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Isolate for x in second equation:   x = -2y + 8

Substitute into first equation for x:

(-2y+8)^2 + 2 y^2 = 8

4y^2 - 32y + 64 + 2y^2 = 8

6y^2 - 32y + 64  = 8

6y^2 - 32y + 56 = 0

3y^2 - 16y + 28 = 0

b^2 - 4ac < 0, so no real solutions for y

second equation giving us:

x = 8 - 2y

From here we can substitute this into the first equation to obtain:

(8 - 2y)^2 +2y^2=8 which when expanded will give us:

4y^2 - 32y + 64 +2y^2 =8 which we can then simplify to:

6y^2 -32y +56 = 0

Or, 3y^2-16y+28=0

Here we see that b^2-4ac <0

Therefore , there is no real value of y which satisfy the equation.

The way I would set about doing it is by making x the subject of the second equation giving us:

x = 8 - 2y

From here we can substitute this into the first equation to obtain:

(8 - 2y)^2 +2y^2=8 which when expanded will give us:

4y^2 - 32y + 64 +2y^2 =8 which we can then simplify to:

6y^2 -32y +56 = 0 From here the quadratic formula can be used to find the values of y (note that there will likely be two) and then from there those values of y can be put into one of the original equations to find the respective values for x