We are given two systems of equations to solve. System 1: $3m - n = 5$ $2m + 5n = 7$ System 2: $5c - d - 11 = 0$ $4d + 3c = -5$

1. Problem Description

We are given two systems of equations to solve.

System 1:

$3m - n = 5$

$2m + 5n = 7$

System 2:

$5c - d - 11 = 0$

$4d + 3c = -5$

2. Solution Steps

System 1:

We can solve for $n$ in the first equation:

$3m - n = 5$

$n = 3m - 5$

Substitute this into the second equation:

$2m + 5(3m - 5) = 7$

$2m + 15m - 25 = 7$

$17m = 32$

$m = \frac{32}{17}$

Now, substitute $m = \frac{32}{17}$ back into $n = 3m - 5$:

$n = 3(\frac{32}{17}) - 5$

$n = \frac{96}{17} - \frac{85}{17}$

$n = \frac{11}{17}$

System 2:

We can rewrite the first equation as:

$5c - d = 11$

$d = 5c - 11$

Substitute this into the second equation:

$4(5c - 11) + 3c = -5$

$20c - 44 + 3c = -5$

$23c = 39$

$c = \frac{39}{23}$

Now, substitute $c = \frac{39}{23}$ back into $d = 5c - 11$:

$d = 5(\frac{39}{23}) - 11$

$d = \frac{195}{23} - \frac{253}{23}$

$d = -\frac{58}{23}$

3. Final Answer

System 1: $m = \frac{32}{17}$, $n = \frac{11}{17}$

System 2: $c = \frac{39}{23}$, $d = -\frac{58}{23}$