[tex]y +1 = 2x~~~...(i)\\\\y^2 = 3x -2~~~...(ii)\\\\\text{From equation (i):}\\\\y +1 = 2x \\\\\implies y = 2x -1~~~....(iii)\\\\\text{Substitute} ~y = 2x-1~ \text{in equation (ii):}\\\\(2x-1)^2 = 3x-2\\\\\implies 4x^2 - 4x +1 = 3x -2\\\\\implies 4x^2 -4x -3x +3 =0\\\\\implies 4x(x-1) -3(x-1)=0\\\\\implies (x-1)(4x-3)=0\\\\\implies x =1 ~\text{or}~ x = \dfrac 34\\\\\text{Substitute x = 1 in equation (iii):}\\\\y = 2 -1 =1\\\\[/tex]
[tex]\text{Substitute}~ x = \dfrac 34 ~\text{in equation (iii):}\\\\\\y = 2\left(\dfrac 34 \right) -1 = \dfrac 32 -1 = \dfrac 12\\\\\text{Hence,}\\\\(x,y) = (1,1)\\\\\text{And}\\\\(x,y) = \left(\dfrac 34 , \dfrac 12 \right)[/tex]
