Solving inequality with absolute values and fraction

Question

I'm new to inequalities and tried to look everywhere looking for an explanation on how to solve such problems with no luck.

Can anyone please explain how to solve:

$(7|a|-1)/(4-|a|)≥3$

Can I use cross multiplication here to get: $7|a|-1 ≥ 3(4-|a|)$ ??

Edit: I don't know how to format the problem here on stachexchange but I added parenthesises to make it clearer

Answer 1

No, you can't cross-multiply without caution.

Hint:

• First, the domain of validity of the inequation is $\mathbf R\smallsetminus\{-4,4\}$.
• Next, ,if $|a|<4$, it is equivalent to $$7|a|-1\ge 12-3|a|\iff |a|\ge\frac{13}{10}.$$
• Last, if $|a|>4$, it is equivalent to $$|a|-1\le 12-3|a|\iff |a|\le\frac{13}{10}.$$

Can you end, taking into account the various conditions?

Answer 2

hint: Solve for $|a|$ first and if $|a| \le r \implies -r \le a \le r$, and if $|a| \ge r \implies a \le -r \text{ or } a \ge r$.

Answer 3

Hint: $7\lvert a\rvert-\lvert a\rvert=6\lvert a\rvert$.

Answer 4

I suppose you wanted to write

$$\frac{7|a|-1}{4-|a|}\geq3$$

Just bring it to the same denominator:

$$\frac{7|a|-1-3(4-|a|)}{4-|a|}\geq0$$

Then evaluate the sign of numerator and denominator. You cannot do what you tried to, because you do not know the sign of $4-|a|$ and multiplying by a negative number reverses the inequality.

Answer 5

Multiply both sides by the nonnegative quantity $(4-|a|)^2$ to get $$(7|a|-1)(4-|a|)\ge 3(4-|a|)^2.$$ Expanding and simplifying gives $$53|a|\ge 10a^2+52,$$ and now you may square both sides (which are both nonnegative) to give $$(53a)^2\ge (10a^2+52)^2.$$ Rearranging and factoring then gives $$(53a-10a^2+52)(53a-10a^2-52)\ge 0.$$ From here I believe you can proceed jollily.

Answer 6

You can't cross multiply because you do not know that $4-|a| > 0$.

You can do 2 cases.

$4-|a| > 0$ or $4 > |a|$ then

$7|a|-1 \ge 3(4-|a|)$

$7|a|- 1 \ge 12-4|a|$

$11|a| \ge 13

So $|a| \ge \frac {13}{11}$ and $\frac {13}{11} \le |a| < 4$

so $\frac {13}{11} \le a < 4$ or $-4 < a \le -\frac {13}{11}$.

Case 2:

$4-|a| < 0$ and $4 < |a|$

then $7|a|-1 \le 3(4-|a|)

$7|a|- 1 \le 12-4|a|$

$11|a| \le 13$

$|a| \le \frac {13}{11}$.

But that contradicts $|a| > 4$ so there are no solutions.

....

Or you can bring to the same side of the inequality

$\frac {7|a|-1}{4-|a|} \ge 3$

$\frac {7|a|-1}{4-|a|} - 3 \ge 0$

$\frac {7|a|-1 -3(4-|a|)}{4-|a|} \ge 0$

$\frac {11|a| -13}{4-|a|} \ge 0$.

so either

Case 1:

$11|a| - 13 \ge 0$ and $4-|a| > 0$

and $\frac {13}{11} \le |a| < 4$ so

so $\frac {13}{11} \le a < 4$ or $-4 < a \le -\frac {13}{11}$.

Case 2:

$11|a|-13 \le 0$ and $4-|a| < 0$

And $a\le \frac {13}{11} $ and $|a| > 4$ which is impossible.