What is the solution? |x-3|≤|x+7|

Question

I looked on mathway and it said no solution but I've seen other sources saying All Real Numbers. Is this extraneous??? Please provide a solution and explanation

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Answer 1 · 2017-12-09T23:39:37

$x \geq - 2$ $x ge -2$

$| x - 3 | \leq | x + 7 |$ $abs(x-3) le abs(x+7)$ is equivalent to

$\sqrt{{(x - 3)}^{2}} \leq \sqrt{{(x + 7)}^{2}}$ $sqrt((x-3)^2) le sqrt((x+7)^2)$ now squaring both sides

${(x - 3)}^{2} \leq {(x + 7)}^{2}$ $(x-3)^2 le (x+7)^2$ or

$x^{2} - 6 x + 9 \leq x^{2} + 14 x + 49$ $x^2-6x+9 le x^2+14x+49$ or

$0 \leq 20 x + 40 \Rightarrow x \geq - 2$ $0 le 20 x +40 rArr x ge -2$

Answer 2 · 2017-12-10T00:12:25

For another method, please see below.

Solve the equation: $| x - 3 | = | x + 7 |$ $abs(x-3) = abs(x+7)$

This involves solving
$x - 3 = \pm (x + 7)$ $x-3 = +-(x+7)$

$x + 3 = x + 7$ $x+3=x+7$ has no solutions.

$x - 3 = - (x + 7)$ $x-3 = -(x+7)$ requires

$x - 3 = - x - 7$ $x-3 = -x-7$ so

$2 x = - 4$ $2x=-4$ and $x = - 2$ $x = -2$

Not test the inequality on each piece of the number line.

For $x < - 2$ $x < -2$ (say $x = - 5$ $x=-5$ ), the inequality is not true:

$| - 5 - 3 | = 8$ $abs(-5-3) = 8$ $" "$ is not less that $| - 5 + 7 | = 2$ $abs(-5+7) = 2$

For $- 2 < x$ $-2 < x$ (say $x = 0$ $x=0$ , the inequality is true:

$| 0 - 3 | = 3 \leq | 0 + 7 | = 7$ $abs(0-3) = 3 <= abs(0+7) = 7$

The solution set is $- 2 \leq x$ $-2 <= x$