Answer :
To determine why the system of equations has two solutions, let's analyze the system:
Given:
[tex]\[ y = 6x^2 + 1 \][/tex]
[tex]\[ y = x^2 + 4 \][/tex]
We are tasked with finding the points where these two equations intersect, i.e., where [tex]\( y = 6x^2 + 1 \)[/tex] and [tex]\( y = x^2 + 4 \)[/tex] are equal. This will help us identify the points of intersection.
1. Set the equations equal to each other:
[tex]\[ 6x^2 + 1 = x^2 + 4 \][/tex]
2. Rearrange the equation to set it to zero:
[tex]\[ 6x^2 + 1 - x^2 - 4 = 0 \][/tex]
[tex]\[ 5x^2 - 3 = 0 \][/tex]
3. Solve for [tex]\(x\)[/tex]:
[tex]\[ 5x^2 = 3 \][/tex]
[tex]\[ x^2 = \frac{3}{5} \][/tex]
[tex]\[ x = \pm\sqrt{\frac{3}{5}} \][/tex]
[tex]\[ x = \pm\frac{\sqrt{15}}{5} \][/tex]
So, we have two solutions for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\sqrt{15}}{5} \][/tex]
[tex]\[ x = -\frac{\sqrt{15}}{5} \][/tex]
4. Find the corresponding [tex]\(y\)[/tex]-values by substituting [tex]\(x\)[/tex] back into either original equation:
Let's use [tex]\( y = 6x^2 + 1 \)[/tex]:
[tex]\[ y = 6\left(\frac{\sqrt{15}}{5}\right)^2 + 1 \][/tex]
[tex]\[ y = 6 \cdot \frac{15}{25} + 1 \][/tex]
[tex]\[ y = 6 \cdot \frac{3}{5} + 1 \][/tex]
[tex]\[ y = \frac{18}{5} + \frac{5}{5} \][/tex]
[tex]\[ y = \frac{23}{5} \][/tex]
This shows that both of our [tex]\(x\)[/tex] values yield the same [tex]\( y \)[/tex]-value:
[tex]\[ y = \frac{23}{5} \][/tex]
Thus, the points of intersection are:
[tex]\[ \left(\frac{\sqrt{15}}{5}, \frac{23}{5}\right) \][/tex]
[tex]\[ \left(-\frac{\sqrt{15}}{5}, \frac{23}{5}\right) \][/tex]
Given these results, we can conclude that the system of equations has two solutions because the graphs of the equations intersect each other at two points. Therefore, the correct statement is:
The graphs of the equations intersect each other at two places.
Given:
[tex]\[ y = 6x^2 + 1 \][/tex]
[tex]\[ y = x^2 + 4 \][/tex]
We are tasked with finding the points where these two equations intersect, i.e., where [tex]\( y = 6x^2 + 1 \)[/tex] and [tex]\( y = x^2 + 4 \)[/tex] are equal. This will help us identify the points of intersection.
1. Set the equations equal to each other:
[tex]\[ 6x^2 + 1 = x^2 + 4 \][/tex]
2. Rearrange the equation to set it to zero:
[tex]\[ 6x^2 + 1 - x^2 - 4 = 0 \][/tex]
[tex]\[ 5x^2 - 3 = 0 \][/tex]
3. Solve for [tex]\(x\)[/tex]:
[tex]\[ 5x^2 = 3 \][/tex]
[tex]\[ x^2 = \frac{3}{5} \][/tex]
[tex]\[ x = \pm\sqrt{\frac{3}{5}} \][/tex]
[tex]\[ x = \pm\frac{\sqrt{15}}{5} \][/tex]
So, we have two solutions for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\sqrt{15}}{5} \][/tex]
[tex]\[ x = -\frac{\sqrt{15}}{5} \][/tex]
4. Find the corresponding [tex]\(y\)[/tex]-values by substituting [tex]\(x\)[/tex] back into either original equation:
Let's use [tex]\( y = 6x^2 + 1 \)[/tex]:
[tex]\[ y = 6\left(\frac{\sqrt{15}}{5}\right)^2 + 1 \][/tex]
[tex]\[ y = 6 \cdot \frac{15}{25} + 1 \][/tex]
[tex]\[ y = 6 \cdot \frac{3}{5} + 1 \][/tex]
[tex]\[ y = \frac{18}{5} + \frac{5}{5} \][/tex]
[tex]\[ y = \frac{23}{5} \][/tex]
This shows that both of our [tex]\(x\)[/tex] values yield the same [tex]\( y \)[/tex]-value:
[tex]\[ y = \frac{23}{5} \][/tex]
Thus, the points of intersection are:
[tex]\[ \left(\frac{\sqrt{15}}{5}, \frac{23}{5}\right) \][/tex]
[tex]\[ \left(-\frac{\sqrt{15}}{5}, \frac{23}{5}\right) \][/tex]
Given these results, we can conclude that the system of equations has two solutions because the graphs of the equations intersect each other at two points. Therefore, the correct statement is:
The graphs of the equations intersect each other at two places.