# For which value(s) of parameter m is there a solution for this system { <mtable columna

For which value(s) of parameter m is there a solution for this system
$\left\{\begin{array}{l}mx+y=m\\ mx+2y=1\\ 2x+my=m+1\end{array}$
when does this system of equations have a solution?
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ryancameron52
Compute the values of $x$ and $y$ dependent on $m$ for the following system, then solve $2x+my=m+1$ (the last equation) to find the values of parameter $m$ for $x$ and $y$:
$\left\{\begin{array}{l}mx+y=m\\ mx+2y=1\end{array}$
So,
$\left\{\begin{array}{l}2mx+2y=2m\\ mx+2y=1\end{array}$
Subtracting two equations, will have:
$mx=2m-1$
- If $m\ne 0$, we may divide by $m$ and get $x=\left(2m-1\right)/m$ and $y=1-m$.
- If $m=0$, the system has no solution.
Putting $x$ and $y$ in the last equation $m\ne 0$, we'll have:
${m}^{3}-3m+2=0$
$\left({m}^{3}-1\right)-3m+3=0$
$\left(m-1\right)\left({m}^{2}+m+1\right)-3\left(m-1\right)=0$
$\left(m-1\right)\left({m}^{2}+m-2\right)=0$
Thus the values of parameter $m$ are $m=1$ or $m=-2$.
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ownerweneuf
Comment to expert's answer. Be careful: If $m=0$ then the equation $mx=2m-1$ gives you $0=-1$, so you have no solution. Ruling out this case you have $x=\frac{2m-1}{m}$ or $\left(2m-1\right)/m$ which is not th esame as $2m-1/m$.