Solve 3x + 5 = -2|x - 1|.

Step-by-step solution

1. 1

   Use the sign of the right side to prune the work. An absolute value is never negative, so $|x-1|\ge0$ and therefore $-2|x-1|\le0$. The left side must match:

   $3x+5\le0\quad\Longrightarrow\quad x\le-\frac53$

   This one line eliminates most of the number line before any case splitting — and in particular guarantees $x<1$.

2. 2

   Reduce to a single case. Since every candidate satisfies $x\le-\tfrac53<1$, the inside of the absolute value is negative, so

   $|x-1|=1-x$

   The case $x\ge1$ never has to be examined at all.

3. 3

   Substitute and expand.

   $3x+5=-2(1-x)=-2+2x$

4. 4

   Solve the resulting linear equation.

   $3x-2x=-2-5\quad\Longrightarrow\quad x=-7$

5. 5

   Confirm the candidate satisfies the case condition. Is $-7\le-\tfrac53$? Yes, $-7<-1.67$ ✓. Had the algebra produced a value above $1$, it would have been an extraneous root and would have to be discarded — that check is not optional.

6. 6

   Verify in the original equation, and confirm uniqueness. Left: $3(-7)+5=-16$. Right: $-2|-7-1|=-2(8)=-16$ ✓. Scanning $6001$ points from $-30$ to $30$ finds $x=-7$ and no other solution ✓, which the sign argument already predicted since $3x+5$ is increasing while $-2|x-1|$ is increasing more slowly on $x<1$.