13. Solve the simultaneous equations. \[ \begin{array}{l}5 x+6 y=9 \\ 3 x-4 y=12\end{array} \]

To solve the simultaneous equations, let's use the method of substitution or elimination. First, we can multiply the first equation by 3 and the second by 5 to align the coefficients of \( x \): 1. \( 15x + 18y = 27 \) (First equation multiplied by 3) 2. \( 15x - 20y = 60 \) (Second equation multiplied by 5) Now, subtract the first modified equation from the second: \[ (15x - 20y) - (15x + 18y) = 60 - 27 \] This simplifies to: \[ -38y = 33 \] Now, solve for \( y \): \[ y = -\frac{33}{38} \] Substituting \( y \) back into one of the original equations (let's use \( 5x + 6y = 9 \)): \[ 5x + 6\left(-\frac{33}{38}\right) = 9 \] \[ 5x - \frac{198}{38} = 9 \] \[ 5x = 9 + \frac{198}{38} \] Converting 9 into a fraction with a denominator of 38: \[ 5x = \frac{342}{38} + \frac{198}{38} = \frac{540}{38} \] Now divide by 5: \[ x = \frac{540}{190} = \frac{108}{38} = \frac{54}{19} \] So the solution to the simultaneous equations is: \[ x = \frac{54}{19}, \quad y = -\frac{33}{38} \]