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

We start with the system \[ \begin{aligned} x + 3y &= 13, \\ x^2 + 3y^2 &= 43. \end{aligned} \] **Step 1.** Express \( x \) in terms of \( y \) from the first equation: \[ x = 13 - 3y. \] **Step 2.** Substitute \( x = 13 - 3y \) into the second equation: \[ (13-3y)^2 + 3y^2 = 43. \] **Step 3.** Expand the square and simplify: \[ (169 - 78y + 9y^2) + 3y^2 = 43. \] \[ 169 - 78y + 12y^2 = 43. \] Subtract \( 43 \) from both sides: \[ 12y^2 - 78y + 126 = 0. \] Divide the entire equation by \( 6 \) to simplify: \[ 2y^2 - 13y + 21 = 0. \] **Step 4.** Solve the quadratic equation \( 2y^2 - 13y + 21 = 0 \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a=2 \), \( b=-13 \), and \( c=21 \). First, compute the discriminant: \[ b^2 - 4ac = (-13)^2 - 4(2)(21) = 169 - 168 = 1. \] Now, substitute into the formula: \[ y = \frac{13 \pm \sqrt{1}}{4}. \] Thus, the two solutions for \( y \) are: \[ y = \frac{13 + 1}{4} = \frac{14}{4} = \frac{7}{2} \quad \text{and} \quad y = \frac{13 - 1}{4} = \frac{12}{4} = 3. \] **Step 5.** Find \( x \) using \( x = 13 - 3y \). For \( y = \frac{7}{2} \): \[ x = 13 - 3\left(\frac{7}{2}\right) = 13 - \frac{21}{2} = \frac{26}{2} - \frac{21}{2} = \frac{5}{2}. \] For \( y = 3 \): \[ x = 13 - 3(3) = 13 - 9 = 4. \] **Final Answer:** \[ \boxed{\left(\frac{5}{2}, \frac{7}{2}\right) \quad \text{and} \quad (4,\,3)} \]