Solve the equation a=h(b1+b2)/2 for b1.
b1 = 2a/h – b2
b1 = a/h – 2b2
b1 = 2h/a – 2b2
b1 = a – hb2

  b1 = 2A/h – b2

Step-by-step explanation:

It is convenient to multiply by the inverse of the coefficient of b1. This gives ...

  2A/h = b1 +b2

Subtracting b2 gives the desired result.

  b1 = 2A/h -b2 . . . . . matches the first selection

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Comment on "solve for"

Usually it works well to "undo" what is done to the variable of interest. The sequence of "undo" should be the reverse of the sequence of operations performed on the variable.

The rules of equality must be obeyed, so any operation performed must be done to both sides of the equation. Knowing this, we can describe the operation without constantly repeating "to both sides of the equation."

Here, the variable b2 is added to b1, then the sum is multiplied by h/2. To undo these operations, first we multiply (both sides of the equation) by 2/h. Then we subtract b2 (from both sides of the equation).

stethere all not eveanp-by-step explanation:

80 the answer is 80

step-by-step explanation: