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September 17, 2014

September 17, 2014

Posted by **Dorinda** on Thursday, August 4, 2011 at 2:21am.

- Algebra -
**Reiny**, Thursday, August 4, 2011 at 7:47amSince both equations are equal to C, you can equate the two expressions

200 + 50h = 300 + 25h

You want the h's on one side and the constants (just the numbers) on the other side.

How do I get rid of the +25h on the right side?

I subtract it, but whatever I do to one side, I must do to the other.

200 + 50h - 25h = 300 + 25h - 2h

200 + 25h = 300

now do the same thing to get rid of the 200 on the left side.

200 - 200 + 25h = 300 - 200

25h = 100

How is the h "hooked on" to the 25 ? - by multiplication, so do the inverse operation....

divide both sides by 25

25h/25 = 100/25

h = 4

After doing a few of these, you can reduce your steps to

200 + 50h = 300 + 25h

50h - 25h = 300 - 200

25h = 100

h = 4

- "sticky" 5 key - Algebra -
**Reiny**, Thursday, August 4, 2011 at 8:12am6th line should have been

200 + 50h - 25h = 300 + 25h - 25h

(but you probably figured that out anyway)

- Algebra -
**Dorinda**, Thursday, August 4, 2011 at 8:36amThank you so much. You've made this problem very clear.

- Algebra -
**rochelle**, Wednesday, September 28, 2011 at 9:41am5+2y-2=6y+5-3y

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