To determine the value of $a$ (or $m$, $k$) for which the given systems of equations have no solution, we need to check when the lines represented by the equations are parallel but not coincident. This happens when the determinant of the coefficient matrix is zero, and the constant terms on the right-hand side don't satisfy the same proportionality.

Let's analyze each system:

a.

1x + 6y &= 3.2, \\ 5x + 1y &= 6.8 - 7m \end{aligned}$$ The coefficient matrix is: $$\begin{pmatrix} 1 & 6 \\ 5 & 1 \end{pmatrix}$$ The determinant is: $$\text{Det} = 1(1) - 6(5) = 1 - 30 = -29 \neq 0$$ Since the determinant is non-zero, this system will always have a solution, so this option is not valid for having no solution. ### b. $$\begin{aligned} 6x - 5y &= 4, \\ 4x + 7.5y &= m \end{aligned}$$ The coefficient matrix is: $$\begin{pmatrix} 6 & -5 \\ 4 & 7.5 \end{pmatrix}$$ The determinant is: $$\text{Det} = 6(7.5) - (-5)(4) = 45 + 20 = 65 \neq 0$$ Since the determinant is non-zero, this system will also always have a solution, so this option is not valid for having no solution. ### c. $$\begin{aligned} 7x + 5y &= 2k, \\ 14x + 4.7y &= m \end{aligned}$$ The coefficient matrix is: $$\begin{pmatrix} 7 & 5 \\ 14 & 4.7 \end{pmatrix}$$ The determinant is: $$\text{Det} = 7(4.7) - 14(5) = 32.9 - 70 = -37.1 \neq 0$$ Again, since the determinant is non-zero, this system will also always have a solution, so this option is not valid for having no solution. ### d. $$\begin{aligned} 3mx + 4y &= 1, \\ 4x + 6y &= -1 \end{aligned}$$ The coefficient matrix is: $$\begin{pmatrix} 3m & 4 \\ 4 & 6 \end{pmatrix}$$ The determinant is: $$\text{Det} = 3m(6) - 4(4) = 18m - 16$$ For the system to have no solution, the determinant should be zero: $$18m - 16 = 0 \Rightarrow m = \frac{16}{18} = \frac{8}{9}$$ If $$m = \frac{8}{9}$$, we need to check if the constant terms lead to a contradiction, which will happen if the equations are not consistent. Thus, $$m = \frac{8}{9}$$ is a candidate for no solution. ### e. $$\begin{aligned} 5mx + 5y &= 5, \\ 5x + 5my &= 5 \end{aligned}$$ The coefficient matrix is: $$\begin{pmatrix} 5m & 5 \\ 5 & 5m \end{pmatrix}$$ The determinant is: $$\text{Det} = 5m(5m) - 5(5) = 25m^2 - 25 = 25(m^2 - 1) = 25(m - 1)(m + 1)$$ For the system to have no solution, the determinant should be zero: $$25(m - 1)(m + 1) = 0$$ So $$m = 1$$ or $$m = -1$$. To have no solution, the right-hand side constant terms should lead to a contradiction when $$m = 1$$ or $$m = -1$$. Thus, these values of $$m$$ are candidates for no solution. ### f. The final option was unclear, as the equation details aren't fully provided. ### Conclusion: - **d.** The system will have no solution if $$m = \frac{8}{9}$$. - **e.** The system will have no solution if $$m = 1$$ or $$m = -1$$. Would you like further details on these, or do you have any questions? --- **Relative Questions:** 1. What is the condition for a system of linear equations to have no solution? 2. How do you interpret the determinant in the context of linear systems? 3. What is meant by a system of equations being inconsistent? 4. How do you check if a system of equations is parallel? 5. What is the relationship between the coefficient matrix and the solutions to a system of linear equations? **Tip:** For a system to have no solution, check if the lines are parallel but not coincident. This occurs when the coefficient matrix has a determinant of zero, but the ratio of the constants does not match the ratio of the coefficients.