The first equation can be solved algebraically,
\(3(x – 7) \geq 9\)

or\(x - 7 \geq 3\)

or \(x \geq 10\)

For 2nd equation:

\(0.25y – 3 \leq 1\)

or \(0.25y \leq 4\)

or \(y\leq 16\)

So we cannot reach to a definite solution for this because x can be of any value from 10 and y can be any value less than or equal to 16

So option is D
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Given: \(3(x – 7) \geq 9\)
Divide both sides of the inequality by \(3\) to get: \(x – 7 \geq 3\)
Add \(7\) to both sides to get: \(x \geq 10\)

Also given: \(0.25y – 3 \leq 1\)
Add \(3\) to both sides to get: \(0.25y \leq 4\)
Multiply both sides by \(4\) to get: \(y \leq 16\)

We now know that \(x \geq 10\) and \(y \leq 16\)
So, for example, it could be the case that x = 12 and y = 12, in which case the two quantities are equal
Or, it could be the case that x = 13 and y = 12, in which case Quantity A is greater

Answer: D

Cheers,
Brent
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\(3(x-7) \geq 9 \text{ ---(1)}\)
\(0.25y-3 \leq 1 \text{ ---(2)}\)

Lets deal with (1) first

\(3(x-7) \geq 9\)

\((x-7) \geq \frac{9}{3}\)

\(x-7 \geq 3\)

\(x \geq 3+7\)

\(x\geq 10\)


Lets now handle (2)

\(0.25y-3 \leq 1\)

\(0.25y \leq 1+3\)

\(0.25y \leq 4\)

\(y \leq \frac{4}{0.25}\)

\(y\leq 16\)

So we now have two ranges.

\(x\) ranges from 10 to infinity

\(y\) ranges from 16 to minus infinity

\(x\) and \(y\) can therefore take any of the following values \(10,11,12,13,14,15,16\)

If \(x = 16\) and \(y= 10\), Quantity A is greater, if it is the other way round, Quantity B is greater. If both \(x\) and \(y\) equal any of the above numbers, both Quantities are Equal. Therefore, there is no discernible relationship between the two quantities.

The correct answer is Choice D.
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