1. \( \frac{16}{3}+\frac{4}{3} y=x \) \( 5 x-2 y=8 \)

Let's solve the system of linear equations: 1. \( \frac{16}{3} + \frac{4}{3} y = x \) 2. \( 5x - 2y = 8 \) **Step 1: Express \( x \) from the first equation** From the first equation: \[ x = \frac{16}{3} + \frac{4}{3} y \] **Step 2: Substitute \( x \) into the second equation** Substitute \( x \) from equation (1) into equation (2): \[ 5\left( \frac{16}{3} + \frac{4}{3} y \right) - 2y = 8 \] **Step 3: Simplify the equation** Distribute the 5: \[ \frac{80}{3} + \frac{20}{3} y - 2y = 8 \] Convert \( 2y \) to thirds to combine like terms: \[ \frac{80}{3} + \frac{20}{3} y - \frac{6}{3} y = 8 \] \[ \frac{80}{3} + \frac{14}{3} y = 8 \] **Step 4: Solve for \( y \)** Subtract \( \frac{80}{3} \) from both sides: \[ \frac{14}{3} y = 8 - \frac{80}{3} \] \[ \frac{14}{3} y = \frac{24}{3} - \frac{80}{3} \] \[ \frac{14}{3} y = -\frac{56}{3} \] Multiply both sides by \( \frac{3}{14} \): \[ y = -4 \] **Step 5: Substitute \( y \) back into the expression for \( x \)** \[ x = \frac{16}{3} + \frac{4}{3} (-4) \] \[ x = \frac{16}{3} - \frac{16}{3} \] \[ x = 0 \] **Solution:** \[ x = 0, \quad y = -4 \]