Answer:
[tex](1, -1)[/tex] and [tex](-\frac{1}{5}, \frac{7}{5})[/tex]
In decimal form this would be (1, -1) and (-0.2, 1.4)
Step-by-step explanation:
Basically here we are trying to solve a system of 2 equations. The first one is a straight line and the second is a circle. This can be done graphically and graph is attached. There will be two points where the line cuts the circle so two solution sets
The equations are
Line Equation [tex]y = 1 - 2x[/tex] (1) and
Circle Equation [tex]x^2+ y^2 = 2[/tex] (2)
[tex]y=1-2x[/tex]
Plug this value of y into the second equation for the circle
[tex]x^2 + (1-2x)^2= 2[/tex]
Apply the perfect squares formula [tex]\left(a-b\right)^2=a^2-2ab+b^2[/tex]
[tex](1-2x)^2 = 1^2-2\cdot \:1\cdot \:2x+\left(2x\right)^2\\= 1-4x+4x^2\\\\[/tex]
So equation (2) becomes
[tex]1-4x+4x^2 +x^2 = 2 \\\\\\5x^2 -4x - 1 = 0[/tex]
Solve using the quadratic formula
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here a is the coefficient of x ², b is the coefficient of x and c is the constant
So
[tex]x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:5\left(-1\right)}}{2\cdot \:5}[/tex]
Now,
[tex]\sqrt{\left(-4\right)^2-4\cdot \:5\left(-1\right)} = \sqrt{\left(-4\right)^2+4\cdot \:5\cdot \:1}[/tex]
= [tex]\sqrt{16+20} = \sqrt{36} = 6[/tex]
Therefore
[tex]x_{1,\:2}=\frac{-\left(-4\right)\pm \:6}{2\cdot \:5}[/tex]
[tex]x_1=\frac{-\left(-4\right)+6}{2\cdot \:5}[/tex] = [tex]\frac{-\left(-4\right)+6}{2\cdot \:5} =[/tex] [tex]\frac{10}{2\cdot \:5}[/tex] [tex]= 1[/tex]
[tex]x_2 = \frac{-\left(-4\right)-6}{2\cdot \:5}[/tex] [tex]= \frac{4-6}{2\cdot \:5}[/tex] = [tex]-\frac{1}{5}[/tex]
To find corresponding values for [tex]y_1 y_2[/tex] plug these values of [tex]x_1, x_2[/tex] into the line equation
[tex]y_1 = 1- 2(1) = -1[/tex]
[tex]y_2 = 1 - 2(-\frac{1}{5}) = 1 + \frac{2}{5} = \frac{7}{5}[/tex]
So the two points of intersection are
[tex](1, -1)[/tex] and ([tex](-\frac{1}{5}, \frac{7}{5})[/tex].
In decimal form this would be (1, -1) and (-0.2, 1.4)
See attached graph
