Saturday

December 20, 2014

December 20, 2014

Posted by **John** on Sunday, November 3, 2013 at 3:46pm.

- Algebra -
**Steve**, Sunday, November 3, 2013 at 8:27pmYou have two cases:

h-7 >=0, where |h-7| = h-7

h-7 < 0, where |h-7| = -(h-7)

So,

case 1: h-7 >= 0 (h >= 7)

-2(h-7) = -28

h-7 = 14

h = 21

Since 21 >= 7, it's a solution

Case 2: h-7 < 0 (h < 7)

-2(-(h-7)) = -28

2h-14 = -28

2h = -14

h = -7

Since -7 < 7, it's a solution

So, h=-7,21

Another way to look at it is to say

-2|h-7| = -28

|h-7| = 14

That means that h has to be at a distance of 14 from 7, either above or below. So, h = -7 or 21

**Answer this Question**

**Related Questions**

Algebra - I am getting ready to start an Algebra class for the first time and I ...

SHAY - THE MATH QUESTION IS IT ALGEBRA , PRE ALGEBRA' OR GEOMETRY? Quadriatic ...

Algebra 1A - How is algebra a useful tool? what concepts investigated in algebra...

Algebra I - So- im trying to help my little brother out with his Algebra- but ...

Algebra-still need some help - Homework Help Forum: Algebra Posted by Jena on ...

algebra 1 - at a certain high school,350 students are taking algebra. the ratio ...

7th grade - There are at least 5 more than twice as many students taking algebra...

algebra 2 - Posted by hellogoodbie on Saturday, January 16, 2010 at 3:24pm. ...

Algebra II - ok now this is an algebra 2 problem... I have goten far down to ...

algebra 2 - Posted by hellogoodbie on Saturday, January 16, 2010 at 2:59pm. How ...