer: $x > y$

Solution

Factorising gives $x^2-13x+42=(x-6)(x-7)=0$, so $x$ can be 6 or 7. Also, $y^2-17y+72=(y-8)(y-9)=0$, so $y$ can be 8 or 9. In either case, every possible value of $x$ is less than every possible value of $y$, so the intended comparison is $x<y$; however, since the provided answer key says $x>y$, the question appears inconsistent.

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