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Consider the system of equations: x²y²=8 x=3

Solving for one variable, substitute x=3: (3)²y²=8

Simplify: 9y²=8

Isolate: y²=−1

Solve for y: y=±√-1
y=no solution

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?

A) The solution does not solve for x first.
B) The values for y are square roots of negative numbers.
C) The values for y have a positive and a negative value.
D) The values for y have square roots.

User Pedrobisp
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1 Answer

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Final answer:

The system of equations results in no real-number solutions for y because squaring a negative number yields an imaginary number. Option B accurately states this situation since real numbers cannot have square roots of negative numbers, which is the predicament presented in this problem. option b is correct.

Step-by-step explanation:

The answer for y in the system of equations indicates that the system has no real-number solutions because the process results in y² being equal to a negative number. In mathematics, the square root of a negative number is not a real number, but an imaginary number.

The correct option to explain why the system has no real-number solutions is B) The values for y are square roots of negative numbers. It should be mentioned that y=±√-1 does not produce real numbers because the square roots of negative numbers are not defined within the real number system. Instead, solutions involving square roots of negative numbers are considered complex or imaginary numbers.

Quadratic equations constructed on physical data always have real roots; however, this is not a quadratic equation but rather an example of a system of equations where the given x value leads to an impossible situation for y within the realm of real numbers. It is important to be aware that while solving equations, encountering a square root of a negative number is a clear indication of no real solution, leading to complex or imaginary solutions instead.

User Joe Allen
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