1. Solve the pairs of simultaneous equations. a) \( x+y=5 ; \quad x-y=3 \) b) \( x-3 y=-5 ; \quad 2 x+3 y=-1 \) c) \( 2 x-3 y=-21 ; \quad 2 x+3 y=1 \) c)

To solve the equations, let's break it down step by step! For a) 1. From the first equation \( x + y = 5 \), we can express \( y \) as \( y = 5 - x \). 2. Substituting into the second equation \( x - (5 - x) = 3 \) gives us \( 2x - 5 = 3 \), leading to \( 2x = 8 \) or \( x = 4 \). 3. Plugging \( x = 4 \) back into \( y = 5 - x \), we find \( y = 1 \). **Solution: \( (4, 1) \)** For b) 1. Rearranging the first equation \( x - 3y = -5 \) gives us \( x = 3y - 5 \). 2. Substitute into the second equation: \( 2(3y - 5) + 3y = -1 \) simplifies to \( 6y - 10 + 3y = -1 \). 3. This leads to \( 9y - 10 = -1 \), so \( 9y = 9 \) or \( y = 1 \). 4. Plugging back, \( x = 3(1) - 5 = -2 \). **Solution: \( (-2, 1) \)** For c) 1. From \( 2x - 3y = -21 \), we can express \( 2x = -21 + 3y \) or \( x = \frac{-21 + 3y}{2} \). 2. Plug this expression into the second equation \( 2\left(\frac{-21 + 3y}{2}\right) + 3y = 1 \) which simplifies to \( -21 + 3y + 3y = 1 \). 3. Thus, \( 6y - 21 = 1 \) leads to \( 6y = 22 \) or \( y = \frac{11}{3} \). 4. Substituting this back, \( 2x - 3\left(\frac{11}{3}\right) = -21 \) leads us to \( 2x - 11 = -21 \) or \( 2x = -10 \) hence \( x = -5 \). **Solution: \( (-5, \frac{11}{3}) \)**