Answer:
(0, - 2 ) and (3, 4 )
Explanation:
Given the equations
3y = 2(x² - 3) , that is
3y = 2x² - 6 → (1)
2x - y = 2 (add y to both sides )
2x = 2 + y ( subtract 2 from both sides )
2x - 2 = y → (2)
Substitute y = 2x - 2 into (1)
3(2x - 2) = 2x² - 6
6x - 6 = 2x² - 6 ( subtract 6x - 6 from bpth sides )
0 = 2x² - 6x ( factor out 2x from each term )
0 = 2x(x - 3)
Equate each factor to zero and solve for x
2x = 0 ⇒ x = 0
x - 3 = 0 ⇒ x = 3
Substitute these values into (2) for corresponding values of y
x = 0 : y = 2(0) - 2 = 0 - 2 = - 2 ⇒ (0, - 2 )
x = 3 : y = 2(3) - 2 = 6 - 2 = 4 ⇒ (3, 4 )