Absoulte value equations?

[Dillinger4]

This should be easy for some of you. Im having trouble getting the second part of the answer for some of these problems.
How to de get the second answer for this?


17 = |7h - 4|
17 + 4 = 7h - 4 + 4

21/7 = 7h/7

3, -1.86 = h

[Dillinger4]

For instance here is another one.

9 = |2p + 7|
9 - 7 = 2p + 7 -7
2/2 = 2p/2

1 = p

Now -8 can be a solution too. The equation can be represented this way also 9 = |2p - (-7)| which makes sense but i still can't fgure out how to get -8.

[snakeeyes1000]

SAMPLE:
9=|x-10|

solutions

9=x-10 -> x=19
or
9=10-x -> x=1

-----------------------

YOUR PROBLEM:
17=|7h-4|

solutions

17=7h-4 -> 7h=21 -> h=3
or
17=4-7h -> -7h=13 -> h= -13/7(or -1.86 for decimal)

Hope that helps.

[Destined Soul]

Ok:

9 = |2p + 7| is actually two eq'ns:

1) 9 = 2p+7
2) 9 = - (2p + 7) = -2p - 7

Solving 1:
9-7 = 2p
2/2 = p = 1

Solving 2:
9+7 = -2p
16 = -2p
-8 = p

I think if your eq'ns are always in the form:

z = |aX + b|

the solutions are

X = (z - Yb)/(Ya), with Y = +1 for sol'n 1, and -1 for sol'n 2.

Destined

[Dillinger4]

Thanks snakeeyes1000. In you example below it seems like you are just swaping.


SAMPLE:
9=|x-10|

solutions

9=x-10 -> x=19
or
9=10-x -> x=1

So if i apply........

17 = |7h - 4|
17 + 4 = 7h - 4 + 4
21/7 = 7h/7

3 = h

or

17 = |7h - 4|
17 = |4 - 7h|
17 - 4 = 4 - 4 -7h
-13/7 = -7h/-7

-1.86 = h

Ok that worked out nicely. Thanks

[Dillinger4]

And it seems that for equations such as 9 = |2p+7| they have to be changed in the form of 9 = |2p - (-7)| then flipped to form 9 = |-7 - (-2p)|

9 = |2p + 7|
9 - 7 = 2p + 7 - 7
2/2 = 2p/2
1 = p

or

9 = |2p+7|
9 = |2p - (-7)|
9 = |-7 - (-2p)|
9 + 7 = -7 + 7 -2p
16/-2 = -2p/-2

-8 = p


So let me get this straight. It the equation is in the form of 17 = |7h - 4| all i have to do is flip it's contents so that it ends up in the form of. 17 =|4 - 7h| but if the equation is in the form of 9 = |2p + 7|. I have to reform as 9 = |2p - (-7)| then flip, 9 =|-7 (-2p)|

[snakeeyes1000]

basically.
if you have more than a binomial in absolut value symbols, fo through and if it's positive, make it negative. vise versa. if you need any more examples done, post em here. glad to help.

[Destined Soul]

Dilenger4 said:
So let me get this straight. It the equation is in the form of 17 = |7h - 4| all i have to do is flip it's contents so that it ends up in the form of. 17 =|4 - 7h| but if the equation is in the form of 9 = |2p + 7|. I have to reform as 9 = |2p - (-7)| then flip, 9 =|-7 (-2p)|

Actually, I think this is way more confusing than it sounds. Why not just take, say,

9 = |2p + 7| and convert this into:

9 = 2p + 7 and
-9 = 2p + 7

This may be easier than negating the inside of the Right-hand side (RHS.) All of that flipping seems quite confusing when it doesn't have to be.

Destined

[snakeeyes1000]

True, you have a point...
I developed the habit of converting the right because often, in more advanced math, the left side will be a polynomial as well; either way works fine, whatever your preference is.

[Dillinger4]

I guess whichever way works. Ill probably stick to using 10 = |2y + 3| ---> 10 = |2y - (-3)|.