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

A. 0

second part: C. The boats' paths do not cross each other

Answer 2

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.)