System of nonlinear equations


Determine the solution of


x²+4x-6y-8=0  (1)


4y-3x-2y=0

We are tasked with solving the following system of nonlinear equations:

1. \( x^2 + 4x - 6y - 8 = 0 \)  (Equation 1)
2. \( 4y - 3x - 2y = 0 \)  (Equation 2)

Let's simplify and solve these equations step by step.

### Step 1: Simplify Equation 2

We simplify the second equation \( 4y - 3x - 2y = 0 \):

\[
(4y - 2y) - 3x = 0
\]

\[
2y - 3x = 0
\]

This simplifies to:

\[
2y = 3x \quad \Rightarrow \quad y = \frac{3}{2}x \quad \text{(Equation 3)}
\]

### Step 2: Substitute Equation 3 into Equation 1

Now, substitute \( y = \frac{3}{2}x \) into the first equation \( x^2 + 4x - 6y - 8 = 0 \):

\[
x^2 + 4x - 6\left(\frac{3}{2}x\right) - 8 = 0
\]

Simplifying the terms:

\[
x^2 + 4x - 9x - 8 = 0
\]

\[
x^2 - 5x - 8 = 0
\]

### Step 3: Solve the quadratic equation

The quadratic equation to solve is:

\[
x^2 - 5x - 8 = 0
\]

We can use the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-8)}}{2(1)}
\]

Simplifying:

\[
x = \frac{5 \pm \sqrt{25 + 32}}{2}
\]

\[
x = \frac{5 \pm \sqrt{57}}{2}
\]

Thus, the two possible solutions for \(x\) are:

\[
x = \frac{5 + \sqrt{57}}{2} \quad \text{or} \quad x = \frac{5 - \sqrt{57}}{2}
\]

### Step 4: Find the corresponding \(y\) values

Using \( y = \frac{3}{2}x \), substitute the two values of \(x\):

1. For \( x = \frac{5 + \sqrt{57}}{2} \):
   \[
   y = \frac{3}{2} \times \frac{5 + \sqrt{57}}{2} = \frac{3(5 + \sqrt{57})}{4}
   \]

2. For \( x = \frac{5 - \sqrt{57}}{2} \):
   \[
   y = \frac{3}{2} \times \frac{5 - \sqrt{57}}{2} = \frac{3(5 - \sqrt{57})}{4}
   \]

### Final Solution:

The two solutions for the system of equations are:

1. \( \left( \frac{5 + \sqrt{57}}{2}, \frac{3(5 + \sqrt{57})}{4} \right) \)
2. \( \left( \frac{5 - \sqrt{57}}{2}, \frac{3(5 - \sqrt{57})}{4} \right) \)

Would you like a detailed breakdown of these solutions or any clarifications?

### Related Questions:
1. How do you solve nonlinear systems of equations in general?
2. What methods are available to solve quadratic equations?
3. How do you solve systems of equations graphically?
4. What is the quadratic formula and when is it used?
5. How do you check if a solution to a nonlinear system is correct?

**Tip**: When solving nonlinear systems, always check if substitution simplifies the process before using more complex methods!

Determine the solution of the system of nonlinear equations: x² + 4x - 6y - 8 = 0 and 4y - 3x - 2y = 0.

Learn how to solve a system of nonlinear equations step-by-step using substitution and the quadratic formula. This problem involves solving the equations x² + 4x - 6y - 8 = 0 and 4y - 3x - 2y = 0. Suitable for high school students in Grades 9-12.

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