2.(a+bc)+cb+ab+ac-3.(ab-cb-ba+bc)

1 answer:

4 0

2(a+bc)=2a+2bc
3.(ab-cb-ba+bc)= 3ab-3cb-3ba+3bc
Since the brackets and multiplication have a higher order than the rest.

Cb=Bc
Ab=ba
Ac=Ca

Because multiplication is commutative.

2a+2bc + cb +ab + ac -(3ab-3cb-3ba+3bc)

2a + 2bc + cb + ab + ac(+ -3ab - - 3cb - -3ba +-3bc)

2a + 2bc + cb + ab + ac -3ab + 3cb + 3ba -3bc

2a + (2bc-3bc +cb + 3cb ) + (ab -3ab + 3ba) + ac

=2a + 3cb + ba+ ac OR 2a +3bc + ab + ac

The Or is just a different way to write it.

Hope this helps

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\\ \quad \\\\

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$\bf \begin{array}{llll}
% right side info
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\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
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