Different ways of solving inequation - different solutions?

Question

After 'solving' the following exercise, I entered the solution in a website that gave me the graph and I noticed that It wasn't the same as the solution it gave me for the 'original' inequation.

The inequation is:

$5-\frac{4-5x}{2}<3\left(2x-1\right)$

And I solved it like:

1) $\frac{10-4+5x}{2}*\frac{1}{2x-1}<3$

2) $\frac{6+5x}{4x-2}-\frac{3(4x-2)}{4x-2}<0$

3) $\frac{12-7x}{4x-2}<0$

Then I solved it and I got:

Solution= $]-\infty ,\frac{1}{2}[ U [\frac{12}{7},+\infty[$

But, it's wrong, it should be: $[\frac{12}{7},+\infty[$

I then did the exercise in a different way:

1) $\frac{6+5x}{2}-3(2x-1)<0$

2)$\frac{6+5x}{2}-\frac{2*3(2x-1)}{2}<0$

3)$\frac{12-7x}{2}<0$

And the result is now the correct one: $[\frac{12}{7},+\infty[$

Can someone explain me why there are these differences between the first way of solving and the second one?

Answer 1

With inequalities you need to be careful because when you multiply for negative value inequality reverses. For this reason the first attempt gives a wrong result and the second was correct.

Note that you can also use first method but you need to distiguish the cases, notably

• for $2x-1>0$

$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies \frac{10-4+5x}{2}\cdot\frac{1}{2x-1}<3$$

• for $2x-1<0$

$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies \frac{10-4+5x}{2}\cdot\frac{1}{2x-1}>3$$

• for $2x-1=0$

$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies 5-\frac{4-5\frac12}{2}<0\implies 5+\frac34<0$$

Answer 2

If your first step you divided by $(2x-1)$, which isn't allowed in inequalities unless you're sure that it's always positive.

Anyways, if one divides both sides of an inequality by a negative number and reverses the inequality (i.e., replaces $<$ with $>$ or vice versa, and similarly with $\le $ and $\ge$), the resulting inequality is equivalent.

So, you can do case breaking for $2x-1 > 0$ and for $2x-1 \le 0$, and solve further.