I. ` 12x^(2)+ 11x+12 = 10x^(2) + 22x`
II. `13y^(2) - 18y+3=9y^(2) - 10y`

Options

A. if`x gt y`
B. if `x ge y`
C. if `x lt y`
D. if `x le y`

Explanation

To solve the given equations step by step, we will tackle each equation separately.

### Step 1: Solve the first equation
The first equation is:
\[ 12x^2 + 11x + 12 = 10x^2 + 22x \]

**Rearranging the equation:**
We will move all terms to one side of the equation:
\[ 12x^2 + 11x + 12 - 10x^2 - 22x = 0 \]
This simplifies to:
\[ 2x^2 - 11x + 12 = 0 \]

### Step 2: Factor the quadratic equation
Now we need to factor the quadratic equation:
\[ 2x^2 - 11x + 12 = 0 \]

**Finding factors:**
We look for two numbers that multiply to \(2 \times 12 = 24\) and add to \(-11\). The numbers \(-8\) and \(-3\) fit this requirement.
So we can rewrite the equation as:
\[ 2x^2 - 8x - 3x + 12 = 0 \]
Grouping the terms:
\[ (2x^2 - 8x) + (-3x + 12) = 0 \]
Factoring by grouping:
\[ 2x(x - 4) - 3(x - 4) = 0 \]
Factoring out the common term:
\[ (2x - 3)(x - 4) = 0 \]

### Step 3: Solve for x
Setting each factor to zero gives us:
1. \(2x - 3 = 0 \Rightarrow x = \frac{3}{2}\)
2. \(x - 4 = 0 \Rightarrow x = 4\)

So the solutions for \(x\) are:
\[ x = \frac{3}{2} \quad \text{and} \quad x = 4 \]

### Step 4: Solve the second equation
The second equation is:
\[ 13y^2 - 18y + 3 = 9y^2 - 10y \]

**Rearranging the equation:**
We will move all terms to one side:
\[ 13y^2 - 18y + 3 - 9y^2 + 10y = 0 \]
This simplifies to:
\[ 4y^2 - 8y + 3 = 0 \]

### Step 5: Factor the quadratic equation
Now we need to factor the quadratic equation:
\[ 4y^2 - 8y + 3 = 0 \]

**Finding factors:**
We look for two numbers that multiply to \(4 \times 3 = 12\) and add to \(-8\). The numbers \(-6\) and \(-2\) fit this requirement.
So we can rewrite the equation as:
\[ 4y^2 - 6y - 2y + 3 = 0 \]
Grouping the terms:
\[ (4y^2 - 6y) + (-2y + 3) = 0 \]
Factoring by grouping:
\[ 2y(2y - 3) - 1(2y - 3) = 0 \]
Factoring out the common term:
\[ (2y - 3)(2y - 1) = 0 \]

### Step 6: Solve for y
Setting each factor to zero gives us:
1. \(2y - 3 = 0 \Rightarrow y = \frac{3}{2}\)
2. \(2y - 1 = 0 \Rightarrow y = \frac{1}{2}\)

So the solutions for \(y\) are:
\[ y = \frac{3}{2} \quad \text{and} \quad y = \frac{1}{2} \]

### Summary of Solutions
The solutions are:
- For \(x\): \(x = \frac{3}{2}\) and \(x = 4\)
- For \(y\): \(y = \frac{3}{2}\) and \(y = \frac{1}{2}\)