SOLUTION: I am currently tryin to solve an absolute value equation with an absolute value on both sides. The equation is |-x+4|=|4x+2| and I cannot solve it because the example I've been giv

Algebra ->  Absolute-value -> SOLUTION: I am currently tryin to solve an absolute value equation with an absolute value on both sides. The equation is |-x+4|=|4x+2| and I cannot solve it because the example I've been giv      Log On


   

Question 212470This question is from textbook intermediate algebra
: I am currently tryin to solve an absolute value equation with an absolute value on both sides. The equation is |-x+4|=|4x+2| and I cannot solve it because the example I've been given to follow is inadequite, the example equation in my book is |x+5|=|2x+1| the book solves it like so,
|x+5|=|2x+1|
x+5=2x+1
5=x+1
4=x


or x+5=-(2x+1)
x+5=-2x-1
3x+5=-1
3x=-6
x=-2
Unfortunately this has put me at a loss because the book only shows this example and offers no explanation for the magical dissapearence of the x the 2 or the 1 in the first solution, nor does it show how the -(2x+1) became -2x-1 or where 3x came from or where -2x even went for that matter in the second solution. Would someone please be so kind to resolve the example with one hundred percent of the steps and not just the important ones so that i can understand where exactly these values are coming from or going too then maybe i will understand it better. This question is from textbook intermediate algebra

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
abs%28-x%2B4%29+=+abs%284x%2B2%29 Remove the absolute-value bars and write the two equations:
-x%2B4+=+4x%2B2 or -x%2B4+=+-%284x%2B2%29
Starting with the first equation:
-x%2B4+=+4x%2B2 Add x to both sides of the equation to get the x's on one side of the equation.
cross%28-x%2Bx%29%2B4+=+x%2B4x%2B2 The x's on the left sides disappear and the x's on the right side get added together.
4+=+5x%2B2 Now subtract 2 from both sides.
4-2+=+5x%2B2-2}
2+=+5x Finally, divide both sides by 5.
2%2F5+=+x or highlight%28x+=+2%2F5%29
For the second equation we have:
-x%2B4+=+-%284x%2B2%29 Apply the "distributive property" to the right side of the equation.
-x%2B4+=+-4x-2 Now add 4x to both sides to get the x's on one side of the equation.
highlight_green%284x-x%2B4%29+=+cross%28-4x%2B4x%29-2 The x's on the right side disappear and the x's on the left side are subtracted.
3x%2B4+=+-2 Subtract 4 from both sides.
3x%2Bcross%284-4%29+=+-2-4 The 4's on the left side disappear.
3x+=+-6 Finally, divide both sides by 3.
highlight%28x+=+-2%29