How do you solve $| 5 x + 2 | \geq | 3 x - 4 |$ $abs(5x+2)>=abs(3x-4)$ ?

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Answer 1 · 2018-03-21T12:10:40

$x \leq - 3$ $x le -3$ and $x \geq \frac{1}{4}$ $x ge 1/4$

This inequality is equivalent to

$\sqrt{{(5 x + 2)}^{2}} \geq \sqrt{{(3 x - 4)}^{2}}$ $sqrt((5x+2)^2) ge sqrt((3x-4)^2)$ now squaring both sides

${(5 x + 2)}^{2} \geq {(3 x - 4)}^{2}$ $(5x+2)^2 ge (3x-4)^2$ or

${(5 x + 2)}^{2} - {(3 x - 4)}^{2} \geq 0$ $(5x+2)^2 - (3x-4)^2 ge 0$ or

$(5 x + 2 - 3 x + 4) (5 x + 2 + 3 x - 4) \geq 0$ $(5x+2-3x+4)(5x+2+3x-4) ge 0$ or

$(2 x + 6) (8 x - 2) \geq 0$ $(2x+6)(8x-2) ge 0$ or

$(x + 3) (x - \frac{1}{4}) \geq 0$ $(x+3)(x-1/4) ge 0$ or

$x \leq - 3$ $x le -3$ and $x \geq \frac{1}{4}$ $x ge 1/4$

How do you solve abs(5x+2)>=abs(3x-4)|5x+2|≥|3x−4|abs(5x+2)>=abs(3x-4)?