Question

Name two solutions for each inequality.

Answer 1

So,

#1: $n \geq 3 \frac{11}{16} + 4 \frac{1}{2}$

Convert to like improper fractions.
$n \geq \frac{59}{16} + \frac{72}{16}$

Add.
$n \geq \frac{131}{16}\ or\ 8 \frac{3}{16}$

So, one solution could be $8 \frac{3}{16}$.

Another solution could by 9.  There is also 10, 11, 12, etc., and all numbers in between.


#2: $k \ \textless \ 6 \frac{2}{5} * 15$

Convert into improper fraction form.
$k \ \textless \ \frac{32}{5} * 15$

Multiply.
$\frac{(2^5)(3)(5)}{5}$

Cross-cancel, and we have our final result.
$(2^5)(3) = 96$
k < 96

96 is not a solution.

95 is a solution.

So is 94, 93, 92, etc, and all numbers in between.