Simplify
a. $ 2 a+(3 a-b+a) $
b. $a+(4 a-a)+(7 a-a) $
ÑÂ. $ 11 b-a+2(3 a-b)
Solution:
a)
$2a + (3a - b + a)$
= $2a + ( 4a - b)$
= $6a - b$
b)
$a + (4a -a) + (7a - a)$
= $a + 3a + 6a$
= $10a$
c)
$11b - a + 2(3a - b)$
$= 11b -a + 2\times3a -2\times b$
= $9b + 5a$
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