We are tasked with solving the following system of simultaneous equations algebraically:

$7x + 5y = 13 \quad \text{(1)}$ $3x - 2y = -11 \quad \text{(2)}$

Let's use the method of substitution or elimination. I'll solve this using elimination for simplicity.

Step 1: Eliminate one variable

To eliminate one variable, we need to make the coefficients of either $x$ or $y$ the same in both equations. Let's eliminate $x$. We can multiply Equation (1) by 3 and Equation (2) by 7 to align the coefficients of $x$.

Multiplying Equation (1) by 3: $3(7x + 5y) = 3(13) \implies 21x + 15y = 39 \quad \text{(3)}$

Multiplying Equation (2) by 7: $7(3x - 2y) = 7(-11) \implies 21x - 14y = -77 \quad \text{(4)}$

Step 2: Subtract the equations

Now subtract Equation (4) from Equation (3) to eliminate $x$:

$(21x + 15y) - (21x - 14y) = 39 - (-77)$ $21x + 15y - 21x + 14y = 39 + 77$ $29y = 116$ $y = \frac{116}{29} = 4$

Step 3: Substitute $y = 4$ into one of the original equations

Let's substitute $y = 4$ into Equation (1):

$7x + 5(4) = 13$ $7x + 20 = 13$ $7x = 13 - 20$ $7x = -7$ $x = \frac{-7}{7} = -1$

Final Answer:

The solution to the system of equations is: $x = -1, \quad y = 4$

Would you like a detailed breakdown or have any questions about this solution? Here are five questions to extend your understanding:

1. What happens if we try to eliminate $y$ instead of $x$?
2. Can you verify the solution by substituting $x = -1$ and $y = 4$ into both original equations?
3. How does the substitution method differ from the elimination method in solving simultaneous equations?
4. What happens when we multiply both equations by negative values instead?
5. Can these equations be solved graphically, and how would the solution appear on a graph?

Tip: When solving systems of equations, it's good practice to check your final answer by substituting it back into both original equations.