Question

How many real solutions does this system of equations have? x2+y2=363x−y+1=0 A. 0 B. 3 C. 2 D. 1

Answer 1

Answer:

Correct option: C -> 2

Step-by-step explanation:

The first equation is:

$x^2+y^2=363$

And the second equation is:

$x-y+1=0$

From the second equation, we have:

$y = x + 1$

Using this value of y in the first equation, we have:

$x^2 + (x+1)^2 = 363$

$x^2 + x^2 + 2x + 1 = 363$

$2x^2 + 2x= 362$

$x^2 + x - 181 = 0$

Calculating the discriminant Delta, we have:

$\Delta = b^2 - 4ac = 1 + 4*181 = 725$

We have $\Delta > 0$, so we have two real values for x, therefore we have two solutions for this system.

Correct option: C.

(If the system of equation is actually:

$x^2+y^2=36$

$3x-y+1=0$

We would have:

$y = 3x + 1$

$x^2+(3x+1)^2=36$

$x^2+9x^2+6x+1=36$

$10x^2+6x-35=0$

$\Delta = 36 + 1400 = 1436$

We also have $\Delta > 0$, so we have two solutions for this system.

Correct option: C.)