Answer:
(0, 1 ) and ([tex]\frac{1}{2}[/tex], [tex]\frac{1}{2}[/tex] )
Step-by-step explanation:
Given the 2 equations
x³ - xy = 0 → (1)
x + y = 1 → (2) ( subtract x from both sides )
y = 1 - x → (3)
Substitute y = 1 - x into (1)
x² - x(1 - x) = 0
x² - x + x² = 0
2x² - x = 0 ← factor out x from each term on the left side
x(2x - 1) = 0
Equate each factor to zero and solve for x
x = 0
2x - 1 = 0 ⇒ 2x = 1 ⇒ x = [tex]\frac{1}{2}[/tex]
Substitute these values into (3) for corresponding values of y
x = 0 : y = 1 - 0 = 1 ⇒ (0, 1 )
x = [tex]\frac{1}{2}[/tex] : y = 1 - [tex]\frac{1}{2}[/tex] = [tex]\frac{1}{2}[/tex] ⇒ ( [tex]\frac{1}{2}[/tex], [tex]\frac{1}{2}[/tex] )