Given that:

\[
\frac{x}{a} + \frac{y}{b} = 0
\]
and the second equation:

\[
ax + by = a^2 + b^2
\]

We will solve this system of equations step by step.

### Step 1: Solve the first equation for \(x\) or \(y\)

Start with:
\[
\frac{x}{a} + \frac{y}{b} = 0
\]
Multiply through by \( ab \) to eliminate the denominators:
\[
b x + a y = 0
\]
So, we have:
\[
bx = -ay \quad \text{(Equation 1)}
\]

### Step 2: Solve for \(x\) in terms of \(y\)
From \( bx = -ay \), solve for \(x\):
\[
x = -\frac{a}{b} y \quad \text{(Equation 2)}
\]

### Step 3: Substitute into the second equation

Now substitute \(x = -\frac{a}{b} y\) into the second equation \( ax + by = a^2 + b^2 \):

\[
a\left(-\frac{a}{b} y\right) + by = a^2 + b^2
\]

Simplify the left-hand side:
\[
-\frac{a^2}{b} y + by = a^2 + b^2
\]

Multiply through by \( b \) to eliminate the denominator:
\[
-a^2 y + b^2 y = b(a^2 + b^2)
\]

### Step 4: Solve for \(y\)

Factor out \( y \) on the left-hand side:
\[
(b^2 - a^2) y = b(a^2 + b^2)
\]

Now solve for \( y \):
\[
y = \frac{b(a^2 + b^2)}{b^2 - a^2}
\]

### Step 5: Solve for \(x\)

Now that we have \( y \), substitute it into Equation 2 \( x = -\frac{a}{b} y \):

\[
x = -\frac{a}{b} \cdot \frac{b(a^2 + b^2)}{b^2 - a^2}
\]

Simplify:
\[
x = -\frac{a(a^2 + b^2)}{b^2 - a^2}
\]

### Final Answers:
\[
x = -\frac{a(a^2 + b^2)}{b^2 - a^2}, \quad y = \frac{b(a^2 + b^2)}{b^2 - a^2}
\]

Let me know if you'd like more explanation on any of the steps!

---

**Relative Questions:**
1. Can these equations be solved using matrices or Cramer's Rule?
2. What happens when \( a = b \)?
3. How does the solution change if the right-hand side of the second equation is different?
4. Can you apply this method to non-linear systems?
5. How do the values of \( a \) and \( b \) affect the solution behavior?

**Tip:** Always check for special cases like \( a = b \) to simplify calculations!

Explore how to solve linear equations involving fractions, illustrated with detailed steps and formulas. This problem involves manipulating equations and solving systems of equations effectively. Suitable for Grades 9-12 students and anyone interested in advanced algebraic problem-solving.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation