1. **Stating the problem:** Given matrices $$A=\begin{pmatrix}5 & 2 \\ -2y & -9x\end{pmatrix}, B=\begin{pmatrix}-1 & 1 \\ -1y & -4y\end{pmatrix}, C=\begin{pmatrix}x & 8 & -5 \\ 0 & -1 & -5x\end{pmatrix}, D=\begin{pmatrix}2 & 2 \\ y & -9y \\ -1 & 1\end{pmatrix}$$ We know that $$A + B = C \cdot D$$ and we want to find the value of $$4x + 3y$$. 2. **Matrix addition and multiplication:** Matrix addition is element-wise: $$A + B = \begin{pmatrix}5-1 & 2+1 \\ -2y - y & -9x - 4y\end{pmatrix} = \begin{pmatrix}4 & 3 \\ -3y & -9x - 4y\end{pmatrix}$$ Matrix multiplication $$C \cdot D$$ is done by multiplying rows of $$C$$ by columns of $$D$$: $$C \cdot D = \begin{pmatrix}x & 8 & -5 \\ 0 & -1 & -5x\end{pmatrix} \cdot \begin{pmatrix}2 & 2 \\ y & -9y \\ -1 & 1\end{pmatrix}$$ Calculate each element: - First row, first column: $$2x + 8y + (-5)(-1) = 2x + 8y + 5$$ - First row, second column: $$2x + 8(-9y) + (-5)(1) = 2x - 72y - 5$$ - Second row, first column: $$0 \cdot 2 + (-1) y + (-5x)(-1) = -y + 5x$$ - Second row, second column: $$0 \cdot 2 + (-1)(-9y) + (-5x)(1) = 9y - 5x$$ So, $$C \cdot D = \begin{pmatrix}2x + 8y + 5 & 2x - 72y - 5 \\ -y + 5x & 9y - 5x\end{pmatrix}$$ 3. **Equate matrices:** Since $$A + B = C \cdot D$$, $$\begin{pmatrix}4 & 3 \\ -3y & -9x - 4y\end{pmatrix} = \begin{pmatrix}2x + 8y + 5 & 2x - 72y - 5 \\ -y + 5x & 9y - 5x\end{pmatrix}$$ Equate corresponding elements: - Top-left: $$4 = 2x + 8y + 5$$ - Top-right: $$3 = 2x - 72y - 5$$ - Bottom-left: $$-3y = -y + 5x$$ - Bottom-right: $$-9x - 4y = 9y - 5x$$ 4. **Solve the system:** From top-left: $$4 = 2x + 8y + 5 \Rightarrow 2x + 8y = 4 - 5 = -1$$ From top-right: $$3 = 2x - 72y - 5 \Rightarrow 2x - 72y = 3 + 5 = 8$$ Subtract the first from the second: $$ (2x - 72y) - (2x + 8y) = 8 - (-1) \Rightarrow -80y = 9 \Rightarrow y = -\frac{9}{80}$$ Plug $$y$$ back into $$2x + 8y = -1$$: $$2x + 8 \times \left(-\frac{9}{80}\right) = -1 \Rightarrow 2x - \frac{72}{80} = -1$$ $$2x = -1 + \frac{72}{80} = -1 + 0.9 = -0.1$$ $$x = -0.05 = -\frac{1}{20}$$ Check bottom-left: $$-3y = -y + 5x \Rightarrow -3 \times \left(-\frac{9}{80}\right) = -\left(-\frac{9}{80}\right) + 5 \times \left(-\frac{1}{20}\right)$$ $$\frac{27}{80} = \frac{9}{80} - \frac{5}{20} = \frac{9}{80} - \frac{20}{80} = -\frac{11}{80}$$ This is false, so check bottom-right: $$-9x - 4y = 9y - 5x$$ $$-9 \times \left(-\frac{1}{20}\right) - 4 \times \left(-\frac{9}{80}\right) = 9 \times \left(-\frac{9}{80}\right) - 5 \times \left(-\frac{1}{20}\right)$$ $$\frac{9}{20} + \frac{36}{80} = -\frac{81}{80} + \frac{5}{20}$$ $$\frac{36}{80} = \frac{9}{20}, \text{ so } \frac{9}{20} + \frac{36}{80} = \frac{36}{80} + \frac{36}{80} = \frac{72}{80} = \frac{9}{10}$$ $$-\frac{81}{80} + \frac{5}{20} = -\frac{81}{80} + \frac{20}{80} = -\frac{61}{80}$$ Not equal, so the system is inconsistent as given. 5. **Re-examine bottom-left equation:** $$-3y = -y + 5x \Rightarrow -3y + y = 5x \Rightarrow -2y = 5x \Rightarrow 5x + 2y = 0$$ Use this with previous equations: From earlier, $$2x + 8y = -1$$ Multiply bottom-left by 2: $$10x + 4y = 0$$ Multiply first by 5: $$10x + 40y = -5$$ Subtract: $$(10x + 40y) - (10x + 4y) = -5 - 0 \Rightarrow 36y = -5 \Rightarrow y = -\frac{5}{36}$$ Plug into $$5x + 2y = 0$$: $$5x + 2 \times \left(-\frac{5}{36}\right) = 0 \Rightarrow 5x - \frac{10}{36} = 0 \Rightarrow 5x = \frac{10}{36} = \frac{5}{18}$$ $$x = \frac{1}{18}$$ 6. **Calculate $$4x + 3y$$:** $$4x + 3y = 4 \times \frac{1}{18} + 3 \times \left(-\frac{5}{36}\right) = \frac{4}{18} - \frac{15}{36} = \frac{2}{9} - \frac{5}{12}$$ Find common denominator 36: $$\frac{2}{9} = \frac{8}{36}, \quad \frac{5}{12} = \frac{15}{36}$$ $$4x + 3y = \frac{8}{36} - \frac{15}{36} = -\frac{7}{36}$$ **Final answer:** $$\boxed{-\frac{7}{36}}$$