system

To find the values of k and m that make the system have infinitely many solutions, we need to find conditions under which the two equations are equivalent. In other words, we want to find values of k and m such that the second equation can be obtained from the first equation by multiplying both sides by a constant and adding a constant.

Let's subtract the first equation from the second equation:

(6a + 2b) - (3a + 2b) = k + 3a + mb - 2

Simplifying, we get:

3a = k + 3a + mb - 2

Rearranging, we get:

2a - mb = k - 2

Now we can use the fact that the system has infinitely many solutions to find values of k and m. Since the system has infinitely many solutions, the equation 2a - mb = k - 2 must be equivalent to one of the given equations. In other words, there must be values of k and m such that:

2a - mb = 2 - (3a + 2b)

Multiplying both sides by -1, we get:

3a + 2b - 2a + mb = -2

Simplifying, we get:

a + 2b + mb = -2

Now we have two equations:

2a - mb = k - 2
a + 2b + mb = -2

We can solve for k and m by eliminating one of the variables, say a or b, from the equations. Let's eliminate a by multiplying the second equation by 2 and subtracting it from the first equation:

2(2a - mb = k - 2) - (a + 2b + mb = -2)

Simplifying, we get:

3a - 5b = 2k - 2

Now we want this equation to be true for all values of a and b, which means the coefficients of a and b on the left-hand side must be equal to 0:

3a - 5b = 0

Equating the coefficients with the previous equation, we get:

2k - 2 = 0

Solving for k, we get:

k = 1

Now substituting k = 1 into the equation 2a - mb = k - 2, we get:

2a - mb = -1

Rearranging, we get:

mb - 2a = 1

We can solve for m by setting a = b = 1:

m - 2 = 1

Solving for m, we get:

m = 3

Therefore, the values of k and m that make the system have infinitely many solutions are k = 1 and m = 3.