How do you solve #-abs(x+1)=-2#?

Question

Hazel Anglin · Accepted Answer

The answer is: #x=1# and #x=-3#

The way to solve it is as follows.

Equal parts of an equation, left and right, can be multiplied by the same non-equal to zero multiplier getting an equivalent equation. Let's multiply them by #-1#: #|x+1|=2#

Now we have to remember the definition of the absolute value of a number. If the number is positive or zero, its absolute value equals to itself: if #Z>=0# then #|Z|=Z#. If the number is negative, its absolute value equals to its opposite (or, not very scientifically, minus this number) if #Z<0# then #|Z|=-Z#.

Applying this to a problem at hand:

CASE 1 Looking for solutions in the area defined by an inequality #x+1>=0# (that is, #x>=-1#) then #|x+1|=x+1# and, using our equation, #x+1=2#, that is, #x=1#. This value is within the area #x>=-1# and, therefore, is the first legitimate solution.

CASE 2 Looking for solutions in the area defined by an inequality #x+1<0# (that is, #x<-1#) then #|x+1|=-(x+1)# and, using our equation, #-(x+1)=2#, that is, #x=-3#. This value is within the area #x<-1# and, therefore, is the second legitimate solution.

We can demonstrate it graphically. Our equation is equivalent to #|x+1|-2=0# Let's draw a graph of a function #y=|x+1|-2# graph{|x+1|-2 [-10, 10, -5, 5]} As you see, it intersects the X-axis (that is, equals to zero) at points #x=-3# and #x=1#. This confirms our solutions.

Ethan Adkins · Answer

undefined To solve the equation $-\lvert x+1 \rvert = -2$, we can first remove the negative sign by multiplying both sides by -1:

$\lvert x+1 \rvert = 2$

Next, we split this into two cases:

1. $x+1 = 2$ when $x+1 \geq 0$
2. $-(x+1) = 2$ when $x+1 < 0$

Solving each case separately:

1. $x+1 = 2$ when $x \geq -1$
   Solving for $x$, we get $x = 1$.

2. $-(x+1) = 2$ when $x < -1$
   Solving for $x$, we get $x = -3$.

So, the solutions to the equation are $x = 1$ and $x = -3$.