# Thread: Absolute value equation help!

1. ## Absolute value equation help!

Hello all! Hoping someone can help me here. I have been trying to figure this out for a while on my own, however it seems I am going in circles and now more confused than before.

The equation is $\displaystyle -5|2+4x|=-32(x+3/4)-|x|+1$. The correct answer is $\displaystyle -11/17$, but I have no I idea how to get there. I have no problem with equations such as $\displaystyle −7+8|−7x− 3|=73$. And I understand that absolute values can be negative or positive. Hoping someone can show me an algorithm to solve problems such as $\displaystyle -5|2+4x|=-32(x+3/4)-|x|+1$.

Thank you!

2. ## Re: Absolute value equation help!

To clarify. I was able to solve it by writing all absolute values as positive numbers. But I am trying to understand why thats the correct answer. What is confusing me the most is finding intervals, identifying intervals and solving for each. I am getting this information from (apologize in advance if for some reason not allowed to post link)
Code:
https://www.symbolab.com/solver/linear-equation-calculator/-5%5Cleft%7C2%2B4x%5Cright%7C%3D-32%5Cleft(x%2B%5Cfrac%7B3%7D%7B4%7D%5Cright)-%5Cleft%7Cx%5Cright%7C%2B1
Looking at the steps, its showing 3 possible equations. Other, simpler absolute value equations I have done in the past indeed have to answers, a positive and negative, and need to solve for both. But those are in form of $\displaystyle |x|+3=5$, that gives me $\displaystyle x+3=5$ and $\displaystyle x+3=(-5)$ and solve for both.

3. ## Re: Absolute value equation help!

Originally Posted by CantMath
Hello all! Hoping someone can help me here. I have been trying to figure this out for a while on my own, however it seems I am going in circles and now more confused than before.

The equation is $\displaystyle -5|2+4x|=-32(x+3/4)-|x|+1$. The correct answer is $\displaystyle -11/17$, but I have no I idea how to get there. I have no problem with equations such as $\displaystyle −7+8|−7x− 3|=73$. And I understand that absolute values can be negative or positive. Hoping someone can show me an algorithm to solve problems such as $\displaystyle -5|2+4x|=-32(x+3/4)-|x|+1$.

Thank you!
No "algorithm" but definitions! |x| is defined to be equal to x if $\displaystyle x\ge 0$ and equal to -x if < 0. The two absolute values here are |2+ 4x| and |x|. $\displaystyle 2+ 4x\ge 0$ if $\displaystyle 4x> -2$, $\displaystyle x\ge -1/2$, 2+ 4x< 0 if x< -1/2. So divide the real number into three regions:
1) $\displaystyle x\ge 0$. Both 2+ 4x and x are non-negative so |2+ 4x|= 2+ 4x and |x|= x. The equation becomes -5(2+ 4x)= -32(x+ 3/4)- x+ 1. Doing the multiplications -10- 20x= -32x- 24- x+ 1. Adding 33x+ 10 to both sides, 13x= -13. That gives x= -1 but -1< 0 so is not a valid solution.

2) $\displaystyle -1/2\ge x< 0$. Now 2x+ 4 is non-negative so |2+ 4x|= 2+ 4x but now |x|= -x. The equation becomes -5(2+ 4x)= -32(x+ 3/4)+ x+ 1. -10- 20x= -32x- 24+ x+ 1. Adding -31x+ 10 to both sides 11x= 11 so x= 1. But now, that is not less than 0 so is not a valid solution.

3) $\displaystyle x< -1/2$. Now both 2x+ 4 and x are negative so |2+ 4x|= -(2+ 4x) and |x|= -x. The equation becomes 5(2+ 4x)= -32(x+ 3/4)+ x+ 1. That is 10+ 20x= -32x- 24+ x+ 1. Adding 31x- 10 to both sides 51x= -13 so x= -13/51. But, again, that is not less than -1/2 so is NOT a valid solution.

There is no value of x satisfying this equation.

4. ## Re: Absolute value equation help!

Originally Posted by CantMath
Hello all! Hoping someone can help me here. I have been trying to figure this out for a while on my own, however it seems I am going in circles and now more confused than before.

The equation is $\displaystyle -5|2+4x|=-32(x+3/4)-|x|+1$. The correct answer is $\displaystyle -11/17$, but I have no I idea how to get there. I have no problem with equations such as $\displaystyle −7+8|−7x− 3|=73$. And I understand that absolute values can be negative or positive. Hoping someone can show me an algorithm to solve problems such as $\displaystyle -5|2+4x|=-32(x+3/4)-|x|+1$.

Thank you!
It isn't necessary, but I would start by putting all the variables on one side and call it $f(x)$ as
$$f(x) = -5|2+4x|+32(x+\frac 3 4)+|x|-1$$
and you are looking for $x$ so that $f(x)=0$. What you want to do is note the points where the absolute value functions are $0$, which are at $x = -\frac 1 2$ and $x=0$. Those are the places where the arguments of the absolute value functions change signs. So you want to consider the three cases: $x < -\frac 1 2$, $-\frac 1 2 < x < 0$, and $x > 0$. Write the formula for $f(x)$ as a three piece formula by correctly removing the absolute values for each interval. Then check where it is $0$.

