How do you solve $8( 2+ 3n ) = 32+ 8n$ ?

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Answer 1 · 2017-11-19T01:27:17

$n=1/2$

Law:
$a*(b+c)=a*b+a*c$

$8*(2+3n)=32+8n$

using the law:
$a*(b+c)=a*b+a*c$
when:
$a=8$
$b=2$
$c=3n$

$=> 8*2+8*3n=32+8n$
$=> 16+24n=32+8n$

minus $8n$ minus $16$

$=> 16+24n-8n-16=32+8n-8n-16$
$=> (16-16)+(24n-8n)=(32-16)+(8n-8n)$
$=> (0)+(32n)=(16)+(0)$
$=> 32n=16$

divided by $32$

$=> (32n)/32=16/32$
$=> (32/32)n=(16/32)$
$=> (1)n=(1/2)$
$=> n=1/2$

How do you solve 8( 2+ 3n ) = 32+ 8n8( 2+ 3n ) = 32+ 8n?