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.