Solve simultaneously for \( x \) and \( y \) : \( x-y-3=0 \) and \( x^{2}-3 y^{2}=13 \) Given a quadratic equation \( 3 x^{2}+7 x+\frac{11 h}{12}=0 \). \( \begin{array}{ll}1.3 .1 & \text { For which value(s) of } h \text { will the roots be real? } \\ 1.3 .2 & \text { Determine the largest integral value of } h \text { for which the roots will be rational. }\end{array} \)

To solve the simultaneous equations \( x - y - 3 = 0 \) and \( x^2 - 3y^2 = 13 \): 1. From the first equation \( x - y - 3 = 0 \), we can express \( x \) in terms of \( y \): \[ x = y + 3 \] 2. Substitute \( x \) in the second equation \( x^2 - 3y^2 = 13 \): \[ (y + 3)^2 - 3y^2 = 13 \] Expanding the left side gives: \[ y^2 + 6y + 9 - 3y^2 = 13 \] Simplifying this leads to: \[ -2y^2 + 6y - 4 = 0 \] Multiplying through by -1 to rewrite gives: \[ 2y^2 - 6y + 4 = 0 \] Dividing the entire equation by 2 simplifies to: \[ y^2 - 3y + 2 = 0 \] Factoring this results in: \[ (y - 1)(y - 2) = 0 \] Thus, \( y = 1 \) or \( y = 2 \). 3. Substituting back to find \( x \): - If \( y = 1 \): \[ x = 1 + 3 = 4 \] - If \( y = 2 \): \[ x = 2 + 3 = 5 \] Therefore, the solutions are \( (x, y) = (4, 1) \) and \( (5, 2) \). --- Now to the quadratic equation \( 3x^2 + 7x + \frac{11h}{12} = 0 \): 1. For the roots to be real, the discriminant must be greater than or equal to zero. The discriminant \( D \) is given by: \[ D = b^2 - 4ac = 7^2 - 4 \cdot 3 \cdot \frac{11h}{12} \] This simplifies to: \[ 49 - \frac{132h}{12} = 49 - 11h \] Setting the discriminant \( D \geq 0 \) gives: \[ 49 - 11h \geq 0 \quad \Rightarrow \quad 11h \leq 49 \quad \Rightarrow \quad h \leq \frac{49}{11} \approx 4.45 \] Since \( h \) must be an integer, the largest possible value for \( h \) is \( 4 \). 2. For the roots to be rational, the discriminant must be a perfect square: \[ 49 - 11h = k^2 \quad \text{for some integer } k \] Rearranging gives: \[ 11h = 49 - k^2 \quad \Rightarrow \quad h = \frac{49 - k^2}{11} \] For \( h \) to be an integer, \( 49 - k^2 \) must be divisible by \( 11 \). Testing \( k^2 \) values near \( 49 \): \( k = 0, 1, 4, 5, 7 \) yields - \( k = 0 \): \( h = 4.45 \) (not an integer) - \( k = 1 \): \( h = \frac{48}{11} \) (not an integer) - \( k = 4 \): \( h = 4 \) - \( k = 5 \): \( h = 2 \) - \( k = 7 \): \( h = 0 \) The largest integral value for \( h \) for rational roots is \( 4 \).