$$ \begin{array}{l} \text { (10) } x^{2}-4 x y-x+4 y=0 \ x+2 y-3=0 \end{array} $$

Question

UpStudy AI · Accepted Answer

To solve the system of equations:

$$
\begin{cases}
x^{2} - 4xy - x + 4y = 0 \quad \text{(Equation 1)} \
x + 2y - 3 = 0 \quad \text{(Equation 2)}
\end{cases}
$$

**Step 1: Solve Equation 2 for $$ x $$**
$$
x + 2y - 3 = 0 \
x = 3 - 2y
$$

**Step 2: Substitute $$ x = 3 - 2y $$ into Equation 1**
$$
(3 - 2y)^2 - 4(3 - 2y)y - (3 - 2y) + 4y = 0
$$

**Expand and Simplify:**
$$
(9 - 12y + 4y^2) - (12y - 8y^2) - 3 + 2y + 4y = 0 \
9 - 12y + 4y^2 - 12y + 8y^2 - 3 + 2y + 4y = 0 \
$$

Combine like terms:
$$
(4y^2 + 8y^2) + (-12y - 12y + 2y + 4y) + (9 - 3) = 0 \
12y^2 - 18y + 6 = 0
$$

**Step 3: Solve the Quadratic Equation for $$ y $$**
$$
12y^2 - 18y + 6 = 0 \
\text{Divide by 6:} \quad 2y^2 - 3y + 1 = 0
$$

Use the quadratic formula:
$$
y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}
$$

Thus, the solutions for $$ y $$ are:
$$
y = 1 \quad \text{or} \quad y = \frac{1}{2}
$$

**Step 4: Find Corresponding $$ x $$ Values**
Using $$ x = 3 - 2y $$:

1. If $$ y = 1 $$:
$$
x = 3 - 2(1) = 1
$$

2. If $$ y = \frac{1}{2} $$:
$$
x = 3 - 2\left(\frac{1}{2}\right) = 3 - 1 = 2
$$

**Final Solutions:**
$$
(x, y) = (1, 1) \quad \text{and} \quad (2, \frac{1}{2})
$$
The solutions are $$ (1, 1) $$ and $$ (2, \frac{1}{2}) $$.

\[ \begin{array}{l} \text { (10) } x^{2}-4 x y-x+4 y=0 \\ x+2 y-3=0 \end{array} \]