# Find all values of x that satisfy this basic inequality.

$(4x-5)/(3x+5) ≥ 3$

I multiply both sides by (3x+5), getting me:

$(4x-5) ≥ 3(3x+5)$

which simplifies to

$(4x-5) ≥ 9x + 15$

after solving for x, I get

$x ≤ -4$

But after testing through Wolframalpha, I am given:

$-4 ≤ x ≤ -5/3$

Which I don't really understand how they got. I tried multiplying the top side of the fraction by the denominator and then expanding the factored form, and then I rearranged everything once I had a quadratic equation, and I got $-5/3$ and $-2$ as answers, but still not what Wolframalpha got.

Would appreciate some help and insight into this.

Thanks.

• The error comes from multiplying with something that can be negative. The best way to do it is to collect all the terms on one side and put on a common denominator: $\frac{4x-5}{3x+5} - 3\frac{3x+5}{3x+5} \geq 0$. Then analyze where the numerator and denominator is positive/negative. – Winther Sep 1 '14 at 23:32
• Your error is that you multiplied both sides of your inequality by $3x + 5$, which is positive some of the time, and negative some of the time. Thus leaving the $\geq$ sign as it was was not permitted. – Dave Sep 1 '14 at 23:32
• When you multiplied both sides by (3x + 5), you forgot to consider the case when (3x + 5) < 0 – user137481 Sep 1 '14 at 23:32

If you multiply by a positive number the $\geq$ stays but when you multiply by a negative number sign changes to $\leq$.So you can take 2 cases when $3x+5\geq0$ and $3x+5<0$.Anyway it's better to go $$\frac{4x-5}{3x+5}-3\geq 0\\\frac{4x-5-9x-15}{3x+5}\geq0\\\frac{-5x-20}{3x+5}\geq0$$
For $x>-\frac{5}{3}$:
$$4x-5 \geq 3(3x+5) \Rightarrow 4x-5 \geq 9x+15 \Rightarrow 5x \geq -20 \Rightarrow x \geq -4$$
For $x<-\frac{5}{3}$:
$$4x-5 \leq 3(3x+5) \Rightarrow 4x-5 \leq 9x+15 \Rightarrow \dots$$