How many solutions does the linear system represented by the equations y = 6x^2 and y = 12x^2y^4 have?

To determine how many solutions the linear system defined by the equations y = 6x² and y = 12x²y⁴ has, we need to analyze the system step-by-step.

1. **Understanding the equations**: Both equations are nonlinear due to the presence of the variables raised to a power greater than one. The first equation is a simple quadratic equation in terms of y, while the second equation involves y raised to the fourth power.

2. **Setting the equations equal to each other**: Since both expressions are equal to y, we can set them equal to each other:

6x² = 12x²y⁴

3. **Rearranging the equation**: We can rearrange this equation by moving all terms to one side:

12x²y⁴ – 6x² = 0

4. **Factoring the equation**: Next, we can factor out the common term, which is 6x²:

6x²(2y⁴ – 1) = 0

This product will equal zero if at least one of the factors is zero.

5. **Finding the solutions**: We can now use the zero-product property to find the solutions:

  • 6x² = 0: This implies x = 0.
  • 2y⁴ – 1 = 0: Rearranging gives us y⁴ = 1/2, leading to y = ±(1/√2) or y = ±√(1/2).

6. **Counting the solutions**: Therefore, we have the following pairs (
x, y):

  • For x = 0 and y = ±(1/√2), we have two unique solutions: (0, √(1/2)) and (0, -√(1/2)).

7. **Conclusion**: In total, the linear system has two solutions:

  • (0, √(1/2))
  • (0, -√(1/2))

Hence, the final answer is: The linear system defined by the equations has two solutions.

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