Linear equation with absolute values

Solve for x over the real numbers:
abs(2 - 3 x) = abs(5 - x) + 1

Split the equation into two possible cases:
2 - 3 x = abs(5 - x) + 1 or 2 - 3 x = -abs(5 - x) - 1

Subtract abs(5 - x) + 1 from both sides:
1 - 3 x - abs(5 - x) = 0 or 2 - 3 x = -abs(5 - x) - 1

Subtract 1 - 3 x from both sides:
-abs(5 - x) = 3 x - 1 or 2 - 3 x = -abs(5 - x) - 1

Multiply both sides by -1:
abs(5 - x) = 1 - 3 x or 2 - 3 x = -abs(5 - x) - 1

Split the equation into two possible cases:
5 - x = 1 - 3 x or 5 - x = 3 x - 1 or 2 - 3 x = -abs(5 - x) - 1

Subtract 5 - 3 x from both sides:
2 x = -4 or 5 - x = 3 x - 1 or 2 - 3 x = -abs(5 - x) - 1

Divide both sides by 2:
x = -2 or 5 - x = 3 x - 1 or 2 - 3 x = -abs(5 - x) - 1

Subtract 3 x + 5 from both sides:
x = -2 or -4 x = -6 or 2 - 3 x = -abs(5 - x) - 1

Divide both sides by -4:
x = -2 or x = 3/2 or 2 - 3 x = -abs(5 - x) - 1

Add abs(5 - x) + 1 to both sides:
x = -2 or x = 3/2 or 3 - 3 x + abs(5 - x) = 0

Subtract 3 - 3 x from both sides:
x = -2 or x = 3/2 or abs(5 - x) = 3 x - 3

Split the equation into two possible cases:
x = -2 or x = 3/2 or 5 - x = 3 x - 3 or 5 - x = 3 - 3 x

Subtract 3 x + 5 from both sides:
x = -2 or x = 3/2 or -4 x = -8 or 5 - x = 3 - 3 x

Divide both sides by -4:
x = -2 or x = 3/2 or x = 2 or 5 - x = 3 - 3 x

Subtract 5 - 3 x from both sides:
x = -2 or x = 3/2 or x = 2 or 2 x = -2

Divide both sides by 2:
x = -2 or x = 3/2 or x = 2 or x = -1

x = -2    or       x = 2

Thanks. I think I get it now.

These double absoutes are always long and tricky.

|3x-2|-1=|5-x|

There are 4 possibilities that might lead to different answers.

$1)\quad If\quad3x-2\ge0\quad and \quad 5-x\ge 0\\ \qquad x\ge 2/3\qquad and \qquad x\le 5\\ \qquad \qquad 2/3\le x\le5\\ then\\ 3x-2-1=5-x\\ \qquad\;\; \quad4x=8\\ \qquad\; \qquad x=2\\ So \;\;x=2\;\;is\;\;a\;\;solution\\~\\ 2)\quad If\quad3x-2\le0\quad and \quad 5-x\le 0\\ \qquad x\le 2/3\qquad and \qquad x\ge 5\\ \qquad No\;\;solution\\ $

$3)\quad If\quad3x-2\ge0\quad and \quad 5-x\le 0\\ \qquad x\ge 2/3\qquad and \qquad x\ge 5\\ \qquad\;\text{The union is }\;\; x\ge 5\\ \qquad3x-2-1=-(5-x)\\ \qquad3x-3=-5+x\\ \qquad2x=-2\\ \qquad x=-1\\ \qquad \text {This is not greater than 5 so there is no solution here}$

$4)\quad If\quad3x-2\le0\quad and \quad 5-x\ge 0\\ \qquad x\le 2/3\qquad and \qquad x\le 5\\ \qquad\;\text{The union is }\;\; x\le 2/3\\ \qquad -3x+2-1=5-x\\ \qquad \qquad -2x=4\\ \qquad \qquad x=-2\\ \qquad \text {This is less than 2/3 so x=-2 is a second solution}\\~\\ \text{So the 2 solutions are } \;\;x=2 \;\;and \;\;x=-2$

After looking at all that effort I and our guest have gone to I expect it would be a lot easier just to solve all the possibilities and then substitute them back in and discard the answers that don't work.

