How do we solve the system of equations: y = 2x^2 and y = 3x + 1?

Solving the System of Equations

We have a system of two equations:

1. y = 2x²
2. y = 3x + 1

To solve for x and y, we can set the two equations equal to each other because they both equal y:

2x2 = 3x + 1

Next, we will rearrange the equation to bring all terms to one side:

2x2 - 3x - 1 = 0

This is a quadratic equation in the form of ax2 + bx + c = 0, where:

• a = 2
• b = -3
• c = -1

We can use the quadratic formula to find the values of x:

x = \frac{-b ± \sqrt{b2 - 4ac}}{2a}

Substituting the values of a, b, and c into the formula:

x = \frac{-(-3) ± \sqrt{(-3)2 - 4(2)(-1)}}{2(2)}

Calculating further:

x = \frac{3 ± \sqrt{9 + 8}}{4}

x = \frac{3 ± \sqrt{17}}{4}

This gives us two possible values for x:

• x1 = \frac{3 + \sqrt{17}}{4}
• x2 = \frac{3 - \sqrt{17}}{4}

Now, we will substitute these values of x back into either original equation to find corresponding y values. Let’s use y = 3x + 1:

For x1:

y1 = 3(\frac{3 + \sqrt{17}}{4}) + 1

y1 = \frac{9 + 3\sqrt{17}}{4} + 1 = \frac{13 + 3\sqrt{17}}{4}

For x2:

y2 = 3(\frac{3 - \sqrt{17}}{4}) + 1

y2 = \frac{9 - 3\sqrt{17}}{4} + 1 = \frac{13 - 3\sqrt{17}}{4}