How would you solve double inequalities like this (1 Viewer)

imagineee · Mar 24, 2024

u can square both sides

Tryingtodowell · Mar 24, 2024

imagineee said:
u can square both sides

oh yeah forgot about that thanks

Aeonium · Mar 24, 2024

Tryingtodowell said:
View attachment 42781

sketch the graph LOL

Luukas.2 · Mar 29, 2024

You can also find the boundary values by solving

$\begin{align*} |3x - 2| &= |x + 4| \\ \\ \textbf{Case 1:} \qquad 3x - 2 = x + 4 \qquad \implies \qquad 2x &= 6 \qquad \implies \qquad x = 3 \\ \\ \textbf{Case 2:} \qquad 3x - 2 = -(x + 4) \qquad \implies \qquad 4x &= -2 \qquad \implies \qquad x = -\frac{1}{2} \end{align*}$

Then, consider values in each of the three possible regions, x < -1/2, -1/2 < x < 3, and x > 3, where the inequality given in the question is true, false, and true (respectively), giving the solution as:

$x \le -\frac{1}{2} \qquad \text{or} \qquad x \ge 3$

Or, in interval notation,

$x \in \left(-\infty,\,-\frac{1}{2}\right]\cup\left[3,\,+\infty\right)$

How would you solve double inequalities like this (1 Viewer)

Tryingtodowell

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imagineee

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Tryingtodowell

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Aeonium

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Luukas.2

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