If |x - 2| + |x - 3| = 7,then x = | vedclass

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(C) The critical points are $x = 2$ and $x = 3$.Case $1$: $x < 2$$|x - 2| = -(x - 2)$ and $|x - 3| = -(x - 3)$$-(x - 2) - (x - 3) = 7$$-x + 2 - x + 3

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$6$ or $-1$

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(C) The critical points are $x = 2$ and $x = 3$.Case $1$: $x $|x - 2| = -(x - 2)$ and $|x - 3| = -(x - 3)$$-(x - 2) - (x - 3) = 7$$-x + 2 - x + 3 = 7$$-2x + 5 = 7$$-2x = 2$$x = -1$ (which satisfies $x Case $2$: $2 \le x $|x - 2| = x - 2$ and $|x - 3| = -(x - 3)$$(x - 2) - (x - 3) = 7$$x - 2 - x + 3 = 7$$1 = 7$ (which is impossible)Case $3$: $x \ge 3$$|x - 2| = x - 2$ and $|x - 3| = x - 3$$(x - 2) + (x - 3) = 7$$2x - 5 = 7$$2x = 12$$x = 6$ (which satisfies $x \ge 3$)Thus,the solutions are $x = 6$ or $x = -1$.

If $|x - 2| + |x - 3| = 7$,then $x =$

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