# Linear Equation question

• Jul 9th 2009, 08:34 AM
allyourbass2212
Linear Equation question
When using the formula $A=\frac{1}{2}(a+b)h$ when solving for b this equals

$\frac{2}{h}(A-\frac{ah}{2})=b$ why do we not distribute the $\frac{2}{h}$ over the parenthesis?

If we were to distribute it I think it would look like this, please correct me if I am mistaken.
$
\frac{A}{1}*\frac{2}{h}-\frac{ah}{h}*\frac{2}{h}=\frac{2A}{h}-a=b$
, if this is correct the book does not list it as an answer choice.

Thank you
• Jul 9th 2009, 08:57 AM
Kasper
Quote:

Originally Posted by allyourbass2212
When using the formula $A=\frac{1}{2}(a+b)h$ when solving for b this equals

$\frac{2}{h}(A-\frac{ah}{2})=b$ why do we not distribute the $\frac{2}{h}$ over the parenthesis?

If we were to distribute it I think it would look like this, please correct me if I am mistaken.
$
\frac{A}{1}*\frac{2}{h}-\frac{ah}{h}*\frac{2}{h}=\frac{2A}{h}-a=b$
, if this is correct the book does not list it as an answer choice.

Thank you

It's probably just them stopping one step short, you could simplify that to what you did (Which we can verify), but it's purely aesthetic.

Though, you did have a small mistake in your simplification, which was probably just a mis-type.

$
\frac{A}{1}*\frac{2}{h}-\frac{ah}{2}*\frac{2}{h}=\frac{2A}{h}-a=b$

It should be $\frac{ah}{2}$ not $\frac{ah}{h}$.

That said, we can verify that your simplification is a correct one by choosing arbitrary values for the equations, as long as we use the same values for both equations, if they put out the same answer, they are equivalent.

For this, we will make $a=2$ , $h=2$ , and $A=10$

Original equation;

$
\frac{2}{h}(A-\frac{ah}{2})=b$

$\frac{2}{2}(10-\frac{4}{2})=b$

$b=8$

$\frac{2A}{h}-a=b$
$\frac{2(10)}{2}-2=b$
$b=8$