Question

Consider the system of equations:

x^2 + y^2 = 8
x = 3

Why does the answer for y indicate that the system has no real-number solutions?

a) The values for y are square roots of negative numbers.
b) The values for y have a positive and a negative value.
c) The values for y have square roots.


Simplify: 9 + y2 = 8
3. Isolate: y²=-1
4. Solve for y: y = -1

Answer 1

The system of equations has no real-number solutions because the equation y^2 = -1 has no real-number solutions.

Step-by-step explanation:

The system of equations is:

x^2 + y^2 = 8

x = 3

To solve this system, we can substitute the value of x from the second equation into the first equation. We get:

(3)^2 + y^2 = 8

9 + y^2 = 8

y^2 = -1

Since the equation y^2 = -1 has no real-number solutions, it means that the system has no real-number solutions.