Question

Please help with the problem below! Thank you

Answer 1

Answer:

$d \le -3$

Step-by-step explanation:

The given inequality is;

$\frac{d-3}{10} \ge \frac{2d+3}{5}+\frac{d+3}{3}$

Clear the fraction by multiply through by 30;

$3(d-3) \ge 6(2d+3)+10(d+3)$

Expand;

$3d-9\ge 12d+18+10d+30$

Group like terms;

$3d-12d-10d \ge 18+30+9$

$-19d \ge 57$

Divide through by -19 and reverse the sign.

$d \le -3$

Answer 2

Answer:

Choice A is correct. d ≤ -3 is the answer.

Step-by-step explanation:

 We have given an inequality:

$\frac{d-3}{10}\geq \frac{2d+3}{5}+\frac{d+3}{5}$

We have to solve the inequality.

$\frac{d-3}{10}\geq \frac{3(2d+3)+5(d+3)}{15}=\frac{6d+9+5d+15}{15}=\frac{11d+24}{15}$

$\frac{d-3}{10}\geq \frac{11d+24}{15}$

$15(d-3)\geq10(11d+24)$

$15d-45\geq110d+240$

15d-110d ≥ 240+45

-95d ≥ 285

d ≤ -3 which is the answer.