Solving for Variables

Jul 2010
2
0
Okay, so I'm having a difficult time understanding how to solve problems such as


Solve for B1: A=1/2(B1+B2)h

Help please!?

Thanks.
 
Jun 2009
68
16
Your first step should be to make B1 not be in the denominator of a fraction. Either take the reciprocal of each side or multiply both sides by that denominator. You could multiply both sides just by (B1 + B2) also. In the end, you'll end up with the same answer regardless.

edit: I'm assuming 2(b1 + b2)h are all in the denominator, though it's a little ambiguous the way you've written it.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
It may help to think of "undoing" what has been done to the variable.

Here, the equation you have is, I suspect, \(\displaystyle A= (\frac{1}{2})(B_1+ B_2)h\). That is NOT what 1005 assumed- as he said, what you wrote is ambiguous. I am assuming only the "2" is in the denominator because that looks to me like the formula for the area of a trapezoid.

Suppose you were given a value of \(\displaystyle B_1\) and the other things on the right and wanted to find A, what would you do? The formula tells you what do do- first add \(\displaystyle B_1\) and \(\displaystyle B_2\) (because they are in parentheses) then multiply by h and then multiply by 1/2 (the order of those two is not really important). To solve for \(\displaystyle B_1\), you do the opposite of each step and in the opposite order! That is first, divide by 1/2 (which is the same as multiply by 2) and, of course, do the same thing to both sides of the equation: multiplying both sides of \(\displaystyle A= (\frac{1}{2})(B_1+ B_2)h\) by 2 gives \(\displaystyle 2A= 2(\frac{1}{2})(B_1+ B_2)h= (B_1+ B_2)h\). Now divide both sides by h: \(\displaystyle \frac{2A}{h}= (B_1+ B_2)h/h= B_1+ B_2. Finally, since the first thing you woud do in calculating A would be to add \(\displaystyle B_1\) and \(\displaystyle B_2\), the last thing you do in solving for \(\displaystyle B_1\) is subtract \(\displaystyle B_2\) from both sides: \(\displaystyle \frac{2A}{h}- B_2= (B_1+ B_2)- B_2= B_1\).\)
 
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