5. ## Re: Absolute value equation help!

Originally Posted by HallsofIvy
3) $\displaystyle x< -1/2$. Now both 2x+ 4 and x are negative so |2+ 4x|= -(2+ 4x) and |x|= -x. The equation becomes 5(2+ 4x)= -32(x+ 3/4)+ x+ 1. That is 10+ 20x= -32x- 24+ x+ 1. Adding 31x- 10 to both sides 51x= -13 so x= -13/51. But, again, that is not less than -1/2 so is NOT a valid solution.

There is no value of x satisfying this equation.
I think you meant $-33$, which does give a valid solution.

6. ## Re: Absolute value equation help!

Originally Posted by HallsofIvy
No "algorithm" but definitions! |x| is defined to be equal to x if $\displaystyle x\ge 0$ and equal to -x if < 0. The two absolute values here are |2+ 4x| and |x|. $\displaystyle 2+ 4x\ge 0$ if $\displaystyle 4x> -2$, $\displaystyle x\ge -1/2$, 2+ 4x< 0 if x< -1/2. So divide the real number into three regions:
1) $\displaystyle x\ge 0$. Both 2+ 4x and x are non-negative so |2+ 4x|= 2+ 4x and |x|= x. The equation becomes -5(2+ 4x)= -32(x+ 3/4)- x+ 1. Doing the multiplications -10- 20x= -32x- 24- x+ 1. Adding 33x+ 10 to both sides, 13x= -13. That gives x= -1 but -1< 0 so is not a valid solution.

2) $\displaystyle -1/2\ge x< 0$. Now 2x+ 4 is non-negative so |2+ 4x|= 2+ 4x but now |x|= -x. The equation becomes -5(2+ 4x)= -32(x+ 3/4)+ x+ 1. -10- 20x= -32x- 24+ x+ 1. Adding -31x+ 10 to both sides 11x= 11 so x= 1. But now, that is not less than 0 so is not a valid solution.

3) $\displaystyle x< -1/2$. Now both 2x+ 4 and x are negative so |2+ 4x|= -(2+ 4x) and |x|= -x. The equation becomes 5(2+ 4x)= -32(x+ 3/4)+ x+ 1. That is 10+ 20x= -32x- 24+ x+ 1. Adding 31x- 10 to both sides 51x= -13 so x= -13/51.
I was afraid I would screw up the arithmetic! As Walagaster says this should be x= -33/51= -11/17 which is less than -1/2 so is a solution!

7. ## Re: Absolute value equation help!

Originally Posted by HallsofIvy
No "algorithm" but definitions! |x| is defined to be equal to x if $\displaystyle x\ge 0$ and equal to -x if < 0. The two absolute values here are |2+ 4x| and |x|. $\displaystyle 2+ 4x\ge 0$ if $\displaystyle 4x> -2$, $\displaystyle x\ge -1/2$, 2+ 4x< 0 if x< -1/2. So divide the real number into three regions:
1) $\displaystyle x\ge 0$. Both 2+ 4x and x are non-negative so |2+ 4x|= 2+ 4x and |x|= x. The equation becomes -5(2+ 4x)= -32(x+ 3/4)- x+ 1. Doing the multiplications -10- 20x= -32x- 24- x+ 1. Adding 33x+ 10 to both sides, 13x= -13. That gives x= -1 but -1< 0 so is not a valid solution.

2) $\displaystyle -1/2\ge x< 0$. Now 2x+ 4 is non-negative so |2+ 4x|= 2+ 4x but now |x|= -x. The equation becomes -5(2+ 4x)= -32(x+ 3/4)+ x+ 1. -10- 20x= -32x- 24+ x+ 1. Adding -31x+ 10 to both sides 11x= 11 so x= 1. But now, that is not less than 0 so is not a valid solution.

3) $\displaystyle x< -1/2$. Now both 2x+ 4 and x are negative so |2+ 4x|= -(2+ 4x) and |x|= -x. The equation becomes 5(2+ 4x)= -32(x+ 3/4)+ x+ 1. That is 10+ 20x= -32x- 24+ x+ 1. Adding 31x- 10 to both sides 51x= -13 so x= -13/51. But, again, that is not less than -1/2 so is NOT a valid solution.

There is no value of x satisfying this equation.
I appreciate the answer. My confusion now, where does the $\displaystyle -1/2$ come from?

8. ## Re: Absolute value equation help!

Originally Posted by HallsofIvy
2) $\displaystyle -1/2\ge x< 0$. Now 2x+ 4 is non-negative so |2+ 4x|= 2+ 4x but now |x|= -x. The equation becomes -5(2+ 4x)= -32(x+ 3/4)+ x+ 1. -10- 20x= -32x- 24+ x+ 1. Adding -31x+ 10 to both sides 11x= 11 so x= 1. But now, that is not less than 0 so is not a valid solution.
You mean Adding 31x + 10 to both sides.

However this still leads to a false conclusion, so no harm done.

-Dan

9. ## Re: Absolute value equation help!

Originally Posted by CantMath
I appreciate the answer. My confusion now, where does the $\displaystyle -1/2$ come from?
We need to know where 2 + 4x is xero to split up the intervals where 2 + 4x switches from negative to positive. This happens when x = -1/2.

-Dan