find general formulas for the solutions of the linear systems
(
x1 + 2x2 = 1
3x1 + 4x2 = 3
and (
x1 + 2x2 = 1
3x1 + 6x2 = 3.
Now suppose a, b, c, d, p, q are real numbers with ad − bc 6= 0.
Find a general formula for the solution of the linear system (
ax1 + bx2 = p
cx1 + dx2 = q.

### Part 1: Finding general formulas for the solutions of the two given linear systems

#### System 1:
\[
\begin{aligned}
x_1 + 2x_2 &= 1 \\
3x_1 + 4x_2 &= 3
\end{aligned}
\]

We can solve this system using substitution or elimination. Let's use **elimination**:

1. Multiply the first equation by 3 to match the coefficient of \(x_1\):
   \[
   3(x_1 + 2x_2) = 3 \quad \Rightarrow \quad 3x_1 + 6x_2 = 3
   \]
   Now the system is:
   \[
   \begin{aligned}
   3x_1 + 6x_2 &= 3 \\
   3x_1 + 4x_2 &= 3
   \end{aligned}
   \]

2. Subtract the second equation from the first:
   \[
   (3x_1 + 6x_2) - (3x_1 + 4x_2) = 3 - 3
   \]
   Simplifying gives:
   \[
   2x_2 = 0 \quad \Rightarrow \quad x_2 = 0
   \]

3. Substitute \(x_2 = 0\) into the first equation:
   \[
   x_1 + 2(0) = 1 \quad \Rightarrow \quad x_1 = 1
   \]

Thus, the solution to the first system is:
\[
x_1 = 1, \quad x_2 = 0
\]

#### System 2:
\[
\begin{aligned}
x_1 + 2x_2 &= 1 \\
3x_1 + 6x_2 &= 3
\end{aligned}
\]

Multiply the first equation by 3:
\[
3(x_1 + 2x_2) = 3 \quad \Rightarrow \quad 3x_1 + 6x_2 = 3
\]
This is exactly the same as the second equation, so both equations are identical, meaning there are **infinitely many solutions**.

To express the solutions, let’s solve for \(x_1\) in terms of \(x_2\). From the first equation:
\[
x_1 = 1 - 2x_2
\]
Thus, the general solution for the second system is:
\[
x_1 = 1 - 2x_2, \quad x_2 \text{ is free (any real number)}
\]

### Part 2: General Formula for a System with \(a, b, c, d, p, q\)

Consider the system:
\[
\begin{aligned}
ax_1 + bx_2 &= p \\
cx_1 + dx_2 &= q
\end{aligned}
\]
where \(ad - bc \neq 0\) (this condition ensures the system has a unique solution).

We can solve this system using **Cramer's Rule**. First, the determinant of the coefficient matrix is:
\[
\Delta = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc
\]
Since \(\Delta \neq 0\), the system has a unique solution given by:
\[
x_1 = \frac{\begin{vmatrix} p & b \\ q & d \end{vmatrix}}{\Delta} = \frac{pd - bq}{ad - bc}
\]
\[
x_2 = \frac{\begin{vmatrix} a & p \\ c & q \end{vmatrix}}{\Delta} = \frac{aq - cp}{ad - bc}
\]

Thus, the general solution for the system is:
\[
x_1 = \frac{pd - bq}{ad - bc}, \quad x_2 = \frac{aq - cp}{ad - bc}
\]

### Summary:
- For the first system: \(x_1 = 1\), \(x_2 = 0\).
- For the second system: \(x_1 = 1 - 2x_2\), where \(x_2\) is a free variable.
- For the general system with \(ad - bc \neq 0\): 
  \[
  x_1 = \frac{pd - bq}{ad - bc}, \quad x_2 = \frac{aq - cp}{ad - bc}
  \]

Would you like more details on any part of the process? Here are some related questions:

1. How does Cramer's rule work for larger systems of equations?
2. What happens when the determinant \(ad - bc = 0\) in a system of equations?
3. How can matrix inversion be used to solve systems of equations?
4. What are the geometric interpretations of solutions to linear systems?
5. How do dependent and independent systems differ?

**Tip:** Inconsistent systems occur when equations represent parallel lines with no intersection points, leading to no solution.

Find general formulas for the solutions of the linear systems: (x1 + 2x2 = 1, 3x1 + 4x2 = 3) and (x1 + 2x2 = 1, 3x1 + 6x2 = 3). Now suppose a, b, c, d, p, q are real numbers with ad − bc ≠ 0. Find a general formula for the solution of the linear system (ax1 + bx2 = p, cx1 + dx2 = q).

Learn how to solve linear systems using Cramer's rule and determinants. This example covers two systems of equations, one with a unique solution and another with infinite solutions, alongside a general solution for any system where ad - bc ≠ 0. Suitable for advanced high school or undergraduate math students.

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