Algebra
posted by Dorinda .
This question has been posted before, unfortunately, I have been out of high school for over 40 years and out of college for more than 20. I am trying to help another old relic enrolled in distance learning (via long distance phone calls) but neither of us seem to know where to start. The question is to solve for: $200+$50h=C and $300+25H=C. I have tried multiplying $200+$50h=C by 3 and multiplying $300+$25h=C by 2. I get $600+$150h=3C and $600+$50h=2C. At this point, I don't know whether to add or what. When I add, I get $1200=$200H=5C. I have no trail markers, not even the book, to see other like examples, to let me know which way to go or if I am going in the right direction. Please help??

Algebra 
Reiny
Since 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
6th line should have been
200 + 50h  25h = 300 + 25h  25h
(but you probably figured that out anyway) 
Algebra 
Dorinda
Thank you so much. You've made this problem very clear.

Algebra 
rochelle
5+2y2=6y+53y
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