Question

How many solutions does the system have?


Y= -2x + 1

Y= -2x^2 + 1

Answer 1

There are two solutions one of them = 1 and the other = zero

Explanation:

∵ y = -2x + 1 and y =
`-2x^(2)` + 1

∴ To find how many solution we will solve them by equating them

∴ -2x + 1 =
`-2x^(2)` + 1

∴ 2
`x^(2)` - 2x + 1 - 1 = 0

∴
`2x^(2)` - 2x = 0⇒ take 2x common factor

∴ 2x (x - 1) = 0

∴ 2x = 0 ⇒ x = 0

∴ x - 1 = 0 ⇒ x = 1

∴ The system has two solutions 1 and zero