To solve the system of equations:
[tex]\[ y = 3x \][/tex]
[tex]\[ y = x^2 - 10 \][/tex]
we need to find the points where these two equations intersect. These intersection points are where the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] satisfy both equations simultaneously.
### Step-by-Step Solution
1. Set the equations equal to each other:
Since both equations equal [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ 3x = x^2 - 10 \][/tex]
2. Rearrange the equation:
Move all terms to one side to set the equation to 0:
[tex]\[ x^2 - 3x - 10 = 0 \][/tex]
3. Solve the quadratic equation:
We solve the quadratic equation [tex]\( x^2 - 3x - 10 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -10 \)[/tex].
[tex]\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{9 + 40}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{49}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm 7}{2}
\][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{3 + 7}{2} = 5
\][/tex]
[tex]\[
x = \frac{3 - 7}{2} = -2
\][/tex]
4. Find the corresponding [tex]\( y \)[/tex] values:
Substitute each [tex]\( x \)[/tex] value back into the first equation [tex]\( y = 3x \)[/tex] to find the corresponding [tex]\( y \)[/tex].
For [tex]\( x = 5 \)[/tex]:
[tex]\[
y = 3(5) = 15
\][/tex]
So, one point of intersection is [tex]\( (5, 15) \)[/tex].
For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = 3(-2) = -6
\][/tex]
So, the other point of intersection is [tex]\( (-2, -6) \)[/tex].
### Conclusion
The points of intersection of the given system of equations are:
[tex]\[ (5, 15) \text{ and } (-2, -6) \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B. (-2,-6) \text{ and } (5,15)} \][/tex]
