To solve the system of equations simultaneously, we have:

\[
xy = 1 \quad \text{(Equation 1)}
\]
\[
4x - y + 3 = 0 \quad \text{(Equation 2)}
\]

**Step 1: Solve Equation 2 for \( y \)**
\[
4x - y + 3 = 0
\]
\[
y = 4x + 3
\]

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

**Step 3: Solve the quadratic equation \( 4x^2 + 3x - 1 = 0 \) using the quadratic formula**
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 4 \), \( b = 3 \), and \( c = -1 \). Substituting these values into the formula:
\[
x = \frac{-3 \pm \sqrt{3^2 - 4(4)(-1)}}{2(4)}
\]
\[
x = \frac{-3 \pm \sqrt{9 + 16}}{8}
\]
\[
x = \frac{-3 \pm \sqrt{25}}{8}
\]
\[
x = \frac{-3 \pm 5}{8}
\]

This gives us two possible values for \( x \):
\[
x_1 = \frac{-3 + 5}{8} = \frac{2}{8} = \frac{1}{4}
\]
\[
x_2 = \frac{-3 - 5}{8} = \frac{-8}{8} = -1
\]

**Step 4: Find the corresponding \( y \) values for each \( x \)**
For \( x = \frac{1}{4} \):
\[
y = 4\left(\frac{1}{4}\right) + 3 = 1 + 3 = 4
\]

For \( x = -1 \):
\[
y = 4(-1) + 3 = -4 + 3 = -1
\]

**Solution:**
The solutions to the system of equations are:
\[
(x, y) = \left(\frac{1}{4}, 4\right) \quad \text{and} \quad (x, y) = (-1, -1)
\]

Would you like more details or have any questions?

---

1. How can you check if the solutions satisfy the original equations?
2. What is the significance of solving simultaneous equations in real-world applications?
3. What other methods can be used to solve simultaneous equations besides substitution?
4. How does the discriminant in the quadratic formula determine the nature of the roots?
5. What happens if the system of equations has no solution?

**Tip:** When solving simultaneous equations, always consider both substitution and elimination methods to find the most efficient solution path.

Learn how to solve the simultaneous equations xy = 1 and 4x - y + 3 = 0 step-by-step using the substitution method and the quadratic formula. This problem involves quadratic equations and is suitable for students in Grades 10-12.

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