We are given two equations:

1. A line equation:
   $3x = 4x - 15$

2. A hyperbola equation:
   $8x^2 - 27y^2 = 45$

Step 1: Simplify the line equation.

Rearrange the line equation $3x = 4x - 15$:

$3x - 4x = -15$

$-x = -15$

$x = 15$

Step 2: Substitute $x = 15$ into the hyperbola equation.

Now that we know $x = 15$, substitute it into the hyperbola equation $8x^2 - 27y^2 = 45$:

$8(15^2) - 27y^2 = 45$

Calculate $15^2 = 225$:

$8(225) - 27y^2 = 45$

$1800 - 27y^2 = 45$

Step 3: Solve for $y^2$.

Rearrange the equation to isolate $y^2$:

$1800 - 45 = 27y^2$

$1755 = 27y^2$

$y^2 = \frac{1755}{27} = 65$

Thus,

$y = \pm \sqrt{65}$

Step 4: Find points A and B.

Since we have $x = 15$ and $y = \pm \sqrt{65}$, the points of intersection $A$ and $B$ are:

$A(15, \sqrt{65}) \quad \text{and} \quad B(15, -\sqrt{65})$

Final Answer:

The points of intersection are $A(15, \sqrt{65})$ and $B(15, -\sqrt{65})$.

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Would you like more details or have any questions?

Here are 5 related questions to extend your learning:

1. How would the solution change if the hyperbola equation was different?
2. What if the line equation had a different slope?
3. Can you find the distance between points A and B?
4. How would you graph the intersection of a line and a hyperbola?
5. What are the general methods to find intersections between conic sections and lines?

Tip: When solving systems of equations involving a line and a curve, always try to substitute one variable into the other equation after simplifying the line equation.