Respuesta :
So we need to solve the following inequality:
[tex]\lvert\frac{3}{x}-2\rvert\leq5[/tex]The first thing to do here is getting rid of the absolute value. Remember what happen when you remove an absolute value in an inequality:
[tex]\begin{gathered} \lvert a\rvert\leq b \\ a\leq b\text{ and }a\ge-b \end{gathered}[/tex]So we'll have two inequalities. The one multiplied by a negative sign must have the other inequality symbol. Taking this into account let's get back to our problem:
[tex]\begin{gathered} \lvert\frac{3}{x}-2\rvert\leq5 \\ \frac{3}{x}-2\leq5\text{ and }\frac{3}{x}-2\ge-5 \end{gathered}[/tex]Let's solve each separately. We take the first one and we add 2 to both sides:
[tex]\begin{gathered} \frac{3}{x}-2\leq5 \\ \frac{3}{x}-2+2\leq5+2 \\ \frac{3}{x}\leq7 \end{gathered}[/tex]Then we multiply x to both sides but first we need to note that if x is negative then we have to change the inequality symbol. So we have two cases for this inequality, x>0 and x<0.
If x>0:
[tex]\begin{gathered} \frac{3}{x}\leq7 \\ \frac{3}{x}\cdot x\leq7x \\ 3\leq7x \end{gathered}[/tex]And we divide by 7:
[tex]\begin{gathered} \frac{3}{7}\leq\frac{7x}{7} \\ \frac{3}{7}\leq x \end{gathered}[/tex]If x<0 we have:
[tex]\begin{gathered} \frac{3}{x}\leq7 \\ \frac{3}{x}\cdot x\ge7x \\ 3\ge7x \end{gathered}[/tex]We divide both sides by 7 and we get:
[tex]\frac{3}{7}\ge x[/tex]Then we solve the other inequality. We add 2 to both sides:
[tex]\begin{gathered} \frac{3}{x}-2+2\ge-5+2 \\ \frac{3}{x}\ge-3 \end{gathered}[/tex]And we can multiply both sides by x. If x>0 we get:
[tex]\begin{gathered} \frac{3}{x}\cdot x\ge-3x \\ 3\ge-3x \end{gathered}[/tex]Then we divide by -3. Remember that multiplying or dividing by a negative number means that we have to change the inequality symbol:
[tex]\begin{gathered} \frac{3}{-3}\leq\frac{-3x}{-3} \\ -1\leq x \end{gathered}[/tex]If x<0 we have:
[tex]\begin{gathered} \frac{3}{x}\cdot x\leq-3x \\ 3\leq-3x \end{gathered}[/tex]Then we divide by -3. Remember that multiplying or dividing by a negative number means that we have to change the inequality symbol:
[tex]\begin{gathered} \frac{3}{-3}\ge-\frac{3x}{-3} \\ -1\ge x \end{gathered}[/tex]Now let's take everything we have. If x<0 we found that:
[tex]\begin{gathered} \frac{3}{7}\ge x \\ -1\ge x \end{gathered}[/tex]Since 3/7>-1 we just need to use the second inequlity:
[tex]-1\ge x[/tex]For x>0 we got:
[tex]\begin{gathered} \frac{3}{7}\leq x \\ -1\leq x \end{gathered}[/tex]Since 3/7>-1 we just need to use the first inequality:
[tex]\frac{3}{7}\leq x[/tex]Then:
[tex]x\leq-1\text{ and }\frac{3}{7}\leq x[/tex]And that's the answer.
