1.3. The substitution technique

We have already seen in example 1.2b that the solution to $\displaystyle \frac{x^2+2x+3}{2x^2+x-6}<0$ is $-2 < x < \frac{3}{2}$.

Follow-up question: solve $\displaystyle \frac{x^2 + 2|x| + 3}{2x^2 + |x| - 6} < 0$.

Notice that $x$ in the original expression has been replaced with $|x|$ (do you know why the $x^2$ term remains $x^2$ and does not need to be changed to $|x|^2$?).

For such follow-up questions, it is not necessary to repeat the earlier steps again. Instead, we can use the previous solutions $-2 < x <\frac{3}{2}$ and replace $x$ with $|x|$ to get $-2 < |x| < \frac{3}{2}$. We then solve this inequality (graphs can come in handy. You may also want to refer to example 1.1 ). This gives the solution $-\frac{3}{2} < x < \frac{3}{2}$.

Other common substitutions

The following examples showcase various common substitutions, starting from $\displaystyle \frac{x^2+2x+3}{2x^2+x-6}<0$. Are you able to identify them?
Hover or click on each question to reveal the substitution and resulting answer.

$\displaystyle \frac{\mathrm{e}^{2x} +2 \mathrm{e}^x +3}{2\mathrm{e}^{2x}+\mathrm{e}^{x}-6}<0$ $ \mathrm{e}^x, \quad x < \ln \frac{3}{2} $

$\displaystyle \frac{ (\ln x)^2 +2 \ln x +3}{2 (\ln x)^2 + \ln x -6}<0$ $ \ln x, \quad \mathrm{e}^{-2} < x < \mathrm{e}^{\frac{3}{2}} $

$\displaystyle \frac{x^2 + 4x + 6}{2(x+1)^2 + x - 5 }<0$ $ x+1, \quad -3 < x < \frac{1}{2} $

$\displaystyle \frac{4x^2 + 4x + 3}{8x^2 + 2x - 6}<0$ $ 2x, \quad -2 < x < \frac{3}{4} $

$\displaystyle \frac{x^2-2x+3}{2x^2 -x -6}<0$ $ -x, \quad -\frac{3}{2}< x < 2 $

$\displaystyle \frac{x^4 + 2x^2 + 3}{2 x^4 + x^2 -6}<0$ $ x^2, \quad -\sqrt{\frac{3}{2}} < x < \sqrt{\frac{3}{2}} $

$\displaystyle \frac{3x^2 + 2x+ 1}{-6x^2 +x +2}<0$ $ \frac{1}{x}, \quad x<-\frac{1}{2} \; \mathrm{or} \; x > \frac{2}{3}$