Question

Solve for x: |x+1|+|x-1|=|2x|​

Answer 1

Answer:

(x+1)+(x-1)=2x

2x=2x

x=1

Answer 2

The solutions are S={1,32}

The solutions are S={1,32}Explanation:

The solutions are S={1,32}Explanation:The equation is

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−∞11323222+∞

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−∞11323222+∞On the first interval (−∞,1)

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−∞11323222+∞On the first interval (−∞,1)−2x+3−x+1=−x+2

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−∞11323222+∞On the first interval (−∞,1)−2x+3−x+1=−x+2⇒, 2x=2

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−∞11323222+∞On the first interval (−∞,1)−2x+3−x+1=−x+2⇒, 2x=2⇒, x=1

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−∞11323222+∞On the first interval (−∞,1)−2x+3−x+1=−x+2⇒, 2x=2⇒, x=1x fits in this interval and the solution is valid

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−∞11323222+∞On the first interval (−∞,1)−2x+3−x+1=−x+2⇒, 2x=2⇒, x=1x fits in this interval and the solution is validOn the second interval (1,32)

The solutions are S={1,32}Explanation:The equation is|2x−3|+|x−1|=|x−2|There are 3 points to consider⎧⎪⎨⎪⎩2x−3=0x−1=0x−2=0⇒, ⎧⎪ ⎪⎨⎪ ⎪⎩x=32x=1x=2There are 4 intervals to consider⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−∞11323222+∞On the first interval (−∞,1)−2x+3−x+1=−x+2⇒, 2x=2⇒, x=1x fits in this interval and the solution is validOn the second interval (1,32)−2x+3+x−1=−