In the following questions, equations I and II are given. Solve both equations and mark the appropriate relation. I. $16x^2 - 32x + 15 = 0$ II. $16y^2 - 48y + 35 = 0$

x > y
x < y
x ≥ y
x ≤ y

Correct answer: x ≤ y

Solution

Equation I factors into $(4x-3)(4x-5)=0$, so $x=\frac34$ or $\frac54$. Equation II factors into $(4y-5)(4y-7)=0$, so $y=\frac54$ or $\frac74$. Thus every possible value of x is less than or equal to every possible value of y, so $x \le y$.

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