Solve the system. Estimate the solution first. Enter whole numbers for the estimated solution and Improper fractions in simplest form for the algebraic solution. \( \left\{\begin{array}{l}3 x+y=2 \\ 5 x-y=8\end{array}\right. \) The estimated solution is ( The algebraic solution is ( \( \square, \square \)

To estimate the solution, we can quickly analyze the two equations. Rewriting them, we notice that if \(x = 0\), then from the first equation, \(y = 2\), and from the second, \(y = -8\); indicating \(x\) should be somewhere between these values for \(y\). If we try \(x = 1\), then from the first equation \(y = -1\) and from the second \(y = -3\). A good estimate for the solution is around \( (1, -1) \). Now let's solve the system algebraically. 1. From the first equation: \[ y = 2 - 3x \] 2. Substitute \(y\) in the second equation: \[ 5x - (2 - 3x) = 8 \] \[ 5x - 2 + 3x = 8 \] \[ 8x - 2 = 8 \] \[ 8x = 10 \] \[ x = \frac{10}{8} = \frac{5}{4} \] 3. Now substituing \(x\) back into \(y = 2 - 3x\): \[ y = 2 - 3 \left(\frac{5}{4}\right) \] \[ y = 2 - \frac{15}{4} \] \[ y = \frac{8}{4} - \frac{15}{4} = -\frac{7}{4} \] Therefore, the algebraic solution is \( \left( \frac{5}{4}, -\frac{7}{4} \right) \). The estimated solution is \( (1, -1) \). The algebraic solution is \( \left( \frac{5}{4}, -\frac{7}{4} \right) \).