How do you solve #3(p-2)+7=32-4(2p+5)#?

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Answer 1 · 2016-07-06T15:57:30

#p=1#

To solve #3(p-2)+7 = 32-4(2p+5)#

Begin by distributing the values outside the parenthesis to the terms inside the parenthesis

#3p-6+7 = 32-8p-20#

Combine like terms on each side of the equation

#3p-6+7 = 32-8p-20#

#3p+1=12-8p#

Now combine the variable terms on the same side of the equation by additive inverse

#3p+8p+1 = 12 cancel(-8p) cancel(+8p)#

#11p+1=12#

Now combine constant terms on the same side of the equation by additive inverse

#11pcancel(+1) cancel(-1)=12-1#

#11p=11#

Now solve for the variable by multiplicative inverse

#(cancel(11)p)/cancel(11) =11/11#

#p=1#