Simple question about moving terms around in an equality

**bkbowser** · October 30th 2009, 09:57 AM

$a/v+b/v + c + d = e + f + g/v+h/v$

Can I just add the additive inverse of the g/v+h/v term (-g/v-h/v i assume) to each side to move it?
c, d, e and f are whole numbers.
Or, should i clear the fractions by multiplying each side by the denominator?

That wouldn't get me a 0 in that position wouldn't it
$g/v+-g/v=g-g/v$ or just $0/v$

**e^(i*pi)** · October 30th 2009, 10:12 AM

Originally Posted by bkbowser

$a/v+b/v + c + d = e + f + g/v+h/v$

Can I just add the additive inverse of the g/v+h/v term (-g/v-h/v i assume) to each side to move it?
c, d, e and f are whole numbers.
Or, should i clear the fractions by multiplying each side by the denominator?

That wouldn't get me a 0 in that position wouldn't it
$g/v+-g/v=g-g/v$ or just $0/v$

You can do either:

Subtracting: $\frac{a}{v}+\frac{b}{v}-\frac{g}{v}-\frac{h}{v} = e+f-c-d$

Multiplying - multiply each term by $v$

$a+b +cv + dv = ev + fv + g+h$

**bkbowser** · October 30th 2009, 10:16 AM

Originally Posted by e^(i*pi)

You can do either:

Subtracting: $\frac{a}{v}+\frac{b}{v}-\frac{g}{v}-\frac{h}{v} = e+f-c-d$

Multiplying - multiply each term by $v$

$a+b +cv + dv = ev + fv + g+h$

Seriously I can do either? I'm dealing with really long decimals in every position so it would just be simpler to move the fractions all onto one side then too multiply.

Thanks

**e^(i*pi)** · October 30th 2009, 10:20 AM

Originally Posted by bkbowser

Seriously I can do either? I'm dealing with really long decimals in every position so it would just be simpler to move the fractions all onto one side then too multiply.

Thanks

Yep. If I take the second equation: $ a+b +cv + dv = ev + fv + g+h $

and put all the v terms on one size: $ a+b -g-h = ev + fv -cv-dv $

Now if I multiply the first equation by v:

$ v\left(\frac{a}{v}+\frac{b}{v}-\frac{g}{v}-\frac{h}{v}\right) = v(e+f-c-d) $

This equals: $a+b-g-h = ev+fv-cv-dv$ which is the same as above. They are just two separate ways to get the same answer

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