[tex]\displaystyle\bf\\\left \{ {{3x^{2} -y=2x} \atop {3x-y=2}} \right. \\\\\\\left \{ {{y=3x-2} \atop {3x^{2} -3x+2-2x=0}} \right. \\\\\\\left \{ {{y=3x-2} \atop {3x^{2} -5x+2=0}} \right. \\\\\\3x^{2} -5x+2=0\\\\D=(-5)^{2} -4\cdot 3\cdot 2=25-24=1\\\\\\x_{1} =\frac{5-1}{6} =\frac{4}{6} =\frac{2}{3} \\\\\\x_{2} =\frac{5+1}{6} =\frac{6}{6} =1\\\\\\y_{1} =3x-2=3\cdot \frac{2}{3} -2=2-2=0\\\\\\y_{2} =3x-2=3\cdot 1-2=3-2=1[/tex]

[tex]\displaystyle\bf\\Otvet \ : \ \Big(\frac{2}{3} \ ; \ 0\Big) \ , \ \Big(1 \ ; \ 1\Big)[/tex]

Объяснение:

3х² - у = 2х

3х - у = 2

3х² - у = 2х

у = 3х - 2

1)

3х² - у = 2х

3х² - (3х - 2) = 2х

3х² - 3х + 2 = 2х

3х² - 3х + 2 - 2х = 0

3х² - 5х + 2 = 0

а = 3; в = -5; с = 2

Д = в² - 4ас

Д = (-5)² - 4*3*2 = 25 - 24 = 1

√Д = √1 = 1

х1 = (-в-√Д)/2а

х1 = (5-1)/2*3 = 4/6 = 2/3

х2 = (-в+√Д)/2а

х2 = (5+1)/2*3 = 6/6 = 1

2)

у = 3х - 2

у = 3*2/3 - 2

у = 6/3 - 2

у = 2 - 2

у1 = 0

у = 3х - 2

у = 3*1 - 2

у = 3 - 2

у2 = 1

Ответ:

(х1, у1) = (2/3; 0)

(х2, у2) = (1; 1)