How to Solve the System of Equations: y = 2x^2 + 3 and y = 3x + 1?

To solve the system of equations given by y = 2x² + 3 and y = 3x + 1, we will set the two equations equal to each other since they both equal y.

Starting with the equations:

• y = 2x² + 3
• y = 3x + 1

Set them equal to each other:

2x² + 3 = 3x + 1

Next, we will rearrange this equation to set it to zero:

2x² – 3x + 3 – 1 = 0

Which simplifies to:

2x² – 3x + 2 = 0

This is a quadratic equation in standard form, which can be solved using the quadratic formula:

x = (-b ± √(b² – 4ac)) / 2a

Here, a = 2, b = -3, and c = 2. Plugging in these values:

• Discriminant: b² – 4ac = (-3)² – 4(2)(2) = 9 – 16 = -7

Since the discriminant is negative, it indicates that there are no real solutions for x; hence, the system of equations has no intersection points on the graph.

In conclusion, the system of equations has no solutions in the real number system.