1. Solve for \( x \) and \( y \) : (a) \( 3 x-14 y=0 \) and \( x-4 y+1=0 \) (b) \( x+y=8 \) and \( 3 x+2 y=21 \) (c) \( y=2 x+1 \) and \( x+2 y+3=0 \) (dd) \( \frac{a}{2}+b=4 \) and \( \frac{a}{4}-\frac{b}{4}=1 \) (e) \( \frac{1}{x}+\frac{1}{y}=3 \) and \( \frac{1}{x}-\frac{1}{y}=11 \)

Let's solve each pair of equations step by step: (a) From \( 3x - 14y = 0 \), we can express \( x \) in terms of \( y \): \( x = \frac{14}{3}y \). Substituting this into the second equation \( x - 4y + 1 = 0 \): \( \frac{14}{3}y - 4y + 1 = 0 \) This simplifies to: \( \frac{14}{3}y - \frac{12}{3}y + 1 = 0 \) \(\frac{2}{3}y + 1 = 0 \Rightarrow y = -\frac{3}{2} \). Substituting \( y \) back into \( x = \frac{14}{3}y \): \( x = \frac{14}{3}(-\frac{3}{2}) = -7 \). So, \( x = -7, y = -\frac{3}{2} \). (b) From \( x + y = 8 \), we can express \( y \): \( y = 8 - x \). Substituting this into the second equation \( 3x + 2y = 21 \): \( 3x + 2(8 - x) = 21 \) This simplifies to: \( 3x + 16 - 2x = 21 \Rightarrow x + 16 = 21 \Rightarrow x = 5 \). Substituting \( x \) back into \( y = 8 - x \): \( y = 8 - 5 = 3 \). So, \( x = 5, y = 3 \). (c) From \( y = 2x + 1 \), substitute \( y \) into the second equation \( x + 2y + 3 = 0 \): \( x + 2(2x + 1) + 3 = 0 \) This expands to: \( x + 4x + 2 + 3 = 0 \Rightarrow 5x + 5 = 0 \Rightarrow 5x = -5 \Rightarrow x = -1 \). Substituting \( x \) back into \( y = 2x + 1 \): \( y = 2(-1) + 1 = -2 + 1 = -1 \). So, \( x = -1, y = -1 \). (d) From \( \frac{a}{2} + b = 4 \), express \( b \): \( b = 4 - \frac{a}{2} \). Substituting into the second equation \( \frac{a}{4} - \frac{b}{4} = 1 \): \( \frac{a}{4} - \frac{1}{4}(4 - \frac{a}{2}) = 1 \). This simplifies to: \( \frac{a}{4} - 1 + \frac{a}{8} = 1 \). Multiplying through by 8 to eliminate the fractions: \( 2a - 8 + a = 8 \Rightarrow 3a - 8 = 8 \Rightarrow 3a = 16 \Rightarrow a = \frac{16}{3} \). Substituting \( a \) back to find \( b \): \( b = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \). So, \( a = \frac{16}{3}, b = \frac{4}{3} \). (e) From \( \frac{1}{x} + \frac{1}{y} = 3 \), multiply through by \( xy \): \( y + x = 3xy \). From \( \frac{1}{x} - \frac{1}{y} = 11 \), multiply through by \( xy \): \( y - x = 11xy \). Now we have the two equations: 1) \( y + x = 3xy \) 2) \( y - x = 11xy \) Adding both equations: \( 2y = 14xy \Rightarrow y = 7xy \) and substituting into either equation yields: Set \( y = 7xy \Rightarrow 1 = 7x \Rightarrow x = \frac