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The way I would set about doing it is by making x the subject of the second equation giving us: x = 8 - 2yFrom 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

07 September 2019

second equation giving us:x = 8 - 2yFrom 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 = 0Or, 3y^2-16y+28=0Here we see that b^2-4ac <0Therefore , there is no real value of y which satisfy the equation.

12 October 2019

Isolate for x in second equation: x = -2y + 8Substitute into first equation for x: (-2y+8)^2 + 2 y^2 = 84y^2 - 32y + 64 + 2y^2 = 86y^2 - 32y + 64 = 86y^2 - 32y + 56 = 03y^2 - 16y + 28 = 0b^2 - 4ac < 0, so no real solutions for y

23 January 2020

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