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: