What is it you want to solve? You can factor it, taking out the common factor, "a". Thus, a + ab = a(1 + b). Or, if you want to convert it to a number, you can only do this if you know the values for "a" and "b". In this case, replace the values, and do the calculation.
That factors to (a + 1)(a + b) a = -1, -b b = -a
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
Commutativity.
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If, as is normal, ab represents a times b, etc then ab + ab + cc = 2ab + c2 which is generally not the same as abc.
A = 1, B = 9
to solve ab=c+a for a:Divide both sides by b, so:a= (c+a)/b
That factors to (a + 1)(a + b) a = -1, -b b = -a
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
yes because ab plus bc is ac
Commutativity.
No - a pos can not donate blood to a negative
3y=ab. Divide by 3 to get y=ab/3
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then the baby will be O plus, or AB plus , I think .
If, as is normal, ab represents a times b, etc then ab + ab + cc = 2ab + c2 which is generally not the same as abc.
associative property