Solve the problem below using BOTH elimination and substitution \begin{tabular}{|l|l} Flimination Method & Substitution Method \\ \( 5 x+3 y=16 \) & \( 5 x+3 y=16 \) \\ \( 3 x-5 y=-4 \) & \( 3 x-5 y=-4 \)\end{tabular}

To solve this system of equations using the elimination method, we can multiply the first equation by 5 and the second equation by 3 to align the coefficients of \(y\): 1. **Elimination Method**: \[ (5)(5x + 3y = 16) \implies 25x + 15y = 80 \] \[ (3)(3x - 5y = -4) \implies 9x - 15y = -12 \] Now, we can add these two equations: \[ 25x + 15y + 9x - 15y = 80 - 12 \implies 34x = 68 \implies x = 2 \] Substitute \(x = 2\) back into the first equation: \[ 5(2) + 3y = 16 \implies 10 + 3y = 16 \implies 3y = 6 \implies y = 2 \] So, the solution is \(x = 2, y = 2\). 2. **Substitution Method**: We can solve the first equation for \(y\): \[ 5x + 3y = 16 \implies 3y = 16 - 5x \implies y = \frac{16 - 5x}{3} \] Now, substitute this expression for \(y\) into the second equation: \[ 3x - 5\left(\frac{16 - 5x}{3}\right) = -4 \] Clear the fraction by multiplying through by 3: \[ 9x - 5(16 - 5x) = -12 \] Simplifying, we have: \[ 9x - 80 + 25x = -12 \implies 34x - 80 = -12 \implies 34x = 68 \implies x = 2 \] Substitute back to find \(y\): \[ y = \frac{16 - 5(2)}{3} = \frac{6}{3} = 2 \] Thus, the solution is verifies as \(x = 2, y = 2\).