Solve the following simultaneous equation: x+y over2=1 3xy-7y=2

The ansa is y= -2/3 or -2
x= 2 or 4/3

X+y=2
--> x=2-y...(1)
y(3x-7)=2
y=2/(3x-7)...(2)
Put (1) in (2)
2
x=2-()
3x-7

2(3x-7)-2
--> x=
3x-7

--> x= (6x-16)/3x-7
x(3x-7)=6x-16
3x^2 -7x = 6x-16

3x^2 -13x+16=0 [Quadratic Equation]

This equation leads to x being an imaginary number based on what you gave. However if you meant
x+y
= 1 & 3x-7y=2
..2
Then,
x+y=2...(1)x7
3x-7y=2...(2)x1

--> 7x + 7y = 14...(3)
3x - 7y = 2(4)
--> 10x=16
--> x=16/10
--> x=8/5 or 1.6
Put x=8/5 in (1)
(8/5)+y=2

8
y=2-()
5

10-8
y=
5

y=2/5
Hence,
{x,y} = {8/5,2/5}

Equatorial triangle and obtuce angle

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