Find the value of y when the arithmetic mean of 3y+1 and 4y +9 is 2y+11. with solution please:)

$We \: \: \: \: know \\ mean \: \: = (sum \: \: \: of \: \: \: the \: \: \: terms)/(number \: \: \: of \: \: \: terms) \\$

$Given \: \: \: here \: \: the \: \: \: mean \\ of \: \: \: 3y + 1 \: \: \: and \: \: \: 4y + 9 \: \: \: is \: \: \: 2y + 11.$

$Therefore \\ ((3y + 1) + (4y + 9))/(2) = 2y + 11 \\ = > (7y + 10)/(2) = 2y + 11 \\ = > 7y + 10 = 2(2y + 11) \\ = > 7y + 10 = 4y + 22 \\ = > 7y - 4y = 22 - 10 \\ = > 3y = 12 \\ = > y = (12)/(3) \\ = > y = 4$

$Your \: \: \: answer \: \: \: is \: \: \: 4.$

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