What am i doing wrong: solve a fraction for a specific variable

B

bobisaka

Junior Member
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Dec 25, 2019
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115
Solve the equation for the variable b.

A = 1/2 h (b+B)
2/1*A = 2/1*1/2 h(b+B)
2A = h(b+B)
2A = bh+Bh
2A-Bh = bh <-----This is where I am stuck.

The only solution i can come up with, is that it is a muliplication for both variables on both sides, which needs to be inversed in order to get b by itself...

2A-Bh / h = bh / h

cross out the h i get:

2A-B = b




The real answer should be:

2A-Bh/h = b
 
bobisaka said:
Solve the equation for the variable b.

A = 1/2 h (b+B)
2/1*A = 2/1*1/2 h(b+B)
2A = h(b+B)
2A = bh+Bh
2A-Bh = bh <-----This is where I am stuck.

The only solution i can come up with, is that it is a muliplication for both variables on both sides, which needs to be inversed in order to get b by itself...

2A-Bh / h = bh / h

cross out the h i get:

2A-B = b




The real answer should be:

2A-Bh/h = b
Click to expand...
You ALMOST had it.

You were here:

2*A - B*h = b*h ....... now divide both sides by 'h' to get

\(\displaystyle \frac{2*A - B*h}{h} = \frac{b*h}{h}\)

\(\displaystyle \frac{2*A - B*h}{h} = b\)

There you have it......
 
Subhotosh Khan said:
2*A - B*h = b*h ....... now divide both sides by 'h' to get

\(\displaystyle \frac{2*A - B*h}{h} = \frac{b*h}{h}\)

\(\displaystyle \frac{2*A - B*h}{h} = b\)
Click to expand...

Why does 'Bh' still contain the 'h'? why does the 'h' variable not get canceled by the 'h' divisor?
but on the right side 'bh' 'h' gets cancelled..

EDIT:

I think i understand now.

The whole point is to get the variable or constant to the opposite side alone. Thus, the reason for the 'h' still be intact and also being a divisor.

But still, could we not take it a step further and cancel that 'h' on the left side? or does it not work?

EDIT 2:

OR

Is it because the opposite side cannot be simplified further, thus the left side is left as an equation?
 
Last edited:
bobisaka said:
The only solution i can come up with, is that it is a muliplication for both variables on both sides, which needs to be inversed in order to get b by itself...

2A-Bh / h = bh / h

cross out the h i get:

2A-B = b

The real answer should be:

2A-Bh/h = b
Click to expand...
Your error is in thinking of it as merely "crossing out the h" rather that as what it is: DIVIDING the ENTIRE left side by h. That means every term.

It doesn't help that you omitted parentheses: (2A-Bh) / h = bh / h.

You can stop there, after "canceling" the h's on the right, which means simply undoing the multiplication, so the answer is b = (2A-Bh) / h, as they apparently left it.

You can optionally simplify; but when you divide (2A - Bh) by h, you divide both terms: (2A)/h - (Bh)/h = (2A)/h - B. So that is the alternative form of the answer.
 
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