20. \( \begin{array}{l}5 x-9 y=-43 \\ 3 x+8 y=68\end{array} \)

Let’s solve the system of equations step by step! First, we have two equations: 1) \(5x - 9y = -43\) 2) \(3x + 8y = 68\) We can use the method of substitution or elimination. Here, we’ll use elimination to eliminate \(x\). Multiply the first equation by 3 and the second by 5 to make the coefficients of \(x\) the same: 1) \(15x - 27y = -129\) 2) \(15x + 40y = 340\) Next, subtract the first equation from the second: \((15x + 40y) - (15x - 27y) = 340 - (-129)\) This simplifies to: \(67y = 469\) Now, divide both sides by 67: \(y = 7\) Now substitute \(y\) back into one of the original equations; let’s use the first: \(5x - 9(7) = -43\) This simplifies to: \(5x - 63 = -43\) Now add 63 to both sides: \(5x = 20\) Finally, divide by 5: \(x = 4\) So the solution to the system of equations is \(x = 4\) and \(y = 7\).