$\displaystyle 5x = 35$

$\displaystyle 5x \div 5 = 35 \div 5$

$\displaystyle x=7$

$\displaystyle 3y = 24$

$\displaystyle 3y \div 3 = 24 \div 3$

$\displaystyle y = 8$

$\displaystyle y > x$, so (B) is greater.

$\displaystyle \frac{2}{3} a = 20$

$\displaystyle \frac{3}{2} \cdot \frac{2}{3} a =\frac{3}{2} \cdot 20$

$\displaystyle a = \frac{60}{2} = 30$

$\displaystyle \frac{3}{5} b = 24$

$\displaystyle \frac{5}{3}\cdot \frac{3}{5} b =\frac{5}{3}\cdot 24$

$\displaystyle b =\frac{120}{3} = 40$

$\displaystyle b > a$, so (B) is the greater quantity.

$\displaystyle y +8.6 < 3.3$

$\displaystyle y +8.6 - 8.6 < 3.3 - 8.6$

$\displaystyle y < 3.3 - 8.6$

$\displaystyle y < -\left ( 8.6-3.3 \right )$

$\displaystyle y < -5.3$

None of choices fit this criterion, so the correct answer is none.

$\displaystyle -9y \geq 18$

$\displaystyle -9y \div (-9) \leq 18\div (-9)$ (Note that the inequality symbol switches here)

$\displaystyle y\leq -2$

Of the elements of $\displaystyle \left \{ -2, -1, 0, 1, 2 \right \}$, only $\displaystyle -2$ fits this criterion.

If $\displaystyle y = 4$, the equation can be rewritten and solved as follows:

$\displaystyle 4xy = 0$

$\displaystyle 4x \cdot 4 = 0$

$\displaystyle 16x = 0$

$\displaystyle 16x \div 16 = 0 \div 16$

$\displaystyle x = 0$

This is the only number that makes this statement true, so the correct choice is "one".

If $\displaystyle x = -8$, then the equation can be rewritten and solved as follows:

$\displaystyle x^{2} + y^{3} = 0$

$\displaystyle (-8)^{2} + y^{3} = 0$

$\displaystyle 64 + y^{3} = 0$

$\displaystyle 64 + y^{3} -64 = 0-64$

$\displaystyle y^{3} = -64$

The only integer that can be cubed to yield the result $\displaystyle -64$ is $\displaystyle -4$, so the correct response is "one".

If $\displaystyle y = 12$, the equation can be restated and solved as follows:

$\displaystyle x^{2} - y^{2} = 0$

$\displaystyle x^{2} - 12^{2} = 0$

$\displaystyle x^{2} -144 = 0$

$\displaystyle x^{2} -144 + 144 = 0 + 144$

$\displaystyle x^{2} = 144$

Both $\displaystyle x = 12$ and $\displaystyle x = -12$ make this true, so both make the original statement true. "Two" is the correct choice.

$\displaystyle y + 3\frac{1}{2} < 2 \frac{1}{4}$

$\displaystyle y + 3\frac{1}{2} - 3\frac{1}{2} < 2 \frac{1}{4} - 3\frac{1}{2}$

$\displaystyle y < 2 \frac{1}{4} - 3\frac{1}{2}$

$\displaystyle y < -\left ( 3\frac{1}{2} - 2 \frac{1}{4} \right )$

$\displaystyle y < - 1 \frac{1}{4}$

Of the elements of the set $\displaystyle \left \{ -2, -1, 0, 1, 2 \right \}$, only $\displaystyle -2$ fits this criterion, making "one" the correct choice.

$\displaystyle \frac{c}{-6 } = 25$

$\displaystyle \frac{c}{-6 }\times (-6) = 25 \times (-6)$

$\displaystyle c= -150$

$\displaystyle \frac{d}{-9} = 20$

$\displaystyle \frac{d}{-9}\times (-9) = 20 \times (-9)$

$\displaystyle d = -180$

Since $\displaystyle 150 < 180$, $\displaystyle -150 > -180$.

$\displaystyle c > d$, so (A) is greater.

If $\displaystyle x = 0$, this can be rewritten as

$\displaystyle 9xy = 0$

$\displaystyle 9 \cdot 0 \cdot y =0$

$\displaystyle 0 y =0$

However, since zero multiplied by any number yields a product of zero, this is a true statement for all values of $\displaystyle y$. This makes "infinitely msny" correct.