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 $a=8$ , $b=12$ .

How do you solve 1/2a+1/3b=8, 3/2a-4/3b=-41/2a+1/3b=8, 3/2a-4/3b=-4?