Inequality sign changing when multiplying with a unknown variable?

Question

This is the inequality: $$\frac{(2x-1)}{(3-5x)} < 2$$ Next step is: $$\frac{(2x-1)}{(3-5x)} \cdot (3-5x) < 2\cdot (3-5x)$$ Now is the sign $<$ or $>$?

The end solution is $x < \frac{7}{12}.$

Answer 1

In general, when multiplying an inequality by an unknown quantity, we can't tell whether the sign changes or not. To solve the problem, then, we must consider both cases, i.e. whether the unknown quantity is positive or negative. In this example, we would do the following:

CASE 1: $3-5x > 0$

\begin{align*} \frac{2x-1}{3-5x} (3-5x) < 2(3-5x) \quad \Rightarrow \quad x < \frac{7}{12} \end{align*}

CASE 2: $3-5x < 0$

\begin{align*} \frac{2x-1}{3-5x} (3-5x) > 2(3-5x) \quad \Rightarrow \quad x > \frac{7}{12} \end{align*}

Note: I have left out the case $3-5x = 0$ because it is in the denominator of a fraction. So, there are in fact two solutions: we require $3-5x>0$ and $x<\frac{7}{12}$, OR we require $3-5x<0$ and $x>\frac{7}{12}$.

However, we can simplify this, since $3-5x>0$ implies $x<\frac{3}{5}$, and similarly $3-5x<0$ implies $x>\frac{3}{5}$. Then since $\frac{3}{5} > \frac{7}{12}$, the two possibilities become:

\begin{align*} x < \frac{3}{5} \text{ and } x < \frac{7}{12} \quad \Rightarrow \quad x < \frac{7}{12} \\ x > \frac{3}{5} \text{ and } x > \frac{7}{12} \quad \Rightarrow \quad x > \frac{3}{5} \\ \end{align*}

So, the solution set is $x < \frac{7}{12}$ OR $x > \frac{3}{5}$.

Answer 2

You must distinguish two cases $$3-5x>0$$ or $$3-5x<0$$ in the case of $$3-5x>0$$ you will get $$2x-1<2(3-5x)$$ and in the second case $$2x-1>2(3-5x)$$

Answer 3

The solution $x<7/12$ is NOT complete. Note that for $x=1$, the inequality holds!!

The given inequality is equivalent to $$\begin{cases}(2x-1) < 2(3-5x)\\ (3-5x)>0\end{cases}\quad\cup\quad \begin{cases}(2x-1) > 2(3-5x)\\ (3-5x)<0\end{cases}.$$ Can you take it from here?

Another way: solve $$0>\frac{(2x-1)}{(3-5x)}-2=\frac{12x-7}{3-5x}.$$ What is the COMPLETE set of solutions?

Answer 4

Remove the denominator multiplying by its square, which is non negative: on the domain of validity od the inequation, \begin{align} \frac{(2x-1)}{(3-5x)} < 2&\iff(2x-1)(3-5x)<2(3-5x)^2 \\&\iff 0<(5x-3)(2(5x-3)+2x-1) \iff (5x-3)(12x-7)>0 \end{align} So it comes down to a quadratic inequation. The set of solutions is $$S=\Bigl(-\infty,\frac7{12}\Bigr)\cup\Bigl(\frac3{5},+\infty\Bigr).$$