+0

# Linear equation with absolute values

+1
594
7
+558

Solve for x:

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

Aug 22, 2017

#1
+2

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

Aug 22, 2017
#2
+558
+1

Thanks. I think I get it now.

Aug 22, 2017
#3
+100803
+1

These double absoutes are always long and tricky.

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

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

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

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

.
Aug 22, 2017
#4
+100803
+1

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

Aug 22, 2017
#5
+178
+1

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.)

Aug 22, 2017
#6
+27795
+1

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

Aug 22, 2017
#7
+100803
0

Thanks Alan,

Graphing is always a good option :)

Melody  Aug 22, 2017