To find the intersection of the two lines given by the equations: $13x - 7y + 20 = 0$ $11x + 10y + 5 = 0$

we need to solve this system of linear equations simultaneously. Here are the steps:

1. Rewrite the equations: $13x - 7y = -20$ $11x + 10y = -5$

2. Use the method of substitution or elimination to find $x$ and $y$. Here, we will use the elimination method.

First, we need to make the coefficients of one of the variables the same in both equations. Let's eliminate $x$.

Multiply the first equation by 11 and the second equation by 13:

$11(13x - 7y) = 11(-20)$ $13(11x + 10y) = 13(-5)$

This gives us:

$143x - 77y = -220$ $143x + 130y = -65$

3. Subtract the first equation from the second to eliminate $x$:

$(143x + 130y) - (143x - 77y) = -65 - (-220)$ $143x + 130y - 143x + 77y = -65 + 220$ $207y = 155$

4. Solve for $y$:

$y = \frac{155}{207} = \frac{5}{9}$

5. Substitute $y = \frac{5}{9}$ back into one of the original equations to solve for $x$. Let's use the first equation:

$13x - 7\left(\frac{5}{9}\right) = -20$ $13x - \frac{35}{9} = -20$ $13x = -20 + \frac{35}{9}$ $13x = -\frac{180}{9} + \frac{35}{9}$ $13x = -\frac{145}{9}$ $x = -\frac{145}{9 \cdot 13}$ $x = -\frac{145}{117}$ $x = -\frac{5}{9}$

So, the intersection point of the two lines is:

$\left( -\frac{5}{9}, \frac{5}{9} \right)$

Would you like more details on the process, or do you have any other questions?

Here are some additional questions for further exploration:

1. How can we verify that the intersection point is correct?
2. What is the graphical interpretation of the solution?
3. Can you find the intersection using matrix methods?
4. How would the solution change if we used substitution instead of elimination?
5. What are the implications if the lines are parallel?
6. What if the coefficients were different, how would that affect the solution?
7. Can we generalize this method to solve for three variables?
8. How does the determinant of the coefficient matrix help in finding the solution?

Tip: Always double-check your solution by substituting the values back into the original equations to ensure they satisfy both equations.