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

$(cancel(11)p)/cancel(11) =11/11$

$p=1$

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