Let's break down and solve the simultaneous equations step by step.

We are given two equations:

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

Step 1: Simplify the first equation

Starting with the first equation:

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

Expand the terms inside the parentheses:

$6 \cdot 2x + 6 \cdot (-1) + 6 \cdot y + 5 = -13$

This simplifies to:

$12x - 6 + 6y + 5 = -13$

Combine like terms:

$12x + 6y - 1 = -13$

Now, add 1 to both sides:

$12x + 6y = -12$

Finally, divide through by 6 to simplify:

$2x + y = -2 \quad \text{(Equation 1)}$

Step 2: Simplify the second equation

Now let's simplify the second equation:

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

Expand both sides:

$3xy + 5y = 6x + 3xy - 26$

Cancel out the $3xy$ terms from both sides:

$5y = 6x - 26$

This simplifies to:

$5y = 6x - 26 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations

We now have the system:

1. $2x + y = -2$
2. $5y = 6x - 26$

Solve for $y$ in the first equation:

From equation 1:

$y = -2 - 2x$

Substitute this into equation 2:

Substitute $y = -2 - 2x$ into equation 2:

$5(-2 - 2x) = 6x - 26$

Expand the left side:

$-10 - 10x = 6x - 26$

Now, combine like terms. First, add 10x to both sides:

$-10 = 16x - 26$

Add 26 to both sides:

$16 = 16x$

Now divide by 16:

$x = 1$

Step 4: Find $y$

Substitute $x = 1$ into the equation $y = -2 - 2x$:

$y = -2 - 2(1) = -2 - 2 = -4$

Final Answer:

The solution to the system of equations is:

$x = 1, \quad y = -4$

Would you like more details on any step?

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Related Questions:

1. How do you solve systems of equations using substitution?
2. What are other methods to solve simultaneous equations, like elimination?
3. How do you check if the solution to a system of equations is correct?
4. Can you solve simultaneous equations graphically?
5. What happens when a system of equations has no solutions?

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