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

Question

asked Jul 1, 2021 33.4k views

2 Answers

Brett Hall · Answer 1 · 2021-07-07T19:35:42+0000

Answer:

Correct option: C -> 2

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.)