Example of a System:

$\text{1. Linear equation: } y = 2x + 1$ $\text{2. Quadratic equation: } y = x^2 + 3x + 2$

Steps to Solve:

1. Substitute the Linear Equation into the Quadratic:

• Since $\ y = 2x + 1$ from the linear equation, substitute this into the quadratic equation: $2x + 1 = x^2 + 3x + 2$

2. Rearrange to Form a Quadratic Equation:

• Move all terms to one side to set the equation to zero: $x^2 + 3x + 2 - 2x - 1 = 0 \quad \Rightarrow \quad x^2 + x + 1 = 0$

3. Solve the Quadratic Equation:

• Solve the quadratic equation $\ x^2 + x + 1 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Here,$\ ( a = 1 ), ( b = 1 ), and ( c = 1 )$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2}$

• Since the discriminant $\ \Delta = -3$ is negative, the solutions are complex: $x = \frac{-1 \pm \sqrt{3}i}{2}$

4. Find the Corresponding $y$ Values:

• Substitute the complex values of $\ x$ back into the linear equation to find the corresponding $\ y$ values: $y = 2\left(\frac{-1 + \sqrt{3}i}{2}\right) + 1 = -1 + \sqrt{3}i + 1 = \sqrt{3}i$
• Similarly, for $( x = \frac{-1 - \sqrt{3}i}{2} ), ( y = -\sqrt{3}i ).$

Interpretation:

• Real Solutions: If the quadratic equation produces real roots, the intersection points of the curves are real and can be plotted on a coordinate plane.
• Complex Solutions: If the quadratic equation produces complex roots, the curves do not intersect on the real plane.

Example with Real Solutions:

Given: $y = 3x + 2 \quad \text{and} \quad y = x^2 + x - 6$

Steps:

5. Substitute $\ ( y = 3x + 2 ) into ( y = x^2 + x - 6 )$: $3x + 2 = x^2 + x - 6$
6. Rearrange: $x^2 - 2x - 8 = 0$
7. Solve using factorization: $(x - 4)(x + 2) = 0 \quad \Rightarrow \quad x = 4 \text{ or } x = -2$
8. Find $\ y$:

• For $\ ( x = 4 ), ( y = 3(4) + 2 = 14 ).$
• For $\ ( x = -2 ), \ y = 3(-2) + 2 = -4$. Solution: The intersection points are $\ (4, 14) \ and \ (-2, -4)$.

Example Problem:

Solve the simultaneous equations:

Step-by-Step Solution:

Step 1: Set the equations equal to each other

Since both equations are equal to $y$, we can set the right-hand sides of both equations equal to each other.

From equations (1) and (2):

Step 2: Rearrange the equation

Now, we need to rearrange the equation to form a standard quadratic equation $ax^2 + bx + c = 0.$

Start by subtracting $2x + 4$ from both sides:

Simplify:

Now, we have the quadratic equation:

Step 3: Solve the quadratic equation

We can solve this quadratic equation by factorizing.

We need to find two numbers that multiply to give $-2$ (the constant term) and add to give $1$ (the coefficient of $x$).

These numbers are 2 and -1, so we can factor the quadratic as:

Step 4: Solve for $x$

Now, set each factor equal to zero:

Solve these:

Step 5: Substitute the $x$-values back into the linear equation

Now that we have two possible values for $x$, we need to substitute each one into the linear equation $y = 2x + 4$ to find the corresponding $y$-values.

• For $x = -2:$

• For x = 1:

Final Answer:

The solutions to the system of equations are:

So, the two points where the quadratic and linear equations intersect are:

Let me know if you'd like more practice or further explanation on any steps!