Question

Hi can anyone solve my math problem

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Answer 1

x = -16/3 or -8/9

Explanation:

The equation can be rewritten as a piecewise linear function with two breakpoints, or three domain regions. Each breakpoint is at the value of x where the argument of the absolute value function is zero: at x=4 and x=-2.

For each absolute value, we have ...

|q| = -q for q < 0

|q| = q for q ≥ 0

Then the three parts of the domain are ...

• x < -2 . . . . . . . . where |x +2| has its vertex
• -2 ≤ x < 4 . . . . . between the vertices
• 4 ≤ x . . . . . . . . . where |x -4| has its vertex

__

Subtracting the right side, the equation becomes ...

1/3|x -4| -2/3x -2|x +2| = 0

Then the three piecewise functions are ...

x < -2

Both absolute value arguments are negated.

1/3(-(x -4)) -2/3x -2(-(x +2)) = 0

x(-1/3 -2/3 +2) +4/3 +4 = 0 . . . . . collect terms

x +5 1/3 = 0 . . . . . . . simplify

x = -5 1/3 . . . . . . . . . subtract 5 1/3. This result is in the domain

__

-2 ≤ x < 4

Only the argument of |x -4| is negated.

1/3(-(x -4)) -2/3x -2(x +2) = 0

x(-1/3 -2/3 -2) +4/3 -4 = 0 . . . . collect terms

-3x -8/3 = 0 . . . . . . . simplify

-3x = 8/3 . . . . . . . add 8/3

x = -8/9 . . . . . divide by -3. This result is in the domain

__

4 ≤ x

Neither absolute value function argument is negated.

1/3(x -4) -2/3x -2(x +2) = 0

x(1/3 -2/3 -2) -4/3 -4 = 0 . . . . . collect terms

-7/3x -8/3 = 0 . . . . . . . . . simplify

-7/3x = 8/3 . . . . . . . . add 8/3

x = -8/7 . . . . . . . . . divide by -7/3. This result is not in the domain.

There is no solution in this region.

__

The solutions are ...

• x = -5 1/3
• x = -8/9

Answer 2

The solutions to this equation are x = -8/9 and x = -16/3.

Explanation:

Simplified Equation

We have this equation here:

• 
  `\displaystyle (1)/(3)|x-4|=(2)/(3) x+2|x+(6)/(3) |`

First, let's simplify the right sight of the equation by simplifying 6/3 to 2.

• 
  `\displaystyle (1)/(3)|x-4|=(2)/(3) x+2|x+2 |`

Multiply both sides of the equation by 3 in order to get rid of the fractions.

• 
  `|x-4|=2x+6|x+2|`

Move the terms on either side of the equation to set them equal to 0.

• 
  `|x-4|-2x-6|x+2|=0`

Now, we can split this equation up into 4 possible cases. In each case, we make the absolute values negative or positive.

Case #1 (first absolute value: positive; second absolute value: positive)

• 
  `(x-4)-2x-6(x+2)=0`

Simplify this equation by distributing -6 inside the parentheses.

• 
  `x-4-2x-6x-12=0`

Combine like terms.

• 
  `-7x-16=0`

Add 16 to both sides of the equation and divide by -7.

• 
  `\displaystyle \boxed{x=-(16)/(7)}`

Case #2 (first absolute value: negative; second absolute value: positive)

• 
  `-(x-4)-2x-6(x+2)=0`

Distribute the negative sign and -6 inside of their respective parentheses.

• 
  `-x+4-2x-6x-12=0`

Combine like terms.

• 
  `-9x-8=0`

Add 8 to both sides of the equation and divide by -9.

• 
  `\displaystyle \boxed{x=-(8)/(9)}`

Case #3 (first absolute value: positive; second absolute value: negative)

• 
  `(x-4)-2x-6[-(x+2)]=0`

Distribute the negative sign inside the parentheses first.

• 
  `x-4-2x-6(-x-2)=0`

Now, distribute -6 inside the parentheses.

• 
  `x-4-2x+6x+12=0`

Combine like terms.

• 
  `5x+8=0`

Subtract 8 from both sides of the equation and divide by 5.

• 
  `\displaystyle \boxed{ x=-(8)/(5)}`

Case #4 (first absolute value: negative; second absolute value: negative)

• 
  `-(x-4)-2x-6[-(x+2)]=0`

Distribute the negative signs inside the parentheses first.

• 
  `-x+4-2x-6(-x-2)=0`

Distribute -6 inside the parentheses.

• 
  `-x+4-2x+6x+12=0`

Combine like terms.

• 
  `3x+16=0`

Subtract 16 from both sides of the equation and divide by 3.

• 
  `\displaystyle \boxed{x=-(16)/(3)}`

Extraneous Solutions

Whenever we solve problems with absolute values, we will always need to check for extraneous solutions.

Definition: These are solutions that may come up while solving but do not actually fit in the domain of the original problem.

Checking for these is tedious, but it will help eliminate wrong answers, so let's plug every "solution" for x that we found back into the original equation.

1) x = -16/7

Substitute this value back into the original equation.

• 
  `\displaystyle (1)/(3)|(-(16)/(7)) -4|=(2)/(3) (-(16)/(7)) +2|(-(16)/(7)) +(6)/(3) |`

If we do all the calculations correctly, we will get:

• 
  `\displaystyle (44)/(21) =-(20)/(21)`

Since this equation is NOT true, this means that x = -16/7 is NOT A SOLUTION.

2) x = -8/9

Substitute this value back into the original equation.

• 
  `\displaystyle (1)/(3)|(-(8)/(9)) -4|=(2)/(3) (-(8)/(9)) +2|(-(8)/(9)) +(6)/(3) |`

If we do all the calculations correctly, we will get:

• 
  `\displaystyle (44)/(27) =(44)/(27)`

Since this equation IS true, this means that x = -8/9 IS A SOLUTION.

3) x = -8/5

Substitute this value back into the original equation.

• 
  `\displaystyle (1)/(3)|(-(8)/(5)) -4|=(2)/(3) (-(8)/(5)) +2|(-(8)/(5)) +(6)/(3) |`

If we do all the calculations correctly, we will get:

• 
  `\displaystyle (28)/(15) =-(4)/(15)`

Since this equation is NOT true, this means that x = -8/5 is NOT A SOLUTION.

4) x = -16/3

Substitute this value back into the original equation.

• 
  `\displaystyle (1)/(3)|(-(16)/(3)) -4|=(2)/(3) (-(16)/(3)) +2|(-(16)/(3)) +(6)/(3) |`

If we do all the calculations correctly, we will get:

• 
  `\displaystyle (28)/(9) =(28)/(9)`

Since this equation IS true, this means that x = -16/3 IS A SOLUTION.

Final Answer

The two true solutions of the double absolute value equation, shown above, are:

• 
  `\displaystyle \boxed{-(8)/(9) } \ \ \& \ \ \boxed{-(16)/(3)}`