1. **State the problem:** Solve the simultaneous equations: $$6(2x - 1 + y) + 5 = -13$$ $$y(3x + 5) = 3x(2 + y) - 26$$ 2. **Simplify the first equation:** Distribute 6: $$6 \times 2x + 6 \times (-1) + 6 \times y + 5 = -13$$ which is $$12x - 6 + 6y + 5 = -13$$ Simplify constants: $$12x + 6y - 1 = -13$$ Add 1 to both sides: $$12x + 6y = -13 + 1$$ $$12x + 6y = -12$$ Divide entire equation by 6: $$\cancel{6}(2x + y) = \cancel{6}(-2)$$ which simplifies to $$2x + y = -2$$ 3. **Simplify the second equation:** Expand both sides: Left side: $$y(3x + 5) = 3xy + 5y$$ Right side: $$3x(2 + y) - 26 = 6x + 3xy - 26$$ Set equation: $$3xy + 5y = 6x + 3xy - 26$$ Subtract $3xy$ from both sides: $$5y = 6x - 26$$ 4. **Express $y$ from the first simplified equation:** $$y = -2 - 2x$$ 5. **Substitute $y$ into the second simplified equation:** $$5(-2 - 2x) = 6x - 26$$ Distribute 5: $$-10 - 10x = 6x - 26$$ Add $10x$ to both sides: $$-10 = 16x - 26$$ Add 26 to both sides: $$16 = 16x$$ Divide both sides by 16: $$\cancel{16} = \cancel{16}x$$ $$x = 1$$ 6. **Find $y$ using $x=1$ in $y = -2 - 2x$:** $$y = -2 - 2(1) = -2 - 2 = -4$$ **Final answer:** $$x = 1, \quad y = -4$$