Let's solve for the points of intersection of the two lines given by the equations:
[tex]\[ y = 3x - 3 \][/tex]
[tex]\[ y = \frac{6}{x} \][/tex]
To find the points of intersection, we need to set the right-hand sides of the equations equal to each other and solve for [tex]\( x \)[/tex]:
[tex]\[ 3x - 3 = \frac{6}{x} \][/tex]
First, we get rid of the fraction by multiplying both sides by [tex]\( x \)[/tex]:
[tex]\[ x(3x - 3) = 6 \][/tex]
[tex]\[ 3x^2 - 3x = 6 \][/tex]
Next, we'll move all terms to one side to set the equation to 0:
[tex]\[ 3x^2 - 3x - 6 = 0 \][/tex]
Now, we can simplify this equation by dividing every term by 3:
[tex]\[ x^2 - x - 2 = 0 \][/tex]
This is a quadratic equation, and we can solve it by factoring:
[tex]\[ x^2 - x - 2 = (x - 2)(x + 1) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
[tex]\[ x + 1 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
Now we need to find the corresponding [tex]\( y \)[/tex] values for each [tex]\( x \)[/tex] by substituting these [tex]\( x \)[/tex] values back into either of the original equations. Let's use [tex]\( y = 3x - 3 \)[/tex].
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 3(2) - 3 = 6 - 3 = 3 \][/tex]
So, one intersection point is [tex]\( (2, 3) \)[/tex].
For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 3(-1) - 3 = -3 - 3 = -6 \][/tex]
So, the other intersection point is [tex]\( (-1, -6) \)[/tex].
Thus, the points of intersection are:
[tex]\[ \{(2, 3), (-1, -6)\} \][/tex]
By comparing these points with the given options:
- A: [tex]\(\{(-2, 3)(1, 6)\}\)[/tex]
- B: [tex]\(\{(-2, -3)(-1, -6)\}\)[/tex]
- C: [tex]\(\{(-2, 3)(1, -6)\}\)[/tex]
- D: [tex]\(\{(2, 3)(-1, -6)\}\)[/tex]
The correct answer is:
D: [tex]\(\{(2, 3), (-1, -6)\}\)[/tex]
