1. **State the problem:** We need to find values of $x$ and $y$ such that the matrices $$\begin{pmatrix}-1 & -5x \\ 2y & 3\end{pmatrix} = \begin{pmatrix}-1 & 1 - 6y \\ 2 - 6x & 3\end{pmatrix}$$ are equal. 2. **Recall the rule for matrix equality:** Two matrices are equal if and only if their corresponding elements are equal. 3. **Set up equations by equating corresponding elements:** - Top-left elements: $-1 = -1$ (always true) - Top-right elements: $-5x = 1 - 6y$ - Bottom-left elements: $2y = 2 - 6x$ - Bottom-right elements: $3 = 3$ (always true) 4. **Write the system of equations:** $$\begin{cases} -5x = 1 - 6y \\ 2y = 2 - 6x \end{cases}$$ 5. **Solve the system:** From the first equation: $$-5x = 1 - 6y \implies 6y = 1 + 5x \implies y = \frac{1 + 5x}{6}$$ Substitute $y$ into the second equation: $$2\left(\frac{1 + 5x}{6}\right) = 2 - 6x$$ Simplify left side: $$\frac{2(1 + 5x)}{6} = 2 - 6x \implies \frac{1 + 5x}{3} = 2 - 6x$$ Multiply both sides by 3: $$1 + 5x = 6 - 18x$$ Bring all terms to one side: $$5x + 18x = 6 - 1 \implies 23x = 5$$ Solve for $x$: $$x = \frac{5}{23}$$ 6. **Find $y$ using $x$ value:** $$y = \frac{1 + 5\left(\frac{5}{23}\right)}{6} = \frac{1 + \frac{25}{23}}{6} = \frac{\frac{23}{23} + \frac{25}{23}}{6} = \frac{\frac{48}{23}}{6} = \frac{48}{23 \times 6} = \frac{48}{138} = \frac{8}{23}$$ 7. **Final answer:** $$x = \frac{5}{23}, \quad y = \frac{8}{23}$$