Question 678974
{{{y+1 = sqrt(5y-1)}}}
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{{{(y+1)^2 = sqrt(5y-1)^2}}}
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{{{y^2+2y + 1 = 5y - 1}}}
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{{{y^2 + 2y - 5y + 1 + 1 = 0}}}
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{{{y^2 - 3y + 2 = 0}}}
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(y-2)(y-1) = 0
y-1=0, so y = 1
y-2=0, so y = 2
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