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Consider the systems of equations below:

System A: 12+y²=17
System B: y=12−75+10
System C: y=−212+9 ,85−y=−17 ,y=−63+5
Determine the number of real solutions for each system of equations.
a) System A has 1 real solution. System B has 2 real solutions. System C has 0 real solutions.
b) System A has 2 real solutions. System B has 1 real solution. System C has 1 real solution.
c) System A has 0 real solutions. System B has 1 real solution. System C has 2 real solutions.
d) System A has 1 real solution. System B has 0 real solutions. System C has 1 real solution.

User Anupam
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7.7k points

1 Answer

4 votes

Final answer:

System A has 2 real solutions, System B has 1 real solution, and System C has 0 real solutions, matching answer choice b.

Step-by-step explanation:

The student is asking about the number of real solutions for each given system of equations: System A, System B, and System C. To solve for the number of real solutions, we must analyze each system separately.

For System A (12+y²=17), by subtracting 12 from both sides and taking the square root, we get two potential solutions for y, which means System A has 2 real solutions.

In System B (y=12−75+10), simplifying right-hand side gives a single value for y, thus System B has 1 real solution.

For System C, there are two equations given (y=−12+9, 85−y=−17, y=−63+5). When simplified, the equations give different values for y, which is not possible for a single value of y, so System C has no real solution (inconsistent).

Therefore, the correct answer is: System A has 2 real solutions, System B has 1 real solution, and System C has 0 real solutions, which corresponds to choice b.

User Garykwwong
by
8.2k points
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