Solve the following systems of equations: б) $$\begin{cases}5x + 8y = -1 \\ x + 2y = 4\end{cases}$$ б) $$\begin{cases}3x + 7y = -5 \\ 5x + 4y = 7\end{cases}$$ б) $$\begin{cases}3x - 2(3y+1) = -2 \\ 2(x+1)-1 = 3y - 1\end{cases}$$ б) $$\begin{cases}\frac{3a}{4} + \frac{3b}{8} = \frac{9}{2} \\ \frac{2a}{3} + \frac{b}{12} = \frac{2}{3}\end{cases}$$ 3) $$\begin{cases}\frac{1}{x} + \frac{1}{y} = 1 \\ \frac{1}{2x} + \frac{2}{y} = 8\end{cases}$$

Let's solve each system of equations step by step: б) $$\begin{cases}5x + 8y = -1 \\ x + 2y = 4\end{cases}$$ From the second equation, we have $$x = 4 - 2y$$. Substituting this into the first equation gives: $$5(4 - 2y) + 8y = -1$$ $$20 - 10y + 8y = -1$$ $$-2y = -21$$ $$y = \frac{21}{2} = 10.5$$ Substituting this back into $$x = 4 - 2y$$, we get: $$x = 4 - 2(10.5) = 4 - 21 = -17$$ So, the solution is $$x = -17$$ and $$y = 10.5$$. Answer: $$x = -17$$, $$y = 10.5$$ б) $$\begin{cases}3x + 7y = -5 \\ 5x + 4y = 7\end{cases}$$ Multiply the first equation by 5 and the second by 3: $$\begin{cases}15x + 35y = -25 \\ 15x + 12y = 21\end{cases}$$ Subtract the second equation from the first: $$23y = -46$$ $$y = -2$$ Substitute this back into $$3x + 7y = -5$$: $$3x + 7(-2) = -5$$ $$3x - 14 = -5$$ $$3x = 9$$ $$x = 3$$ So, the solution is $$x = 3$$ and $$y = -2$$. Answer: $$x = 3$$, $$y = -2$$ б) $$\begin{cases}3x - 2(3y+1) = -2 \\ 2(x+1)-1 = 3y - 1\end{cases}$$ Simplify the equations: $$\begin{cases}3x - 6y - 2 = -2 \\ 2x + 2 - 1 = 3y - 1\end{cases}$$ $$\begin{cases}3x - 6y = 0 \\ 2x + 1 = 3y - 1\end{cases}$$ $$\begin{cases}3x = 6y \\ 2x - 3y = -2\end{cases}$$ From the first equation, $$x = 2y$$. Substitute this into the second equation: $$2(2y) - 3y = -2$$ $$4y - 3y = -2$$ $$y = -2$$ Substitute this back into $$x = 2y$$: $$x = 2(-2) = -4$$ So, the solution is $$x = -4$$ and $$y = -2$$. Answer: $$x = -4$$, $$y = -2$$ б) $$\begin{cases}\frac{3a}{4} + \frac{3b}{8} = \frac{9}{2} \\ \frac{2a}{3} + \frac{b}{12} = \frac{2}{3}\end{cases}$$ Multiply the first equation by 8 and the second by 12 to eliminate fractions: $$\begin{cases}6a + 3b = 36 \\ 8a + b = 8\end{cases}$$ From the second equation, $$b = 8 - 8a$$. Substitute this into the first equation: $$6a + 3(8 - 8a) = 36$$ $$6a + 24 - 24a = 36$$ $$-18a = 12$$ $$a = -\frac{12}{18} = -\frac{2}{3}$$ Substitute this back into $$b = 8 - 8a$$: $$b = 8 - 8(-\frac{2}{3}) = 8 + \frac{16}{3} = \frac{24 + 16}{3} = \frac{40}{3}$$ So, the solution is $$a = -\frac{2}{3}$$ and $$b = \frac{40}{3}$$. Answer: $$a = -\frac{2}{3}$$, $$b = \frac{40}{3}$$ 3) $$\begin{cases}\frac{1}{x} + \frac{1}{y} = 1 \\ \frac{1}{2x} + \frac{2}{y} = 8\end{cases}$$ Let $$u = \frac{1}{x}$$ and $$v = \frac{1}{y}$$. Then the system becomes: $$\begin{cases}u + v = 1 \\ \frac{u}{2} + 2v = 8\end{cases}$$ Multiply the first equation by -1/2: $$\begin{cases}-\frac{u}{2} - \frac{v}{2} = -\frac{1}{2} \\ \frac{u}{2} + 2v = 8\end{cases}$$ Add the two equations: $$\frac{3}{2}v = \frac{15}{2}$$ $$v = 5$$ Substitute this back into $$u + v = 1$$: $$u + 5 = 1$$ $$u = -4$$ So, $$u = -4$$ and $$v = 5$$. Since $$u = \frac{1}{x}$$ and $$v = \frac{1}{y}$$, we have: $$x = \frac{1}{u} = \frac{1}{-4} = -\frac{1}{4}$$ $$y = \frac{1}{v} = \frac{1}{5}$$ So, the solution is $$x = -\frac{1}{4}$$ and $$y = \frac{1}{5}$$. Answer: $$x = -\frac{1}{4}$$, $$y = \frac{1}{5}$$