Solving Linear Equations: Additional Practice (Set 1)

More than one solution means an identity. Add $4$ to both sides to get $mx = 6x$, so the coefficients must match.

$m = 6$

With $m = 6$ the equation is $6x - 4 = 6x - 4$, true for all $x$. Any other $m$ gives the single solution $x = 0$.

The expression $\frac{x - 6}{x - 6}$ equals $1$ for every $x$ where it is defined, but it is undefined at $x = 6$.

$x \neq 6 \;\Rightarrow\; \frac{x - 6}{x - 6} = 1$

So the equation holds for every real number except the excluded value $6$.

Both sides are the identical expression, defined for every $x$ except where the denominator is zero.

$x - 2 = 0 \;\Rightarrow\; x = 2 \text{ is excluded}$

So the equation holds for all real numbers except $2$.

Notice $2x - 2 = 2(x - 1)$, so the right side simplifies.

$\frac{10}{2x - 2} = \frac{10}{2(x - 1)} = \frac{5}{x - 1}$

Both sides are the same expression, defined for all $x \neq 1$. The solution set is every real number except $1$.

Exclude $x = 4$. Multiplying by $x - 4$ gives $x + 1 = 5$, so $x = 4$.

$x = 4 \text{ is the excluded value}$

The only candidate is extraneous, so nothing survives and the solution set is $\varnothing$.

A contradiction reduces to a false numerical statement. Subtract $4x$ from both sides of the third.

$4x + 5 = 4x - 5 \;\Rightarrow\; 5 = -5$

That is impossible, so it is a contradiction. The first is conditional and the other two are identities.

Exclude $x = 0$. The term $\frac{2}{x}$ appears on both sides, so subtract it from each (a reversible move where $x \neq 0$).

$3 = x$

The candidate $x = 3$ is allowed and checks in the original, so the solution set is $\{3\}$.

The left side $0 \cdot x$ is $0$ for every $x$, so it can never equal a nonzero number.

$0 \cdot x = 7 \;\Rightarrow\; 0 = 7 \text{ (impossible)}$

So $b = 7$ gives no solution. When $b = 0$ the equation is $0 = 0$, an identity with every real number as a solution.

Exclude $x = -1$ and $x = 1$, then multiply both sides by $(x + 1)(x - 1)$.

$4(x - 1) = 2(x + 1) \;\Rightarrow\; 4x - 4 = 2x + 2 \;\Rightarrow\; x = 3$

The candidate $x = 3$ is allowed and checks in the original, so the solution set is $\{3\}$.

Multiplying by $x - 2$ is the move whose factor can be zero (at $x = 2$), so it can add an extraneous root.

$x - 2 = 0 \;\Rightarrow\; x = 2$

The other moves use constants and are reversible for every $x$, so they cannot change the solution set.

The denominators are constants, so multiply both sides by $6$ (equivalently, cross multiply).

$2(x + 2) = 3(x - 4) \;\Rightarrow\; 2x + 4 = 3x - 12 \;\Rightarrow\; x = 16$

Nothing is excluded, so the solution set is $\{16\}$.

Multiplying by an expression adds candidates exactly at the values where that expression is zero, because the step is not reversible there.

$x + 4 = 0 \;\Rightarrow\; x = -4$

At $x = -4$ the multiplier is zero, so $x = -4$ can appear as an extraneous root and must be checked.