To find the values of \( k \) and \( m \) for which the system of equations has infinitely many solutions, we first need to simplify the given equations.
The system is:
- \( a + 2b = -3 \)
- \( 4a + 2b = k - a - mb \)
We can rearrange the second equation as follows:
\[ 4a + 2b + a + mb = k \]
Combine \( a \) terms and \( b \) terms:
\[ (4a + a) + (2b + mb) = k \quad \Rightarrow \quad 5a + (2 + m)b = k \]
So, we now have the equivalent system:
- \( a + 2b = -3 \) (Equation 1)
- \( 5a + (2 + m)b = k \) (Equation 2)
For the system to have infinitely many solutions, the second equation must be a multiple of the first equation. This implies that the coefficients of corresponding terms must be proportional.
We can express the first equation in standardized form to check the conditions. Let's use Equation 1:
\[ a + 2b = -3 \]
If we multiply each term in this equation by some constant factor \( t \):
\[ ta + 2tb = -3t \]
We desire this to equate to Equation 2:
\[ 5a + (2 + m)b = k \]
Setting the coefficients proportional, we have:
- \( t = 5 \) (from \( ta = 5a \))
- \( 2t = 2 + m \) (from \( 2tb = (2+m)b \))
First, substituting \( t = 5 \) in
\[ 2t = 2 + m \]
gives:
\[ 2(5) = 2 + m \quad \Rightarrow \quad 10 = 2 + m \quad \Rightarrow \quad m = 8 \]
Next, substituting \( t = 5 \) into \( -3t = k \):
\[ -3(5) = k \quad \Rightarrow \quad k = -15 \]
Thus, the values of \( k \) and \( m \) are:
\[ \boxed{-15} ; \text{and} ; \boxed{8} \]