Solve the equation $(m^2-m-2)x=m^2+4m+3$ Here's how I solve it
I think that m is the variable (am I right?).
Then
$$m^2x-mx-2x-m^2-4m-3=0$$
$$m^2(x-1)-m(x+4)-(2x+3)=0$$
$$D=x^2+8x+16+4(x-1)(2x+3)$$
 $$=x^2+8x+16+4(2x^2-2x+3x-3)$$
 $$=9x^2+12x+4$$
 $$=(3x+2)^2$$
$$m=\frac{x+4\pm (3x-2))}{2(x-1)}$$
$$m_1=\frac{4x+2}{2x-2}$$
$$m_2=\frac{-2x+6}{2x-2}$$
Is this right?
I don't know if the tag is right, so please don't be based on it.
 A: $m^2-m-2=(m-2)(m+1)$ and $m^2+4m+3=(m+1)(m+3)$
Clearly $m+1=0$ is a solution
Else $(m-2)x=m+3\implies x=\dfrac{m+3}{m-2}=1+\dfrac5{m-2}$
If $x$ is an integer, $(m-2)$ must divide $5$
A: In case $m$ is not a variable but a parameter, the following method will do.
To solve the following equation for $x$ and every possible $m$ $(x,m \in \mathbb R)$
$$ g(m)x=f(m) $$
you ultimately need to divide both sides by $g(m)$
$$ x = \frac{f(m)}{g(m)}$$
But first, you need to check when $g(m) = 0$. If $m$ is such that $g(m) = 0 = f(m)$ then you get
$$0 *x = 0$$
which is true for every $x$. If $m$ is such that $g(m) = 0 \neq f(m)$ then you get$$0 * x = f(m) \neq 0$$ which is false for every x.
Now that you sure what you get when $g(m)= 0$ you can divide by it. Typically, after the division you can simplify the fraction.

In your case, the answer may look like that:
for $m = -1$, $x \in \mathbb R$
for $m = 1$, the equation has no roots
for any other $m$, $x=\dfrac{m+3}{m-2}$
A: $$(m^2-m-2)x=m^2+4m+3$$
$$(m-2)(m+1)x=(m+1)(m+3)$$
$$m=-1\,\,\,\,\,\text{or}\,\,\,\,\,(m-2)x=(m+3)$$
A: Assuming you want to solve for $m$, the equation can immediately be rewritten as:
$(m-2)(m+1)x = (m+3)(m+1)$
Now it should be obvious that $m=-1$ is a solution as both LHS and RHS become zero (and equal) for this value of $m$.
Now consider the case $m \neq -1$. It is now permissible to divide both sides by $m+1$, simplifying the algebra.
$$(m-2)x = m+3 \implies mx - 2x = m+3 \implies m(x-1) = 2x + 3 \implies m = \frac{2x+3}{x-1}$$
and it should be clear that $x \neq 1$ for this solution to be valid.
Hence your solution set is:
$m=-1$ or $\displaystyle m = \frac{2x+3}{x-1}, (x \neq 1)$
