Question

Please need help in algebra quick

Answer 1

After solving $|8-3p|\geq 2$ we get $p\geq \frac{10}{3}\,\, or\,\,p\leq 2$

So, Option A is correct.

Step-by-step explanation:

We need to solve the expression:

$|8-3p|\geq 2$

For solving, we apply the rule that:

if |u| ≥ a and a > 0 then, u ≤ -a and u ≥ a

Applying the rule:

$|8-3p|\geq 2\\8-3p\leq -2 \,\, and\,\,8-3p\geq 2$

$8-3p-8\leq -2-8 \,\, or\,\,8-3p-8\geq 2-8\\-3p\leq -10 \,\, or\,\,-3p\geq -6\\Multiply\,\,both\,\,sides\,\,by\,\,-1 (reverse\,\,the\,\,inequality\\3p\geq 10\,\, or\,\,3p\leq 6\\p\geq \frac{10}{3}\,\, or\,\,p\leq\frac{6}{3}\\p\geq\frac{10}{3}\,\, or\,\,p\leq 2$

After solving $|8-3p|\geq 2$ we get $p\geq \frac{10}{3}\,\, or\,\,p\leq 2$

So, Option A is correct.

Keywords:  Absolute rule

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