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#