Question

find the integral zero of 3y^3 -4y^3+4y^2-4y+1

Answer 1

Answer:

So (1,8/3) is a solution. For x =2, 4×2 + 3y = 12 ⇒ 3y = 12 - 8 ⇒ 3y = 4 ⇒ y ...

The given equation is 4x + 3y = 12For x=0, 4×0 + 3y = 12⇒ 3y = 12 - 0⇒ 3y = 12⇒ y = ...

given that 4x+3y = 12if y=0, 4x+3(0)=12 4x=12 x=12/4 x=3 x=3,y=0if y=1,4x+3(1)=12 4x+3=12 ...