We are given two equations:

1. The curve: $y = 2xy + 5$
2. The line: $2x + 5y = 1$

We are tasked with finding the points of intersection $A$ and $B$.

Step-by-Step Solution:

Step 1: Express $y$ in terms of $x$ from the line equation

From the line equation $2x + 5y = 1$, we can solve for $y$:

$5y = 1 - 2x$ $y = \frac{1 - 2x}{5}$

Step 2: Substitute this expression for $y$ into the curve equation

The curve equation is $y = 2xy + 5$. Substituting $y = \frac{1 - 2x}{5}$ into the curve equation:

$\frac{1 - 2x}{5} = 2x \left( \frac{1 - 2x}{5} \right) + 5$

Step 3: Simplify the equation

First, distribute $2x$ in the right-hand side:

$\frac{1 - 2x}{5} = \frac{2x(1 - 2x)}{5} + 5$ $\frac{1 - 2x}{5} = \frac{2x - 4x^2}{5} + 5$

Now, multiply through by 5 to eliminate the denominators:

$1 - 2x = 2x - 4x^2 + 25$

Step 4: Rearrange the terms

Rearrange the equation:

$0 = -4x^2 + 4x + 24$ $4x^2 - 4x - 24 = 0$

Step 5: Solve the quadratic equation

Divide through by 4 to simplify:

$x^2 - x - 6 = 0$

Factor the quadratic:

$(x - 3)(x + 2) = 0$

So, $x = 3$ or $x = -2$.

Step 6: Find the corresponding $y$-coordinates

Substitute these $x$-values into the line equation $y = \frac{1 - 2x}{5}$:

• For $x = 3$: $y = \frac{1 - 2(3)}{5} = \frac{1 - 6}{5} = \frac{-5}{5} = -1$

• For $x = -2$: $y = \frac{1 - 2(-2)}{5} = \frac{1 + 4}{5} = \frac{5}{5} = 1$

Final Answer:

The points of intersection are:

• $A(3, -1)$
• $B(-2, 1)$

Would you like further details or clarification on any part of the solution?

Related Questions:

1. How do you solve systems of non-linear equations algebraically?
2. How can you verify if two curves intersect at a certain point?
3. What techniques are useful in factoring quadratic equations?
4. How do substitution and elimination methods differ in solving systems of equations?
5. What are common types of curves and their properties?

Tip: Always check if a quadratic equation can be factored easily before using the quadratic formula—it can save time!