for what values of k does the linear system below have:

a) infinite solutions?
b) one solution?
c) no solution?

2/3x+y=16
kx+3y=48

To have an infinite number of solutions, both equations are really the same equation

if we multiply the first by 3,
2x + 3y = 48
comparing with the 2nd:
kx + 3y = 48
there would be an infinite number of solutions if k = 2

One solution, k ≠ 2

to have no solution, the two lines would have to be parallel, that is, they must have the same slope
so (2/3) / 1 = k/3
k = 2, but we found that if k = 2, they are one and the same line, with an infinite number of solutions.
Thus no value of k exists for which there would be no solution

a) For the linear system to have infinitely many solutions, the two equations must represent the same line. In other words, when the ratio of the coefficients of x and y in both equations is equal.

Comparing the coefficients, we have (2/3) / k = 1 / 3. Solving this equation for k gives us k = (2/3) / (1/3) = 2. Therefore, the linear system has infinitely many solutions when k = 2.

b) For the linear system to have one solution, the two equations must represent two distinct lines that intersect at a single point. In other words, when the ratio of the coefficients of x and y in both equations is not equal.

Comparing the coefficients, we have (2/3) / k ≠ 1 / 3. Therefore, the linear system has one solution for all values of k except k = 2.

c) For the linear system to have no solution, the two equations must represent two parallel lines that do not intersect.

Comparing the coefficients, we have (2/3) / k = 1 / 3. Since k can only equal 2 for infinite solutions, there is no value of k for which the linear system has no solution.

To determine the values of k for which the linear system has infinite solutions, one solution, or no solutions, we can use the concept of determinant.

First, let's write the given system of equations in matrix form:

```
[ 2/3 1 ] [ x ] [ 16 ]
[ k 3 ] * [ y ] = [ 48 ]
```

We can calculate the determinant of the coefficient matrix [A] = [2/3 1; k 3], denoted as |A|, to determine the solution behavior.

a) Infinite Solutions: If |A| = 0 and the determinant of the augmented matrix [A|b] ≠ 0, the system has an infinite number of solutions.

b) One Solution: If |A| ≠ 0, the system has exactly one solution.

c) No Solution: If |A| = 0 and the determinant of the augmented matrix [A|b] = 0, the system has no solution.

To proceed, we need to calculate the determinant |A| and the determinant of the augmented matrix |A|_aug.
Let's calculate them step by step:

Step 1: Calculate the determinant |A|:
|A| = (2/3 * 3) - (1 * k)
|A| = 2 - k/3

Step 2: Calculate the determinant |A|_aug:
|A|_aug = |A| * (48) - (2/3 * 16 * 3)
|A|_aug = (2 - k/3)(48) - (2/3 * 16 * 3)
|A|_aug = 96 - 16k/3 - 32
|A|_aug = 64 - 16k/3

Now we can analyze the cases:

a) Infinite Solutions: |A| = 0 and |A|_aug ≠ 0
For infinite solutions, we need the determinant |A| to be equal to zero while the determinant |A|_aug is not zero.
Thus, setting |A| = 0, we can solve the equation:
2 - k/3 = 0
k/3 = 2
k = 6

b) One Solution: |A| ≠ 0
If the determinant |A| is not zero, there will be one unique solution.
So, for one solution, we need |A| ≠ 0. Therefore, any value of k except k = 6 will result in one solution.

c) No Solution: |A| = 0 and |A|_aug = 0
For no solution, both the determinant |A| and the determinant |A|_aug need to be zero.
Setting |A| = 0 and |A|_aug = 0, we can solve the equations:
2 - k/3 = 0
64 - 16k/3 = 0
Solving these equations, we find k = 6/5.

In summary:

a) Infinite solutions: k = 6
b) One solution: any value of k except k = 6
c) No solution: k = 6/5

To determine the values of k for which the given linear system has different types of solutions, we need to use the concept of determinants and the method of solving linear systems.

The given linear system is:
1) (2/3)x + y = 16
2) kx + 3y = 48

a) Infinite Solutions:
A linear system has infinite solutions when the two equations are dependent, meaning that one equation can be expressed in terms of the other.

To check for infinite solutions, we need to compare the slope of the two equations. In slope-intercept form, equation 1 can be written as:
y = -(2/3)x + 16

By comparing the slopes, we see that the slope of equation 2 is also -(2/3). Therefore, for the system to have infinite solutions, the slopes must be equal.

The slope of equation 2 is given by: (coefficient of x) / (coefficient of y) = k/3.
So, for infinite solutions, k/3 must equal -(2/3).

Hence, k = -2.

b) One Solution:
A linear system has one solution when the two equations are independent, meaning that the slopes are different and the lines intersect at a single point.

To check for one solution, we need to calculate the determinant of the coefficient matrix (a matrix containing the coefficients of x and y).

The determinant of the coefficient matrix is given by:
(det) = | 2/3 1 |
| k 3 |

To have one solution, the determinant must be non-zero, meaning that the coefficient matrix is invertible.

Using the formula for a 2x2 determinant, we can calculate:
(det) = (2/3 * 3) - (1 * k) = 2 - k/3

For one solution, (det) ≠ 0. So we have:
2 - k/3 ≠ 0

Solving this equation, we have:
2 ≠ k/3
6 ≠ k

Hence, k ≠ 6.

c) No Solution:
A linear system has no solution when the two equations are independent, meaning that the slopes are different and the lines are parallel.

If k is not equal to -2 (from part a) or 6 (from part b), then the slopes of the two lines will not be equal. Therefore, the lines will never intersect, leading to no solution.

In summary:
a) For infinite solutions, k = -2.
b) For one solution, k ≠ 6.
c) For no solution, k can be any value except -2 or 6.