TASK 1 If: $$ \begin{array}{l} x^{2}-3 x y+2 y^{2}=0 \ \therefore(x-2 y)(x-y)=0 \ \therefore x=2 y \text { of } x=y \ \therefore \frac{x}{y}=\frac{2}{1} \text { of } \frac{x}{y}=\frac{1}{1} \end{array} $$ A: For cach equation: $$ \begin{array}{l} x^{2}-2 x y+y^{2}=0 \ 5 x^{2}-13 x y+8 y^{2}=0 \ 8 x^{2}-21 x y+13 y^{2}=0 \end{array} $$ 1. Solve for $$ x $$ in terms of $$ y $$. 2. Find the value of the ration $$ \frac{x}{y} $$.

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To solve the equations given in Task 1, we will follow the steps outlined in the problem. We will solve each equation for $$ x $$ in terms of $$ y $$ and then find the ratio $$ \frac{x}{y} $$.

### 1. Solve for $$ x $$ in terms of $$ y $$

#### Equation 1: $$ x^{2} - 2xy + y^{2} = 0 $$

This is a quadratic equation in $$ x $$. We can use the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

where $$ a = 1 $$, $$ b = -2y $$, and $$ c = y^2 $$.

Calculating the discriminant:

$$
b^2 - 4ac = (-2y)^2 - 4(1)(y^2) = 4y^2 - 4y^2 = 0
$$

Since the discriminant is zero, there is one solution:

$$
x = \frac{2y}{2} = y
$$

#### Equation 2: $$ 5x^{2} - 13xy + 8y^{2} = 0 $$

Using the quadratic formula again with $$ a = 5 $$, $$ b = -13y $$, and $$ c = 8y^2 $$:

Calculating the discriminant:

$$
b^2 - 4ac = (-13y)^2 - 4(5)(8y^2) = 169y^2 - 160y^2 = 9y^2
$$

Now, applying the quadratic formula:

$$
x = \frac{13y \pm \sqrt{9y^2}}{2 \cdot 5} = \frac{13y \pm 3y}{10}
$$

This gives us two solutions:

1. $$ x = \frac{16y}{10} = \frac{8y}{5} $$
2. $$ x = \frac{10y}{10} = y $$

#### Equation 3: $$ 8x^{2} - 21xy + 13y^{2} = 0 $$

Using the quadratic formula with $$ a = 8 $$, $$ b = -21y $$, and $$ c = 13y^2 $$:

Calculating the discriminant:

$$
b^2 - 4ac = (-21y)^2 - 4(8)(13y^2) = 441y^2 - 416y^2 = 25y^2
$$

Now, applying the quadratic formula:

$$
x = \frac{21y \pm \sqrt{25y^2}}{2 \cdot 8} = \frac{21y \pm 5y}{16}
$$

This gives us two solutions:

1. $$ x = \frac{26y}{16} = \frac{13y}{8} $$
2. $$ x = \frac{16y}{16} = y $$

### 2. Find the value of the ratio $$ \frac{x}{y} $$

Now we can summarize the ratios for each equation:

1. From the first equation: 
   $$
   \frac{x}{y} = 1
   $$

2. From the second equation: 
   $$
   \frac{x}{y} = \frac{8}{5} \quad \text{and} \quad \frac{x}{y} = 1
   $$

3. From the third equation: 
   $$
   \frac{x}{y} = \frac{13}{8} \quad \text{and} \quad \frac{x}{y} = 1
   $$

### Summary of Results

- For the first equation: $$ \frac{x}{y} = 1 $$
- For the second equation: $$ \frac{x}{y} = \frac{8}{5} $$ or $$ 1 $$
- For the third equation: $$ \frac{x}{y} = \frac{13}{8} $$ or $$ 1 $$

Thus, the values of the ratio $$ \frac{x}{y} $$ for each equation are $$ 1 $$, $$ \frac{8}{5} $$, and $$ \frac{13}{8} $$.
For each equation: 1. $$ x = y $$ → $$ \frac{x}{y} = 1 $$ 2. $$ x = \frac{8}{5}y $$ or $$ x = y $$ → $$ \frac{x}{y} = \frac{8}{5} $$ or $$ 1 $$ 3. $$ x = \frac{13}{8}y $$ or $$ x = y $$ → $$ \frac{x}{y} = \frac{13}{8} $$ or $$ 1 $$ So, the ratios $$ \frac{x}{y} $$ are 1, $$ \frac{8}{5} $$, and $$ \frac{13}{8} $$.

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