Inverse function of y = abs(x) - 3x

My first thought was to make 3 cases:
x > 0: x = -y/2
x = 0: x = -y/3
x < 0: x = -y/4

The answer, however, according to my book is:
x = 1/8(|y| - 3y)

Can someone explain how this answer can be found?

Let's look at functions of the form y = a|x| + bx

If x > 0, then y = (a + b)x
If x = 0, then y = 0
If x < 0, then y = (-a + b)x

Consider your solution:

If x > 0 then x = -y/2
If x = 0, then x = 0
If x < 0, then x = -y/4

We can rewrite this as

If y < 0 then x = -y/2
If y = 0, then x = 0
If y > 0, then x = -y/4

So, an answer of the form x = a|y| + by would require

a + b = -1/4
-a + b = -1/2

2b = -3/4
b = -3/8

a = -1/4 - b = -1/4 + 3/8 = 1/8

So x = (1/8) |y| - (3/8)y = (|y| - 3y)/8

If you redefine y you get y=

-x-3= -4x, x<0

x-3x=-2x, x ≥ 0

The inverse is

x= -4y then y=-x/4, x<0

x=-2y then y=-x/2, x ≥ 0

For example

For x<0

x=-2, f(-2)=8

f^-1(8)=-(8)/4= -2

For x x ≥ 0

at x=2, f(2)=-4

f^-1(-4)= -(-4)/2=2

So the answer 1/8(|y| -3y) is the same as f^-1(x)=

-x/4 , x<0

-x/2, x ≥ 0

The answer in your book is the same as yours. It is just different notation because
1/8 (|y| - 3y) = -y/2 when y < 0 (y < 0 when x > 0)
1/8 (|y| - 3y) = -y/4 when y > 0 (e.g. when x < 0)