The given system of simultaneous equations is:

$6(2x - 1 + y) + 5 = -13$ $y(3x + 5) = 3x(2 + y) - 26$

Let's solve these step-by-step.

Step 1: Simplify the First Equation

Expand the equation: $6(2x - 1 + y) + 5 = -13$ $12x - 6 + 6y + 5 = -13$ Combine like terms: $12x + 6y - 1 = -13$ Add 1 to both sides: $12x + 6y = -12$ Divide the whole equation by 6: $2x + y = -2$ So, the first equation simplifies to: $y = -2 - 2x$

Step 2: Substitute $y$ in the Second Equation

The second equation is: $y(3x + 5) = 3x(2 + y) - 26$ Substitute $y = -2 - 2x$ into this equation: $(-2 - 2x)(3x + 5) = 3x(2 + (-2 - 2x)) - 26$

Step 3: Expand and Simplify

Expand both sides of the equation:

1. Left side: $(-2 - 2x)(3x + 5)$ $-2(3x + 5) - 2x(3x + 5) = -6x - 10 - 6x^2 - 10x$ $= -6x^2 - 16x - 10$

2. Right side: $3x(2 + (-2 - 2x)) - 26$ $3x(2 - 2x) - 26 = 6x - 6x^2 - 26$

So the equation becomes: $-6x^2 - 16x - 10 = 6x - 6x^2 - 26$

Step 4: Simplify Further

Rearrange the equation to combine like terms: $-6x^2 - 16x - 10 + 6x^2 - 6x + 26 = 0$ $-22x + 16 = 0$ $x = \frac{16}{22} = \frac{8}{11}$

Step 5: Solve for $y$

Substitute $x = \frac{8}{11}$ into the simplified equation $y = -2 - 2x$: $y = -2 - 2\left(\frac{8}{11}\right)$ $y = -2 - \frac{16}{11} = -\frac{22}{11} - \frac{16}{11} = -\frac{38}{11}$

Solution

The solution to the system of equations is: $x = \frac{8}{11}, \quad y = -\frac{38}{11}$

Would you like more details on any of these steps or have any questions?

Here are five related questions to consider:

1. How can we verify the solution of these simultaneous equations?
2. What are the different methods to solve simultaneous equations?
3. How do substitution and elimination methods compare in solving systems of equations?
4. Can these types of equations always be solved algebraically, or are there cases where they require numerical solutions?
5. How can graphing be used to solve and interpret solutions to simultaneous equations?

Tip: When solving simultaneous equations, it's always good to double-check your work by substituting the solution back into the original equations to ensure they hold true.