There are two approaches to this problem, one using rearranging methods and another using matrix methods
Using the first method, rearrangingFor infinite solutions, we need the two equations to give the same line, so that means
same gradients and
same y interceps.
For no solutions we need the lines to be parallel but not to have the same equation, that is
same gradients but
different y intercepts.
For a unique solution, we need the lines to cross once, so that is we need to have
different gradients.
a) So lets rearrange the lines into
form.
y & =6<br />\\ \therefore y & =\frac{3}{m-2}x-\frac{6}{m-2}<br />\end{alignedat}<br /> )
So lets now compare the gradient and y intercepts.
For infinite solutions we have
Same gradients
 & =15<br />\\ m^{2}-2m-15 & =0<br />\\ \left(m-5\right)\left(m+3\right) & =0<br />\\ m & =-3,\:5<br />\end{alignedat})
Same y intercepts
The values of
that satisfy both conditions is
For no solutions we want the same gradient and different y intercepts, we already know that
for the same gradients, and for different y interecepts we will have
, so that leaes us with
for no solutions.
b) Now for the restrictions, we know that we need to have different gradients, so that will be (using the same solving from above but using a
sign)
. Now we need to solve the equations.
 & =2-\frac{6}{m-2}<br />\\ x\left(\frac{m^{2}-2m-15}{5\left(m-2\right)}\right) & =\frac{2m-4-6}{m-2}<br />\\ x & =\frac{10\left(m-5\right)}{m^{2}-2m-15}<br />\\ x & =\frac{10\left(m-5\right)}{\left(m-5\right)\left(m+3\right)}<br />\\ x & =\frac{10}{m+3},\: m\neq5<br />\\ y & =-2+\frac{m}{5}\left(\frac{10}{m+3}\right)<br />\\ & =-2+\frac{10\left(m+3-3\right)}{5\left(m+3\right)}<br />\\ & =-2+\frac{10\left(m+3\right)}{5\left(m+3\right)}-\frac{30}{5\left(m+3\right)}<br />\\ & =\frac{-6}{m+3}<br />\\ \therefore x=\frac{10}{m+3}, & \: y=\frac{-6}{m+3}<br />\end{alignedat})
Using the second method, matriciesThe second method makes use of matricies. So lets put the equations into matrix form.
If
=0)
we have
infinite solutions or
no soultions.
If
 \neq 0)
we have
a unique solutionNow if the determinate of the first matrix is 0, then we will either have infinite solutions or not solutions ot the system of equations (we can check which we get but substituting in the values we get and checking if we get the same equation).
 & =0<br />\\ m\left(2-m\right)-\left(-5\right)\left(3\right) & =0<br />\\ 2m-m^{2}+15 & =0<br />\\ m^{2}-2m-15 & =0<br />\\ \left(m-5\right)\left(m+3\right) & =0<br />\\ m & =-3,\:5<br />\end{alignedat})
Now subbing in, if we get the same equation then we get infinite solutions, if we get different equations we have no solutions.
y & =6<br />\\ 3x-5y & =6....[2]<br />\end{alignedat})
i.e. different equations, so that is we have no solutions for
y & =6<br />\\ 3x-3y & =6<br />\\ x-y & =2....[2]<br />\end{alignedat}<br /> )
That is we get the same equation, so we have inifinite solutions for
b) Now for there to be a solution, the determinate of the matrix has to
not be zero, that is solving the above with a
sign will give
Then solve like in the previous method.
Hope that helps
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