Solve the simultaneous equations?

2x + y = 8 (solve byusing substitution)
3x^2 + xy = 1

2x + y = 8
y = 8 - 2x

3x^2 + xy = 1
3x^2 + x(8 - 2x) = 1
3x^2 + 8x - 2x^2 = 1
3x^2 - 2x^2 + 8x - 1 = 0
x^2 + 8x - 1 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 8
c = -1

x = [-8 ±√(64 + 4)]/2
x = [-8 ±√68]/2
x = [-8 ±√(2^2 * 17)]/2
x = [-8 ±2√17]/2
x = -4 ±√17

2x + y = 8
2(-4 ±√17) + y = 8
-8 ±2√17 + y = 8
y = 8 + 8 ±2√17
y = 16 ±2√17

∴ x = -4 ±√17 , y = 16 ±2√17

a = -4
b = 1

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3x² + x (8 - 2x) = 1
3x² + 8x - 2x² = 1
x² + 8x - 1 = 0
x = [ - 8 ± √ (64 + 4) ] / 2
x = [ - 8 ± √ (68) ] / 2
x = [ - 8 ± 2√17 ] / 2
x = - 4 ± √17

2x+y=8
-2x -2x
y = 8-2x

3x^2 + xy = 1
3x^2 + x(8-2x) = 1
3x^2 + (8x - 2x^2) = 1
0 = x^2 + 8x - 1

Since (x+4)(x+4)=x^2+8x+16
x^2+8x = (x+4)^2-16

So 0 = x^2 + 8x - 1 = (x+4)^2 - 16 -1
(x+4)^2-17=0
(x+4)^2 = 17
(x+4) = sqrt(17)
-> x = -4 + 1 sqrt(17)

y = 8 - 2x
y = 8 - 2(sqrt(17)-4)
y = 8 - 2 sqrt(17) + 8
-> y = 16 -2 sqrt(17)

2x + y = 8
3x^2 + xy = 1
y = 8-2x
3x^2 +x(8-2x) =1
3x^2 -2x^2 +8x -1 =0
x^2 +8x -1 =0
Discriminant = (8)^2 +4(1) = 68= (2 sqrt(17) )^2
x1 = (-8 - 2sqrt(17))/2 = -4 -sqrt(17)
x2 = -4 +sqrt(17)
y1 = 8 -2x1 = 8 -2(-4 -sqrt(17)
y1 = 16 +2sqrt(17)
y2 = 16 -2sqrt(17)