Question

How do you solve `(b+6)/(4b^2)+3/(2b^2)=(b+4)/(2b^2)` and check for extraneous solutions?

Answer 1

b = 4

Explanation:

`(b+6)/(4b^2)+3/(2b^2)=(b+4)/(2b^2)`
multiply both sides by `4b^2`
`4b^2*(b+6)/(4b^2)+4b^2*3/(2b^2)=4b^2(b+4)/(2b^2)`
`b+6+6=2b+8`
`b+12=2b+8`
`b-2b=8-12`
`-b=-4`
`b=4`
check:
substitute b=4
`((4)+6)/(4(4)^2)+3*(4(4^2))/(2(4^2)`=`(4+4)/(2(4^2)`
`10/64+3/32=8/32`
`5/32+3/32=8/32`
`8/32=8/32`