4. algebraically and check: \[ \begin{array}{l}3 x+2 y=6 \\ +5 x-3 y=-28\end{array} \]

To solve the system of equations, let's first express the two equations clearly: 1) \(3x + 2y = 6\) 2) \(5x - 3y = -28\) We can use the substitution or elimination method. Here, we'll use elimination. First, we'll multiply the first equation by 3 to align the coefficients of \(y\): 1) \(9x + 6y = 18\) 2) \(5x - 3y = -28\) Next, we can multiply the second equation by 2: 1) \(9x + 6y = 18\) 2) \(10x - 6y = -56\) Now, we can sum these two equations: \[ (9x + 6y) + (10x - 6y) = 18 - 56 \] This simplifies to: \[ 19x = -38 \implies x = -2 \] Now, substituting \(x = -2\) back into the first equation: \[ 3(-2) + 2y = 6 \implies -6 + 2y = 6 \implies 2y = 12 \implies y = 6 \] Thus, the solution is \(x = -2\) and \(y = 6\). To check, we can substitute \(x\) and \(y\) back into the second equation: \[ 5(-2) - 3(6) = -10 - 18 = -28 \] Both equations are satisfied with these values!