The values of x and y in the simultaneous equations
`xy+3y^(2)-x+4y-7=0`
`2xy+y^(2)-2x-2y+1=0` are

Options

A. (3,4)
B. (2,-3)
C. (1,5)
D. (2,3)

Answer

To solve the simultaneous equations:

1. **Equations:**
\[ \text{(1)} \quad xy + 3y^2 - x + 4y - 7 = 0 \]
\[ \text{(2)} \quad 2xy + y^2 - 2x - 2y + 1 = 0 \]

2. **Multiply Equation (1) by 2 and subtract Equation (2):**
\[ 2(xy + 3y^2 - x + 4y - 7) - (2xy + y^2 - 2x - 2y + 1) = 0 \]
Expanding this gives:
\[ 2xy + 6y^2 - 2x + 8y - 14 - 2xy - y^2 + 2x + 2y - 1 = 0 \]
Simplifying:
\[ (2xy - 2xy) + (6y^2 - y^2) + (-2x + 2x) + (8y + 2y) - 14 - 1 = 0 \]
This results in:
\[ 5y^2 + 10y - 15 = 0 \]

3. **Divide the entire equation by 5:**
\[ y^2 + 2y - 3 = 0 \]

4. **Factor the quadratic equation:**
\[ (y + 3)(y - 1) = 0 \]
Thus, the solutions for \(y\) are:
\[ y + 3 = 0 \quad \Rightarrow \quad y = -3 \]
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]

5. **Substituting values of \(y\) back to find \(x\):**

**For \(y = -3\):**
Substitute \(y = -3\) into Equation (1):
\[ x(-3) + 3(-3)^2 - x + 4(-3) - 7 = 0 \]
This simplifies to:
\[ -3x + 27 - x - 12 - 7 = 0 \]
Combining like terms:
\[ -4x + 8 = 0 \quad \Rightarrow \quad 4x = 8 \quad \Rightarrow \quad x = 2 \]

**For \(y = 1\):**
Substitute \(y = 1\) into Equation (1):
\[ x(1) + 3(1)^2 - x + 4(1) - 7 = 0 \]
This simplifies to:
\[ x + 3 - x + 4 - 7 = 0 \]
Combining like terms:
\[ 0 = 0 \]
This means \(x\) can take any value when \(y = 1\).

6. **Final Solutions:**
The pairs \((x, y)\) are:
\[ (2, -3) \quad \text{and} \quad (x, 1) \text{ (where \(x\) can be any value)} \]