How do you solve #a(a+3)+a(a-6)+35=a(a-5)+a(a+7)#?

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Answer 1 · 2017-12-31T16:26:00

#a=7#

#a(a+3)+a(a-6)+35=a(a-5)+a(a+7)#

#a^2+3a+a^2-6a+35=a^2-5a+a^2+7a# (distribute)

#2a^2-3a+35=2a^2+2a# (combine like terms)

#35=5a#

#a=7#

Answer 2 · 2017-12-31T19:31:36

#a=7#

#a(a+3)+a(a-6)+35=a(a-5)+a(a+7)#

Let's start by using the distributive property

#(a)(a) + (a)(3) + (a)(a)+(a)(-6)+35 = (a)(a)+(a)(-5)+(a)(a)+(a)(7)#

#2a^2 - 3a + 35 =2a^2+2a#

Then subtract #color(red)(2a^2)# from both sides

#cancel(2a^2) - 3a + 35 - cancelcolor(red)(2a^2)= cancel(2a^2) +2a - cancelcolor(red)(2a^2)#

#-3a + 35 = 2a#

Subtract #2a# from both sides

#-3a + 35 - 2a = 2a - 2a#

#-5a + 35 = 0#

Subtract #35# from bot sides

#-5a + 35 - 35 = 0 - 35

#-5a = -35#

Divide both sides by #-5#

#(-5a)/-5 = (-35)/-5#

#a = 7#