How do you solve #1/2a+1/3b=8, 3/2a-4/3b=-4#?

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Answer 1 · 2016-12-31T13:41:53

Solution is #a=8#, #b=12#.

The two given are equations are

#1/2a+1/3b=8# ...................................(1) and

#3/2a-4/3b=-4# ...................................(2)

Multiplying (1) by #3# we get

#3/2a+3/3b=24# ...................................(3)

and subtracting (2) from (3), we get

#3/3b-(-4/3b)=24-(-4)#

or #3/3b+4/3b=24+4#

or #7/3b=28# i.e.

#b=28xx3/7=4cancel28xx3/(1cancel7)=12#

Putting #b=12# in (1), we get

#1/2a+1/3xx12=8#

or #1/2a+4=8#

or #1/2a=8-4=4#

i.e. #a=8#

Hence, solution is #a=8#, #b=12#.