\( \begin{array}{l}\frac{2}{3} x-\frac{1}{2} y=3 \\ \frac{1}{2} x+y=5\end{array} \) Which of the following systems is equivalent to the given system? \( 2 x-3 y=3 \) and \( x+2 y=5 \) \( 4 x-3 y=18 \) and \( x+2 y=10 \) \( 2 x-y=18 \) and \( x+2 y=10 \)

To find an equivalent system of equations, we can manipulate the original equations in different ways. Let's first start by clearing the fractions in the first equation: Multiply the entire first equation by 6 (the least common multiple of the denominators 3 and 2): \[ 6 \left(\frac{2}{3} x - \frac{1}{2} y\right) = 6(3) \implies 4x - 3y = 18 \] For the second equation: Multiply the entire second equation by 2 (to eliminate the fraction): \[ 2 \left(\frac{1}{2} x + y\right) = 2(5) \implies x + 2y = 10 \] Now we have transformed the system into: \[ 4x - 3y = 18 \quad \text{and} \quad x + 2y = 10 \] Now we can see which given system matches: 1. \( 2x - 3y = 3 \) and \( x + 2y = 5 \) 2. \( 4x - 3y = 18 \) and \( x + 2y = 10 \) 3. \( 2x - y = 18 \) and \( x + 2y = 10 \) The second system \( 4x - 3y = 18 \) and \( x + 2y = 10 \) is indeed equivalent to the original equations. So, the correct answer is: **The equivalent system is: \( 4x - 3y = 18 \) and \( x + 2y = 10 \).**