28. Solve the equations. \( \begin{array}{r}x+3 y=13 \\ x^{2}+3 y^{2}=43\end{array} \)

To solve the equations \( x + 3y = 13 \) and \( x^2 + 3y^2 = 43 \), we can start with the first equation. From \( x + 3y = 13 \), we can express \( x \) in terms of \( y \): \[ x = 13 - 3y \] Now, we can substitute this expression for \( x \) into the second equation: \[ (13 - 3y)^2 + 3y^2 = 43 \] Expanding \( (13 - 3y)^2 \): \[ 169 - 78y + 9y^2 + 3y^2 = 43 \] Combining like terms gives us: \[ 169 - 78y + 12y^2 = 43 \] Subtracting 43 from both sides results in: \[ 12y^2 - 78y + 126 = 0 \] Now, we can simplify this quadratic equation by dividing all terms by 6: \[ 2y^2 - 13y + 21 = 0 \] Next, we can factor the quadratic: \[ (2y - 3)(y - 7) = 0 \] Setting each factor to zero gives us the solutions: \[ 2y - 3 = 0 \quad \Rightarrow \quad y = \frac{3}{2} \] \[ y - 7 = 0 \quad \Rightarrow \quad y = 7 \] Now we can substitute these values back into the expression for \( x \). For \( y = \frac{3}{2} \): \[ x = 13 - 3\left(\frac{3}{2}\right) = 13 - \frac{9}{2} = \frac{26 - 9}{2} = \frac{17}{2} \] For \( y = 7 \): \[ x = 13 - 3(7) = 13 - 21 = -8 \] Thus, the pairs \( (x, y) \) that solve the original equations are: 1. \(\left( \frac{17}{2}, \frac{3}{2} \right)\) 2. \((-8, 7)\)