g) \( \quad-4 \leq \frac{2}{2} \leq 8 \) Solve the pairs of simultaneous equatio a) \( 3 x+5 y=5 ; 3 x+y=9 \)

To solve the equations \( 3x + 5y = 5 \) and \( 3x + y = 9 \), we can start with the second equation and express \( y \) in terms of \( x \). From \( 3x + y = 9 \), we can rearrange it to get \( y = 9 - 3x \). Now, we'll substitute \( y \) into the first equation: \[ 3x + 5(9 - 3x) = 5 \] Expanding gives: \[ 3x + 45 - 15x = 5 \] Combining like terms results in: \[ -12x + 45 = 5 \] Subtracting 45 from both sides, we have: \[ -12x = 5 - 45 \] \[ -12x = -40 \] Dividing both sides by -12 yields: \[ x = \frac{40}{12} = \frac{10}{3} \] Next, we can find \( y \) by plugging \( x \) back into the equation \( y = 9 - 3x \): \[ y = 9 - 3(\frac{10}{3}) = 9 - 10 = -1 \] Thus, the solution to the system of equations is \( x = \frac{10}{3} \) and \( y = -1 \).