How do you solve $3(3-2b)-3(3b-6)=-48$ using the distributive property?

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Answer 1 · 2017-10-02T15:32:00

$b = 5$

$3(3-2b)-3(3b-6)=- 48$

Distributive property:

$a(b-c) = ab - ac$

$=> 3\times 3 - 3\times 2b - 3\times 3b + 3\times 6= -48$

Applying BODMAS rule:

$=> 9 -6b- 9b +18 =- 48$

$=> 9 +18 -6b- 9b =- 48$

$=> 27 -15b = -48$

By transposing,

$=> -15b = -48 -27$

$=> -15b = -75$

$=> b = -75/-15$

$=> b = 5$

How do you solve 3(3-2b)-3(3b-6)=-483(3-2b)-3(3b-6)=-48 using the distributive property?