To get rid of the 1/2 how is it done?

2d+4c=1/2 4b+2a

is it

2/1(2d+4c)=2/1(1/2 4b+2a)

so

4d+8c=4b+2a

or

2/1(2d+4c)= (2/1)1/2 4b+2a

so

4d+8c=4b+2a

or something different?

Thanks for any help you have.

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- Nov 28th 2008, 09:16 AMbrentwoodbcmultiply by reciprocal
To get rid of the 1/2 how is it done?

2d+4c=1/2 4b+2a

is it

2/1(2d+4c)=2/1(1/2 4b+2a)

so

4d+8c=4b+2a

or

2/1(2d+4c)= (2/1)1/2 4b+2a

so

4d+8c=4b+2a

or something different?

Thanks for any help you have. - Nov 28th 2008, 09:49 AMTheEmptySet
Hello bentwoodbc,

I guess I'm unsure what you mean. Is this the problem

$\displaystyle 2d+4c=\frac{1}{2}(4b+2a)$

if so you can use the distributive law to simplify

$\displaystyle 2d+4c=\frac{1}{2}\cdot 4b+\cdot \frac{1}{2} 2a$

$\displaystyle 2d+4c=2b+a$

if it is just $\displaystyle \frac{1}{2}4b=2b$

Good luck.