`x^(2) + x-20= 0, y^(2)-y-30=0`

Options

A. if `x gt y`
B. if `x gt= y`
C. if `x lt y`
D. if `x lt= y`

Answer

To solve the equations \(x^2 + x - 20 = 0\) and \(y^2 - y - 30 = 0\), we will first find the values of \(x\) and \(y\) by factoring each quadratic equation.

### Step 1: Solve for \(x\) in the equation \(x^2 + x - 20 = 0\)

1. We need to factor the quadratic equation. We are looking for two numbers that multiply to \(-20\) (the constant term) and add up to \(1\) (the coefficient of \(x\)).
2. The numbers that satisfy this are \(5\) and \(-4\) because:
- \(5 \times (-4) = -20\)
- \(5 + (-4) = 1\)
3. We can rewrite the equation as:
\[ x^2 + 5x - 4x - 20 = 0 \]
4. Now, we can group the terms:
\[ (x^2 + 5x) + (-4x - 20) = 0 \]
5. Factor by grouping:
\[ x(x + 5) - 4(x + 5) = 0 \]
6. Factor out the common term \((x + 5)\):
\[ (x + 5)(x - 4) = 0 \]
7. Set each factor to zero:
\[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

### Step 2: Solve for \(y\) in the equation \(y^2 - y - 30 = 0\)

1. Similarly, we need to factor this quadratic equation. We are looking for two numbers that multiply to \(-30\) and add up to \(-1\).
2. The numbers that satisfy this are \(5\) and \(-6\) because:
- \(5 \times (-6) = -30\)
- \(5 + (-6) = -1\)
3. We can rewrite the equation as:
\[ y^2 - 6y + 5y - 30 = 0 \]
4. Now, we can group the terms:
\[ (y^2 - 6y) + (5y - 30) = 0 \]
5. Factor by grouping:
\[ y(y - 6) + 5(y - 6) = 0 \]
6. Factor out the common term \((y - 6)\):
\[ (y - 6)(y + 5) = 0 \]
7. Set each factor to zero:
\[ y - 6 = 0 \quad \Rightarrow \quad y = 6 \]
\[ y + 5 = 0 \quad \Rightarrow \quad y = -5 \]

### Step 3: Compare the values of \(x\) and \(y\)

Now we have the values:
- For \(x\): \(x = -5\) and \(x = 4\)
- For \(y\): \(y = 6\) and \(y = -5\)

### Step 4: Determine the relationship between \(x\) and \(y\)

1. The possible values of \(x\) are \(-5\) and \(4\).
2. The possible values of \(y\) are \(6\) and \(-5\).
3. We compare:
- If \(x = 4\), then \(4 < 6\) (so \(x < y\)).
- If \(x = -5\), then \(-5 < 6\) (so \(x < y\)).
- If \(x = -5\), then \(-5 = -5\) (so \(x = y\)).
4. Therefore, we can conclude that \(y\) is greater than or equal to \(x\).

### Final Conclusion

The correct option is that \(y\) is greater than or equal to \(x\).

---