10. \( \left\{\begin{array}{l}6 x-3 y=12 \\ 4 x+3 y=24\end{array}\right. \)

To solve the system of equations, we can use the method of substitution or elimination. First, let's use the elimination method. We have the following equations: 1) \(6x - 3y = 12\) 2) \(4x + 3y = 24\) We can add both equations to eliminate \(y\). Adding both equations gives us: \( (6x - 3y) + (4x + 3y) = 12 + 24 \) Combining like terms results in: \(10x = 36\) Now, solving for \(x\), we get: \(x = \frac{36}{10} = 3.6\) Now, substitute \(x = 3.6\) back into either of the original equations to find \(y\). We'll use the first equation: \(6(3.6) - 3y = 12\) \(21.6 - 3y = 12\) Subtract 21.6 from both sides: \(-3y = 12 - 21.6\) \(-3y = -9.6\) Now, divide by -3: \(y = \frac{-9.6}{-3} = 3.2\) So the solution to the system of equations is \(x = 3.6\) and \(y = 3.2\).