In Exercises 23-36, solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination: 23. \( x + 2y = 7 \) \( 2x + y = 8 \) 25. \( -x + 2y = 1.5 \) \( 2x - 4y = 3 \)

Let's solve both systems of linear equations, starting with exercise 23.

Exercise 23:

\( x + 2y = 7 \) (1) \( 2x + y = 8 \) (2)

Although both Gaussian elimination with back-substitution and Gauss-Jordan elimination can be used, for these examples I'll use Gaussian elimination with back-substitution.

Step 1: Write down the augmented matrix of the system.

``` | 1 2 | 7 | | 2 1 | 8 | ```

Step 2: Transform the matrix into an upper triangular form. We can multiply the first equation by 2 and subtract it from the second equation to eliminate x from the second equation.

2*(1) - (2) yields:

``` | 1 2 | 7 | | 0 -3 | -6 | ```

Step 3: Solve for y using the second equation.

\( -3y = -6 \) Divide both sides by -3 to isolate y.

\( y = 2 \)

Step 4: Back-substitute y into one of the original equations to solve for x. Using equation (1):

\( x + 2(2) = 7 \)

\( x + 4 = 7 \)

Subtract 4 from both sides.

\( x = 7 - 4 \)

\( x = 3 \)

So, the solution for exercise 23 is \( x = 3 \), \( y = 2 \).

Now, let's solve exercise 25:

Exercise 25:

\( -x + 2y = 1.5 \) (1) \( 2x - 4y = 3 \) (2)

Step 1: Write down the augmented matrix.

``` | -1 2 | 1.5 | | 2 -4 | 3 | ```

Step 2: Make the coefficients of x opposite by multiplying equation (1) by 2.

``` | -2 4 | 3 | ```

Now when we add this to equation (2) we will eliminate x.

``` | -2 4 | 3 | | 2 -4 | 3 | ------------- | 0 0 | 6 | ```

This yields a contradiction since 0 cannot equal 6. This suggests that there is no solution for the system, meaning the lines are parallel and do not intersect.

The solution for exercise 25 is that there is no solution as the system of equations is inconsistent.

Extra:

In mathematics, systems of equations are a set of equations with multiple variables that you are trying to find solutions for that satisfy all equations simultaneously. Gaussian elimination and Gauss-Jordan elimination are two popular methods for solving systems of linear equations.

Gaussian elimination transforms the system into an upper triangular matrix, then uses back-substitution to find the solutions. Gauss-Jordan elimination further simplifies the matrix to reduce it to row-echelon form, from which the solutions can be read directly.

These methods are very useful as they can be applied to any size of the system and can also reveal if a system has no solution or an infinite number of solutions. When practicing these methods, it's essential to perform operations carefully, as small mistakes can significantly affect the final answer.