Solving Specified Variables...

Hi everyone,

                 How would you solve these?

        1. V=1/3(h+c)-2, solve for h?

         2. A=1/2+3/5hb, solve for b?

         3. A= x-3/6-x + h, solve for h?

Problem 1:

V+2 = 1/3(h+c)

3(V+2)= H+C

H=3(V+2) -C

Problem 2:

A-1/2 = (3/5)hb  I multiply both sides by 5, then divide by 3.

(5/3)(A-1/2) = hb

((5/3)(A-1/2))/h = b

Make sense?  Hopefully I was right about the "hb". 

The third problem is confusing.  You wrote:

3. A= x-3/6-x + h, solve for h?

Does that mean:

3. A = x - x - (3/6) + h

Because if it does, the answer is very simple.  x - x = 0  and 3/ 6 = 1/2 so,

A = (1/2) + h

h = A - (1/2)

Trying again!  I'll use this: * to mean multiplication.

A= x-3/6-x + h  The entire mess of 6-x+h can be multiplied to the other side, giving us:

A * (6-x+h) = x-3  In other words, A is multiplied by all three of those guys.  Multiplying A through gives us:

A*6 - A*x + A*h = x-3  Now we can add A*x to both sides:

A*6 + A*h = x-3 + A*x  Subtract A*6 from both sides.

A*h = x-3 + A*x - A*6  Now all we have to do is divide both sides by A to get h by itself.  So you have to take the whole disaster on the right and divide it by A.

h = (x-3 + A*x - A*6)/A

Make more sense?  Hopefully?    I don't know if your teacher wants you to simplify or not.  You could divide everything by A...

I presume this is Calculus...???...