Answer:
[tex] x = \frac{12 \pm \sqrt{(-12)^2 -4(5)(-7)}}{2(5)}[/tex]
And we got for the solution:
[tex] x_1= 2.885 , x_2 = -0.485[/tex]
And the value sof y are using the function y =2x-3:
[tex] y_1=2.77, y_2=-3.97[/tex]
Step-by-step explanation:
For this case we have this function given:
[tex] y = 2x-3[/tex] (1)
And the circle with center the origin and radius 4 is given by;
[tex] x^2 +y^2 = 16[/tex] (2)
We can solve fro y from the last equation and we got:
[tex] y = \pm \sqrt{16-x^2}[/tex] (3)
Now we can set equal equations (3) and (1) and we got:
[tex] 2x-3 = \sqrt{16-x^2}[/tex]
[tex] (2x-3)^2=16-x^2[/tex]
[tex] 4x^2 -12x +9 = 16-x^2[/tex]
[tex] 5x^2 -12x -7=0[/tex]
And using the quadratic equation we got:
[tex] x = \frac{12 \pm \sqrt{(-12)^2 -4(5)(-7)}}{2(5)}[/tex]
And we got for the solution:
[tex] x_1= 2.885 , x_2 = -0.485[/tex]
And the value sof y are using the function y =2x-3:
[tex] y_1=2.77, y_2=-3.97[/tex]
