Rewrite without absolute value for the given conditions: y=|x−3|+|x+2|−|x−5|, if x<−2

y=-x-4 given that x is less than -2.

Step-by-step explanation:

|x-3|=x-3 when x-3 is positive, . So solving by adding 3 on both sides gives

|x-3|=-(x-3) when x-3 is negative, . So solving by adding 3 on both sides gives

is included in the last one since these values are less than 3. This means for our problem with given condition that |x-3|=-(x-3)=-x+3.

|x+2|=x+2 when x+2 is positive, . So solving by subtracting 2 on both sides gives

|x+2|=-(x+2) when x+2 is negative, . So solving by subtracting 2 on both sides gives

is included in the last one since these values are less than -2. This means for our problem with given condition that |x+2|=-(x+2)=-x-2.

|x-5|=x-5 when x-5 is positive, . So solving by adding 5 on both sides gives

|x-5|=-(x-5) when x-5 is negative, . So solving by adding 5 on both sides gives

is included in the last one since these values are less than 5. This means for our problem with given condition that |x-5|=-(x-5)=-x+5.

Lets put it altogether now.

y=|x−3|+|x+2|−|x−5|, if x<−2

y=-x+3+-x-2-(-x+5)

y=-x+3-x-2+x-5

y=-x-x+x+3-2-5

y=-x-4

c.

step-by-step explanation: