Is it necessary to check for an extraneous solution for the equation below? How do you know? ( I attached the image with equation)

No, it is not is not necessary because we don't have any division by zero issues that will come up. If we had a variable in the denominator, then it would be a different story.

If you're curious about solving this equation, then you would first multiply both sides by the LCD 15 to clear out the fractions.

(6w-1)/5 - 3w/1 = (12w-16)/15

15*[  (6w-1)/5 - 3w/1  ] = 15*[ (12w-16)/15 ]

15*[  (6w-1)/5 ] - 15*[ 3w/1  ] = 15*[ (12w-16)/15 ]

3(6w-1) - 15(3w) = 12w-16

18w-3 - 45w = 12w-16

-27w-3 = 12w-16

-3+16 = 12w+27w

13 = 39w

39w = 13

w = 13/39

w = 1/3

To check this answer, you plug it back into the original equation. You should get the same thing on both sides when you simplify fully. Again we don't have to worry about extraneous solutions because we don't have any division by zero issues.

on friday there were 15 birds

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step-by-step explanation: