10. Solve the system of equations using the method of your choice. \[ \begin{array}{r}-2 x+4 y=10 \\ 6 x-3 y=42\end{array} \]

To solve the system of equations, let's first simplify each equation. From the first equation, \(-2x + 4y = 10\), we can isolate \(y\): \[ 4y = 2x + 10 \quad \Rightarrow \quad y = \frac{1}{2}x + \frac{5}{2} \] Now, plug this expression for \(y\) into the second equation \(6x - 3y = 42\): \[ 6x - 3\left(\frac{1}{2}x + \frac{5}{2}\right) = 42 \] Distributing the \(-3\): \[ 6x - \frac{3}{2}x - \frac{15}{2} = 42 \] To combine like terms, multiply everything by \(2\) to eliminate the fraction: \[ 12x - 3x - 15 = 84 \quad \Rightarrow \quad 9x - 15 = 84 \] Adding \(15\) to both sides: \[ 9x = 99 \quad \Rightarrow \quad x = 11 \] Now substitute \(x = 11\) back into the expression for \(y\): \[ y = \frac{1}{2}(11) + \frac{5}{2} = \frac{11 + 5}{2} = \frac{16}{2} = 8 \] Thus, the solution to the system of equations is: \[ (x, y) = (11, 8) \]