In questions two equations numbered I and II are given you have to solve both equations and give answer
`I. 3x^2-13x-10=0 II. 3y^2+10y-8=0`

Options

A. if `xgty`
B. if`xgey`
C. if`xlty`
D. if relation between x and y can not be established

Answer

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve Equation I
The first equation is:
\[ 3x^2 - 13x - 10 = 0 \]

To solve this quadratic equation, we can use the factorization method. We need to find two numbers that multiply to \(3 \times (-10) = -30\) and add to \(-13\).

The numbers that satisfy this are \(-15\) and \(2\). Thus, we can rewrite the equation as:
\[ 3x^2 - 15x + 2x - 10 = 0 \]

Now, we can group the terms:
\[ (3x^2 - 15x) + (2x - 10) = 0 \]

Factoring out common terms:
\[ 3x(x - 5) + 2(x - 5) = 0 \]

Now, factor out \((x - 5)\):
\[ (3x + 2)(x - 5) = 0 \]

Setting each factor to zero gives us:
1. \(3x + 2 = 0 \Rightarrow x = -\frac{2}{3}\)
2. \(x - 5 = 0 \Rightarrow x = 5\)

### Step 2: Solve Equation II
The second equation is:
\[ 3y^2 + 10y - 8 = 0 \]

Again, we will use the factorization method. We need to find two numbers that multiply to \(3 \times (-8) = -24\) and add to \(10\).

The numbers that satisfy this are \(12\) and \(-2\). Thus, we can rewrite the equation as:
\[ 3y^2 + 12y - 2y - 8 = 0 \]

Now, we can group the terms:
\[ (3y^2 + 12y) + (-2y - 8) = 0 \]

Factoring out common terms:
\[ 3y(y + 4) - 2(y + 4) = 0 \]

Now, factor out \((y + 4)\):
\[ (3y - 2)(y + 4) = 0 \]

Setting each factor to zero gives us:
1. \(3y - 2 = 0 \Rightarrow y = \frac{2}{3}\)
2. \(y + 4 = 0 \Rightarrow y = -4\)

### Summary of Solutions
From the equations, we have:
- For \(x\): \(x = -\frac{2}{3}\) or \(x = 5\)
- For \(y\): \(y = \frac{2}{3}\) or \(y = -4\)

### Step 3: Establish the Relationship Between x and y
Now, we need to compare the values of \(x\) and \(y\):
1. If \(x = -\frac{2}{3}\) and \(y = \frac{2}{3}\), then \(x < y\).
2. If \(x = 5\) and \(y = \frac{2}{3}\), then \(x > y\).
3. If \(x = -\frac{2}{3}\) and \(y = -4\), then \(x > y\).
4. If \(x = 5\) and \(y = -4\), then \(x > y\).

Since we have different relationships depending on the values of \(x\) and \(y\), we conclude that the relationship between \(x\) and \(y\) cannot be established.

### Final Answer
The relationship between \(x\) and \(y\) cannot be established.

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