Simplifying an inequality: $4x(x-2) \lt 2(2x-1)(x-3)$

Question

I have: $$4x(x-2) \lt 2(2x-1)(x-3)$$ For the last part, do I multiply both things in $()$ by two then solve them like I normally would? If I solve them and then multiply will it work the same? Is that an appropriate plan?

Answer 1

You can simply rearrange the inequality to find the $x$ satisfying said inequality. For instance, \begin{align*} 4x(x-2) &< 2(2x-1)(x-3) \\ 4x^2-8x &< 4x^2-14x+6 \\ 6x &< 6 \\ x &< 1. \end{align*} Since all of our steps are reversible, the inequality is satisfied by those $x<1$.

Answer 2

$$4x(x-2) \lt2(2x-1)(x-3)$$ $$2x^2 - 4x \lt 2x^2-6x-x+3$$ $$2x^2\lt2x^2-3x+3$$ $$0\lt-3x+3$$ $$-3\lt-3x$$ $$\text{Multiplying by }-1\text{ reverses inequality}$$ $$3\gt 3x$$ $$1\gt x$$