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Translate inequalities involving absolute values to inequalities without absolute values

by RoRi

Write the following inequalities in equivalent forms without the absolute values.

a1 is equivalent to b2:

    \[|x| < 3 \iff -3 < x < 3. \]

a2 is equivalent to b5:

    \[|x-1| < 3 \iff -3 < x-1 < 3 \iff -2 < x < 4.\]

a3 is equivalent to b7:

    \[|3-2x| < 1 \iff -1 < 3-2x < 1 \iff 1 > 2x -3 > -1 \iff 4 > 2x > 2 \iff 1 < x < 2.\]

a4 is equivalent to b10:

    \[|1+2x| \leq 1 \iff -1 \leq 1+2x \leq 1 \iff -2 \leq 2x \leq 0 \iff -1 \leq x \leq 0.\]

a5 is equivalent to b3:

    \[|x-1| > 2 \iff x-1>2 \text{ or } x-1 < -2. \quad \text{Thus, } x > 3 \text{ or } x< -1.\]

a6 is equivalent to b8:

    \[|x+2| \geq 5 \iff x+2 \geq 5 \text{ or } x+2 \leq -5. \quad \text{Thus, } x\geq 3 \text{ or } x \leq -7. \]

a7 is equivalent to b9:

    \begin{align*} |5 - x^{-1}|< 1 &\iff -1 < 5-\frac{1}{x} < 1 \\ &\iff 1 > \frac{1}{x} - 5 > -1 \\ &\iff 6 > \frac{1}{x} > 4 \\ &\iff \frac{1}{6} < x < \frac{1}{4}. \end{align*}

a8 is equivalent to b4:

    \begin{align*} |x-5| < |x+1| &\iff (x-5)^2 < (x+1)^2 \\ &\iff x^2 - 10x + 25 < x^2 + 2x + 1 \\ &\iff -12x < -24 \iff x > 2. \end{align*}

a9 is equivalent to b6:

    \begin{align*} |x^2 - 2| \leq 1 &\iff -1 \leq x^2 - 2 \leq 1 \\ &\iff 1 \leq x^2 \leq 3 \\ &\iff -\sqrt{3} \leq x \leq -1 \text{ or } 1 \leq x \leq \sqrt{3}. \end{align*}

a10 is equivalent to b1: x < x^2 - 12 < 4x \iff 0 < x^2 - x - 12 and x^2 -4x - 12 < 0. Thus, we must have 0 < (x-4)(x+3) and (x-6)(x+2) < 0. The first of these inequalities requires (x-4) and (x+3) to both be positive or both be negative. Thus, x < -3 or x > 4. The second requires (x-6) and (x+2) to have opposite signs. So, -2 < x < 6. Combining these restrictions on x we have 4 < x < 6 (b1).

2 comments

  1. Jacare says:

    Commenting to point out a minor typo. For a5, the solution number is b3, not b5.

    Thanks a lot for these solutions. They are a big help for me to go through this book.

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