Answer:
(a) No value of [tex]m[/tex] will enable this system of equations to have an infinite number of solutions.
(b) As long as [tex]m \ne 4[/tex] and [tex]m \ne (-3)[/tex], this system of equations will have one unique solution.
Step-by-step explanation:
Given a system of two linear equations with two variables:
[tex]\begin{aligned} & a_{1}\, x + b_{1}\, y = c_{1} \\ & a_{2}\, x + b_{2}\, y = c_{2}\end{aligned}[/tex].
The determinant of this system of equations would be:
[tex]\displaystyle a_{1}\, b_{2} - a_{2}\, b_{1}[/tex].
The number of possible solutions depend on the value of the determinant and whether [tex]c_{1} = c_{2}[/tex]:
- If the determinant is non-zero: one unique solution.
- If the determinant is zero and [tex]c_{1} \ne c_{2}[/tex]: no solutions.
- If the determinant is zero and [tex]c_{1} = c_{2}[/tex]: infinitely many solutions.
In this question, the determinant of this system of equations is:
[tex](2)\, (6) - (m - 1)\, m[/tex].
Simplify to obtain:
[tex]-m^{2} + m + 12[/tex].
[tex]- (m - 4)\, (m + 3)[/tex].
As long as [tex]m \ne 4[/tex] and [tex]m \ne (-3)[/tex], this determinant will be non-zero, and the system will have one unique solution.
When either [tex]m = 4[/tex] or [tex]m = (-3)[/tex], the determinant will be zero. However, since [tex]m + 17 \ne -7[/tex] for both values, the system will have no solutions in these two cases. Hence, there would be no value of [tex]m[/tex] that enable to system to have infinitely many solutions.
