Solve |x - 2| + |x - 1| = x - 3 | vedclass

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(D) We need to solve the equation $|x - 2| + |x - 1| = x - 3$. Case $1$: $x < 1$. In this interval,$|x - 2| = -(x - 2) = -x + 2$ and $|x - 1| = -(x -

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(D) We need to solve the equation $|x - 2| + |x - 1| = x - 3$. Case $1$: $x In this interval,$|x - 2| = -(x - 2) = -x + 2$ and $|x - 1| = -(x - 1) = -x + 1$. The equation becomes $(-x + 2) + (-x + 1) = x - 3$,which simplifies to $-2x + 3 = x - 3$. Solving for $x$,we get $3x = 6$,so $x = 2$. However,$x = 2$ is not in the interval $x Case $2$: $1 \le x In this interval,$|x - 2| = -(x - 2) = -x + 2$ and $|x - 1| = x - 1$. The equation becomes $(-x + 2) + (x - 1) = x - 3$,which simplifies to $1 = x - 3$. Solving for $x$,we get $x = 4$. However,$x = 4$ is not in the interval $1 \le x Case $3$: $x \ge 2$. In this interval,$|x - 2| = x - 2$ and $|x - 1| = x - 1$. The equation becomes $(x - 2) + (x - 1) = x - 3$,which simplifies to $2x - 3 = x - 3$. Solving for $x$,we get $x = 0$. However,$x = 0$ is not in the interval $x \ge 2$,so there is no solution in this case. Since no case yields a valid solution,the equation has no solution.

Solve $|x - 2| + |x - 1| = x - 3$

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