We are solving the system of equations:

1. $x(y + 3) = 4$
2. $3y(x - 4) = 5$

Step 1: Expand and simplify both equations

Equation 1: $x(y + 3) = 4$ Expanding: xy + 3x = 4 \tag{1}

Equation 2: $3y(x - 4) = 5$ Expanding: 3yx - 12y = 5 \tag{2}

Step 2: Express one variable in terms of the other

From Equation (1), solve for $x$ in terms of $y$: $xy + 3x = 4$ Factor out $x$: $x(y + 3) = 4$ x = \frac{4}{y + 3} \quad \text{(as long as y \neq -3)} \tag{3}

Step 3: Substitute $x$ from Equation (3) into Equation (2)

Substitute $x = \frac{4}{y + 3}$ into $3yx - 12y = 5$: $3y \left( \frac{4}{y + 3} \right) - 12y = 5$ Simplify: $\frac{12y}{y + 3} - 12y = 5$ Combine into a single fraction: $\frac{12y - 12y(y + 3)}{y + 3} = 5$ Simplify the numerator: $12y - 12y^2 - 36y = -12y^2 - 24y$ Thus: $\frac{-12y^2 - 24y}{y + 3} = 5$

Step 4: Solve for $y$

Multiply through by $y + 3$ (as long as $y \neq -3$): $-12y^2 - 24y = 5(y + 3)$ Expand the right-hand side: $-12y^2 - 24y = 5y + 15$ Rearrange to set the equation to 0: $-12y^2 - 24y - 5y - 15 = 0$ $-12y^2 - 29y - 15 = 0$ Divide through by $-1$ to simplify: 12y^2 + 29y + 15 = 0 \tag{4}

Step 5: Solve the quadratic equation

The quadratic equation is: $12y^2 + 29y + 15 = 0$ Use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 12$, $b = 29$, $c = 15$. Substituting: $y = \frac{-29 \pm \sqrt{29^2 - 4(12)(15)}}{2(12)}$ $y = \frac{-29 \pm \sqrt{841 - 720}}{24}$ $y = \frac{-29 \pm \sqrt{121}}{24}$ $y = \frac{-29 \pm 11}{24}$

Now calculate the two solutions:

1. $y = \frac{-29 + 11}{24} = \frac{-18}{24} = \frac{-3}{4}$
2. $y = \frac{-29 - 11}{24} = \frac{-40}{24} = \frac{-5}{3}$

Step 6: Solve for $x$ for each value of $y$

Substitute each $y$ value into Equation (3): $x = \frac{4}{y + 3}$.

Case 1: When $y = \frac{-3}{4}$: $x = \frac{4}{\frac{-3}{4} + 3} = \frac{4}{\frac{-3}{4} + \frac{12}{4}} = \frac{4}{\frac{9}{4}} = 4 \cdot \frac{4}{9} = \frac{16}{9}$

Case 2: When $y = \frac{-5}{3}$: $x = \frac{4}{\frac{-5}{3} + 3} = \frac{4}{\frac{-5}{3} + \frac{9}{3}} = \frac{4}{\frac{4}{3}} = 4 \cdot \frac{3}{4} = 3$

Final Answer:

The solutions are:

1. $x = \frac{16}{9}, y = \frac{-3}{4}$
2. $x = 3, y = \frac{-5}{3}$

Do you want more details or explanations for any step?

Related Questions:

1. How do you verify the solutions in the original equations?
2. How does factoring help solve quadratic equations more efficiently?
3. What happens if $y = -3$ in this problem? Why is it excluded?
4. How can substitution be applied in solving systems of equations in general?
5. What other methods could be used to solve this system (e.g., elimination)?

Tip:

Always double-check your solutions by substituting them back into the original equations to ensure they satisfy both equations.