Recognizing the No-Solution Case Algebraically Consider the system of equations:
x2 + y2 = 8
x=3
Solving for one variable:
Work is shown to solve for the system of equations.
Why does the answer for y indicate that the system has
no real-number solutions?
1. Substitute:
(3)2 + y2 = 8
* The solution does not solve for x first.
O The values for y are square roots of negative
numbers.
The values for y have a positive and a negative
value.
O The values for y have square roots.
2. Simplify:
9 + y2 = 8
3. Isolate:
y²=-1
4. Solve fory:
y = -1
No solution