Solve this system algebraically. \( \begin{array}{l}7 x-2 y=4 \\ 5 y+3 x=10 \\ \{(-41 / 40,-41 / 58)\} \\ \{(40 / 41,58 / 41)\} \\ \{(41 / 40,41 / 58)\} \\ \{(-40 / 41,-58 / 41)\}\end{array} \)

To solve the system of equations algebraically, we can start by rewriting the equations for clarity: 1) \( 7x - 2y = 4 \) 2) \( 3x + 5y = 10 \) Now, we can solve one of the equations for one variable. Let's solve equation (1) for \( y \): From \( 7x - 2y = 4 \), we get: \( 2y = 7x - 4 \) \( y = \frac{7x - 4}{2} \) Next, we substitute this expression for \( y \) into equation (2): \( 3x + 5\left(\frac{7x - 4}{2}\right) = 10 \) Multiplying through by 2 to eliminate the fraction: \( 6x + 5(7x - 4) = 20 \) \( 6x + 35x - 20 = 20 \) \( 41x - 20 = 20 \) \( 41x = 40 \) \( x = \frac{40}{41} \) Now we can substitute \( x \) back into our expression for \( y \): \( y = \frac{7\left(\frac{40}{41}\right) - 4}{2} \) \( y = \frac{\frac{280}{41} - 4}{2} \) Since 4 is \( \frac{164}{41} \): \( y = \frac{\frac{280 - 164}{41}}{2} = \frac{\frac{116}{41}}{2} = \frac{58}{41} \) Thus, the solution to the system is: \( \left(\frac{40}{41}, \frac{58}{41}\right) \) And from your options, the correct answer is \( \{(40 / 41, 58 / 41)\} \).