9. \( \left\{\begin{array}{l}x y=6 \\ 2 x-y=1\end{array}\right. \)
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To solve the system of equations \( \left\{\begin{array}{l}x y=6 \\ 2 x-y=1\end{array}\right. \), we can start by expressing \( y \) from the second equation: \( y = 2x - 1 \). Now, substitute this expression for \( y \) into the first equation: \[ x(2x - 1) = 6 \] This simplifies to: \[ 2x^2 - x - 6 = 0 \] Factoring gives us: \[ (2x + 3)(x - 2) = 0 \] Setting each factor to zero, we find \( x = -\frac{3}{2} \) and \( x = 2 \). Now substituting these values back into \( y = 2x - 1 \): For \( x = 2 \): \[ y = 2(2) - 1 = 3 \] For \( x = -\frac{3}{2} \): \[ y = 2(-\frac{3}{2}) - 1 = -4 \] So the solutions are \( (2, 3) \) and \( \left(-\frac{3}{2}, -4\right) \). Notice that the method of substitution provides a clear path to the solution, but be on the lookout for signs! A small slip can lead to a big mistake – always double-check your work before finalizing answers. And remember: systems of equations can pop up in areas like economics, engineering, and even game development. Understanding how to solve them not only boosts your math skills but also prepares you for real-world scenarios!
