# Algebra

posted by on .

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 - ,

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 - ,

6th line should have been

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

(but you probably figured that out anyway)

• Algebra - ,

Thank you so much. You've made this problem very clear.

• Algebra - ,

5+2y-2=6y+5-3y