№62 б 2x^2-xy+3y^2-7x=12y-1 x-y=-1

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Let's substitute y=x+1 into the first equation:

2x^2 - x(x + 1) + 3(x + 1)^2 - 7x = 12(x + 1) - 1
2x^2 - x^2 - x + 3(x^2 + 2x + 1) - 7x = 12x + 12 - 1
2x^2 - x^2 - x + 3x^2 + 6x + 3 - 7x = 12x + 11
4x^2 - x - 4 = 12x + 11
4x^2 - 13x - 15 = 0

Now, we can solve this quadratic equation for x:

x = [-(-13) ± √((-13)^2 - 44(-15))] / 2*4
x = [13 ± √(169 + 240)] / 8
x = [13 ± √409] / 8

Therefore, the solutions for x are:

x = (13 + √409) / 8 or x = (13 - √409) / 8

Now, we can substitute these values back into the second equation to find the corresponding y values.