a + 2b = 5, 13a-6=8；

To solve the system of equations:

1. \( a + 2b = 5 \) \quad (Equation 1)
2. \( 13a - 6 = 8 \) \quad (Equation 2)

First, let's simplify Equation 2:

\[ 13a - 6 = 8 \] Adding 6 to both sides:

\[ 13a = 14 \] Dividing both sides by 13:

\[ a = \frac{14}{13} \]

Now, we will substitute the value of \( a \) back into Equation 1 to find \( b \):

\[ \frac{14}{13} + 2b = 5 \] Subtracting \( \frac{14}{13} \) from both sides:

\[ 2b = 5 - \frac{14}{13} \] To perform the subtraction, convert 5 to a fraction with a denominator of 13:

\[ 5 = \frac{65}{13} \]

Now, we can subtract:

\[ 2b = \frac{65}{13} - \frac{14}{13} = \frac{51}{13} \]

Now divide both sides by 2:

\[ b = \frac{51}{26} \]

So, the solution to the system of equations is:

\[ a = \frac{14}{13}, \quad b = \frac{51}{26} \]