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To determine the number of solutions for the given system of equations, we need to find the points where the graphs of the two equations intersect. Let’s go through this step-by-step without using any external code, relying solely on algebraic methods and logical reasoning.
The equations given are:
1. [tex]\( y = 6x^2 + 1 \)[/tex]
2. [tex]\( y = x^2 + 4 \)[/tex]
To find the points of intersection, we set the two equations equal to each other and solve for [tex]\( x \)[/tex]:
[tex]\[ 6x^2 + 1 = x^2 + 4 \][/tex]
First, we simplify this equation by combining like terms:
[tex]\[ 6x^2 + 1 - x^2 - 4 = 0 \][/tex]
[tex]\[ 5x^2 - 3 = 0 \][/tex]
Next, we 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]
We have two solutions for [tex]\( x \)[/tex]: [tex]\( x = -\frac{\sqrt{15}}{5} \)[/tex] and [tex]\( x = \frac{\sqrt{15}}{5} \)[/tex].
For these [tex]\( x \)[/tex]-values, we can find the corresponding [tex]\( y \)[/tex]-values by substituting them back into one of the original equations. It doesn’t matter which one we use, since both equations will give us the same [tex]\( y \)[/tex]-values at the points of intersection.
Substituting [tex]\( x = \frac{\sqrt{15}}{5} \)[/tex] into [tex]\( y = 6x^2 + 1 \)[/tex]:
[tex]\[ y = 6\left(\frac{\sqrt{15}}{5}\right)^2 + 1 = 6\left(\frac{15}{25}\right) + 1 = 6\left(\frac{3}{5}\right) + 1 = \frac{18}{5} + 1 = \frac{18}{5} + \frac{5}{5} = \frac{23}{5} \][/tex]
Substituting [tex]\( x = -\frac{\sqrt{15}}{5} \)[/tex] into [tex]\( y = 6x^2 + 1 \)[/tex] will yield the same [tex]\( y \)[/tex]:
[tex]\[ y = 6\left(-\frac{\sqrt{15}}{5}\right)^2 + 1 = 6\left(\frac{15}{25}\right) + 1 = 6\left(\frac{3}{5}\right) + 1 = \frac{18}{5} + 1 = \frac{18}{5} + \frac{5}{5} = \frac{23}{5} \][/tex]
Therefore, the intersection points are [tex]\(\left(-\frac{\sqrt{15}}{5}, \frac{23}{5}\right)\)[/tex] and [tex]\(\left(\frac{\sqrt{15}}{5}, \frac{23}{5}\right)\)[/tex].
Hence, we observe that the two graphs intersect each other at two distinct points. This means the system of equations has two solutions, confirming the last statement:
"The graphs of the equations intersect each other at two places."
Thus, the correct choice is:
"The graphs of the equations intersect each other at two places."
The equations given are:
1. [tex]\( y = 6x^2 + 1 \)[/tex]
2. [tex]\( y = x^2 + 4 \)[/tex]
To find the points of intersection, we set the two equations equal to each other and solve for [tex]\( x \)[/tex]:
[tex]\[ 6x^2 + 1 = x^2 + 4 \][/tex]
First, we simplify this equation by combining like terms:
[tex]\[ 6x^2 + 1 - x^2 - 4 = 0 \][/tex]
[tex]\[ 5x^2 - 3 = 0 \][/tex]
Next, we 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]
We have two solutions for [tex]\( x \)[/tex]: [tex]\( x = -\frac{\sqrt{15}}{5} \)[/tex] and [tex]\( x = \frac{\sqrt{15}}{5} \)[/tex].
For these [tex]\( x \)[/tex]-values, we can find the corresponding [tex]\( y \)[/tex]-values by substituting them back into one of the original equations. It doesn’t matter which one we use, since both equations will give us the same [tex]\( y \)[/tex]-values at the points of intersection.
Substituting [tex]\( x = \frac{\sqrt{15}}{5} \)[/tex] into [tex]\( y = 6x^2 + 1 \)[/tex]:
[tex]\[ y = 6\left(\frac{\sqrt{15}}{5}\right)^2 + 1 = 6\left(\frac{15}{25}\right) + 1 = 6\left(\frac{3}{5}\right) + 1 = \frac{18}{5} + 1 = \frac{18}{5} + \frac{5}{5} = \frac{23}{5} \][/tex]
Substituting [tex]\( x = -\frac{\sqrt{15}}{5} \)[/tex] into [tex]\( y = 6x^2 + 1 \)[/tex] will yield the same [tex]\( y \)[/tex]:
[tex]\[ y = 6\left(-\frac{\sqrt{15}}{5}\right)^2 + 1 = 6\left(\frac{15}{25}\right) + 1 = 6\left(\frac{3}{5}\right) + 1 = \frac{18}{5} + 1 = \frac{18}{5} + \frac{5}{5} = \frac{23}{5} \][/tex]
Therefore, the intersection points are [tex]\(\left(-\frac{\sqrt{15}}{5}, \frac{23}{5}\right)\)[/tex] and [tex]\(\left(\frac{\sqrt{15}}{5}, \frac{23}{5}\right)\)[/tex].
Hence, we observe that the two graphs intersect each other at two distinct points. This means the system of equations has two solutions, confirming the last statement:
"The graphs of the equations intersect each other at two places."
Thus, the correct choice is:
"The graphs of the equations intersect each other at two places."
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