solve this simultaneously please! 3x+2y=14 and 2xy+5 =21, I need the working please!?

3x+2y=14

so, 3x = 14-2y
x = 14-2y/3.--------> 1
substituting value of x in 2nd eqn.
2 { y (14-2y/3 ) } + 5 = 21
so, 2 (14y-2y^2/3 ) + 5 = 21
28y-4y^2/3 + 5 = 21
28y-4y^2 /3 = 16
28y-4y^2 = 48
so, 4y^2-28y+48 =0
y^2-7y+12 = 0
so, y^2-4y-3y+12=0
y ( y-4 ) - 3 (y-4) =0
(y-3 ) , ( y-4 ) = 0

so y=3 or y=4

taking y as 3 in eqn 3x +2y =14
so 3x +6 = 14
so, 3x =8
so, y = 8/3

taking y as 4 in eqn 3x+2y = 14
so , 3x+8= 14
so, 3x= 6
so, x = 2

so x= 8/3 or x= 2 .

so solution set is ( x, y ) { (8/3, 2) ( 3, 4) }

3x + 2y = 14
2xy + 5 = 21
xy = 8
Solutions:
x = 2, y = 4
x = 8/3, y = 3

3x+2y=14 ----> ♡
2xy+5=21 ----> ◇

♡+◇

3x+2y+2xy+5 = 35

3x+2y+2xy = 30

3x+2(y+xy)= 30

X=2
Y=4

x=77