We need to solve for $x$ in the following four equations, where $m$ is a real parameter: a) $m(mx + 1) = 2(2x - 1)$ b) $m^3x - m^2 - 4 = 4m(x - 1)$ c) $m^2x + 4 = m(x + 4)$ d) $10 + 3(x - 2) = mx + 3$

1. Problem Description

We need to solve for $x$ in the following four equations, where $m$ is a real parameter:

a) $m(mx + 1) = 2(2x - 1)$

b) $m^3x - m^2 - 4 = 4m(x - 1)$

c) $m^2x + 4 = m(x + 4)$

d) $10 + 3(x - 2) = mx + 3$

2. Solution Steps

a) $m(mx + 1) = 2(2x - 1)$

$m^2x + m = 4x - 2$

$m^2x - 4x = -2 - m$

$x(m^2 - 4) = -(m + 2)$

$x(m - 2)(m + 2) = -(m + 2)$

If $m \neq 2$ and $m \neq -2$, then $x = \frac{-(m + 2)}{(m - 2)(m + 2)} = \frac{-1}{m - 2} = \frac{1}{2 - m}$.

If $m = -2$, then $x(4 - 4) = -(-2 + 2)$, so $0 = 0$, which means $x$ can be any real number.

If $m = 2$, then $x(4 - 4) = -(2 + 2)$, so $0 = -4$, which is impossible, so there are no solutions.

b) $m^3x - m^2 - 4 = 4m(x - 1)$

$m^3x - m^2 - 4 = 4mx - 4m$

$m^3x - 4mx = m^2 - 4m + 4$

$x(m^3 - 4m) = (m - 2)^2$

$xm(m^2 - 4) = (m - 2)^2$

$xm(m - 2)(m + 2) = (m - 2)^2$

If $m \neq 0$, $m \neq 2$, and $m \neq -2$, then $x = \frac{(m - 2)^2}{m(m - 2)(m + 2)} = \frac{m - 2}{m(m + 2)}$.

If $m = 0$, then $x(0) = (0 - 2)^2$, so $0 = 4$, which is impossible, so there are no solutions.

If $m = 2$, then $x(2)(0)(4) = 0$, so $0 = 0$, which means $x$ can be any real number.

If $m = -2$, then $x(-2)(-4)(0) = (-2 - 2)^2$, so $0 = 16$, which is impossible, so there are no solutions.

c) $m^2x + 4 = m(x + 4)$

$m^2x + 4 = mx + 4m$

$m^2x - mx = 4m - 4$

$x(m^2 - m) = 4(m - 1)$

$xm(m - 1) = 4(m - 1)$

If $m \neq 0$ and $m \neq 1$, then $x = \frac{4(m - 1)}{m(m - 1)} = \frac{4}{m}$.

If $m = 0$, then $x(0) = 4(-1)$, so $0 = -4$, which is impossible, so there are no solutions.

If $m = 1$, then $x(1)(0) = 4(0)$, so $0 = 0$, which means $x$ can be any real number.

d) $10 + 3(x - 2) = mx + 3$

$10 + 3x - 6 = mx + 3$

$4 + 3x = mx + 3$

$3x - mx = 3 - 4$

$x(3 - m) = -1$

If $m \neq 3$, then $x = \frac{-1}{3 - m} = \frac{1}{m - 3}$.

If $m = 3$, then $x(0) = -1$, so $0 = -1$, which is impossible, so there are no solutions.

3. Final Answer

a) If $m \neq 2$ and $m \neq -2$, then $x = \frac{1}{2 - m}$. If $m = -2$, then $x$ can be any real number. If $m = 2$, then there are no solutions.

b) If $m \neq 0$, $m \neq 2$, and $m \neq -2$, then $x = \frac{m - 2}{m(m + 2)}$. If $m = 0$ or $m = -2$, then there are no solutions. If $m = 2$, then $x$ can be any real number.

c) If $m \neq 0$ and $m \neq 1$, then $x = \frac{4}{m}$. If $m = 0$, then there are no solutions. If $m = 1$, then $x$ can be any real number.

d) If $m \neq 3$, then $x = \frac{1}{m - 3}$. If $m = 3$, then there are no solutions.