|3x-2|-1=|5-x|

So

1) solve   3x-2-1=5-x   then check the answer works

2) solve    3x-2-1=-5+x   then check the answer works

3) solve    -3x+2-1=5-x   then check the answer works

4) solve    -3x+2-1=-5+x   then check the answer works

This is a REALLY difficult question to answer (Atleast in my knowledge), but I will try my best:

Input:

Solve for x:

|3x-2|-1=|5-x|

Intepretation: Sove for $x$ in the function $|3x-2|-1=|5-x|$

Step 1. Rewrite the expressions:

$\sqrt{\left(3x-2\right)^2}-1=\sqrt{\left(5-x\right)^2}$

Step 2. Square both sides:

$({\sqrt{\left(3x-2\right)^2}-1})^2=(\sqrt{\left(5-x\right)^2})^2$

Step 3. Expand x2:

$(3x-2)^2-2\sqrt{\left(3x-2\right)^2}+1=(5-x)^2$

$9x^4-12x+5-2\sqrt{\left(3x-2\right)^2}=x^2-10x+25$

Step 4. Move the corresponding terms:

$8x^2-2x-20=2\sqrt{\left(3x-2\right)^2}$

Step 5. Square both sides again:

$(8x^2-2x-20)^2=(2\sqrt{\left(3x-2\right)^2})^2$

$64x^4-16x^3-160x^2-16x^3+4x^2+40x-160x^2+40x+400)=4(3x-2)^2$

Step 6. Simplify:

$64x^4-32x^3-316x^2+80x+400=36x^2-48x+16$

Step 7. Move the terms again:

$64x^4-32x^3-352x^2+128x+384=0$

Step 8. Divide by their greatest common divisor (i.e. $32$)

$2x^4-x^3-11x^2+4x+12=0$

Step 9. Factorize the function:

Now we use the rational root theorem:

For a polynomial $ax^n+bx^{n-1}+cx^{n-2}+...+yx^1+z$

The possible rational roots will be located at $± \frac{all\space\space factors\space of\space a}{all\space factors\space of\space z}$

Now find the factors of both $a$ and $z$

9.1 The factors of $12=z$ is: $1,2,3,4,6,12$

The factors of $2=a$ is: $1,2$

9.2 Plug the factors in:

$±\frac{1,2,3,4,6,12}{1,2}$

9.3 Expand & List them out:

$±1,2,3,4,6,12,1,\frac{3}{2},\frac{1}{2},2,3,6$

9.4 Cancel & Reorganize the repeated possible roots:

$±\frac{1}{2},1,2,\frac{3}{2},3,4,6,12$

9.5 Plug all the numbers above into the equation

$2x^4-x^3-11x^2+4x+12=0$

Is it $±\frac{1}{2}$? No.

Is it $±1$? $1$ doesn't, while $-1$ does

We can deduct that $(x+1)$ is a factor of the polynomial:

9.6 Perform polynomial long division:

Solve for $\frac{2x^4-x^3-11x^2+4x+12}{x+1}=(2x^3-3x^2-8x+12)$

9.7 Factor $(2x^3-3x^2-8x+12)$

Factor $x^2$ from $2x^3-3x^2$, $-4$ from $-8x+12$

$=x^2(2x-3)-4(2x-3)$

Merge the terms:

$=(x^2-4)(2x-3)$

9.71 Factor $(x^2-4)$

Use the law: $a^2-b^2=(a+b)(a-b)$ with $a=1$ and $b=2$

$=(x+2)(x-2)$

Therefore:

$2x^4-x^3-11x^2+4x+12=(x+1)(x+2)(x-2)(2x-3)$

Step 10.  Find the possible roots of $x$

$x+1=0, x+2=0, x-2=0, 2x-3=0$

$x=-1\space or\space -2\space or\space 2\space or\space \frac{3}{2}$

Step 11. Final Step:

Plug the four $x$ above into $|3x-2|-1=|5-x|$

And making sure the left and the right side is the same:

In the end: $x=-2,2$

Therefore the final solution for $|3x-2|-1=|5-x|$ is:

$x=±2$

(There must be an easier way to solve this.)

I always think it is worth drawing (or sketching) a graph for these inequality questions.  It can help cut down the algebraic work: