Answer:
Given the equations:
[tex]y = \sqrt{1-x^2}[/tex] ....[1]
[tex]y = 2x-1[/tex] .....[2]
Taking square both sides in [1] we have;
[tex]y^2 = 1-x^2[/tex] ....[3]
Substitute the equation [2] into [3] we have;
[tex](2x-1)^2 = 1-x^2[/tex]
Using identity rule: [tex](a-b)^2 = a^-2ab+b^2[/tex]
then;
[tex]4x^2-4x+1 = 1-x^2[/tex]
Add [tex]x^2[/tex] both sides we have;
[tex]5x^2-4x+1 = 1[/tex]
Subtract 1 from both sides we have;
[tex]5x^2-4x=0[/tex]
⇒[tex]x(5x-4) = 0[/tex]
By zero product property we have;
x = 0 and 5x-4 = 0
⇒x= 0 and [tex]x = \frac{4}{5} = 0.8[/tex]
For x = 0,
[tex]y=\sqrt{1-0} = 1[/tex]
y = 2(0)-1 = -1
Since, the value of x = 0 doesn't satisfy the equations.
Therefore, the values of x is 0.8
