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Solve the system of equations. y=x-5; y=x^2-5x+3

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loneguy

[tex]y = x-5~~....(i)\\\\y =x^2-5x+3~~.....(ii)\\\\\text{Substitute}~ y= x-5 ~\text{in equation (ii):}\\\\x-5 = x^2 -5x +3\\\\\implies x^2 -5x -x +3 +5 =0\\\\\implies x^2 - 6x + 8 =0\\\\\implies x^2 --4x -2x +8 =0\\\\\implies x(x-4) -2(x-4) =0\\\\\implies (x-4)(x-2)= 0\\\\\implies x =4~~ \text{or}~~ x = 2\\\\\text{Substitute x = 4 in equation (i):} \\\\y = 4-5 = -1\\\\\text{Substitute x = 2 in equation (i):} \\\\y = 2-5 = -3\\\\\text{Hence,}~ (x,y) = (4,-1)~ \text{and}~ (x,y) = (2,-3)}[/tex]

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