Solving Linear Equations: Practice

Collect the variable on one side: $(k - 3)x = 5$. This has no solution exactly when the coefficient is zero while the right side is not.

$k - 3 = 0 \;\Rightarrow\; k = 3$

Then the equation reads $0 \cdot x = 5$, which is impossible, so the solution set is $\varnothing$.

Exclude $x = 3$ and $x = -3$. Multiply through by $(x - 3)(x + 3)$.

$(x + 3) + (x - 3) = 6 \;\Rightarrow\; 2x = 6 \;\Rightarrow\; x = 3$

The only candidate is the excluded value $x = 3$, so it is extraneous. The solution set is $\varnothing$.

Exclude $x = 4$, then multiply both sides by $x - 4$.

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

The single candidate is the excluded value, so it is extraneous and no solution remains.

Losing the solution $x = 0$ points to a step that is illegal at $x = 0$. Dividing by $x$ assumes $x \neq 0$.

$\text{divide by } x \;\Rightarrow\; \text{loses } x = 0$

The other moves are reversible for every $x$ and preserve the whole solution set.

Exclude $x = 3$, then multiply both sides by $x - 3$.

$2x = 2(x - 3) + 6 \;\Rightarrow\; 2x = 2x$

The cleared equation is true for every $x$, but the original excludes $x = 3$. So the solution set is every real number except $3$.

For $x \neq 2$, multiplying both sides by $x - 2$ gives $k = 3$, a condition on $k$ alone.

$k = 3$

If $k = 3$ the equation holds for every allowed $x$ (infinitely many solutions); if $k \neq 3$ it holds for none. So it has a solution only when $k = 3$.

The new equation is zero when either factor is zero.

$(x - 1)(x + 2) = 0 \;\Rightarrow\; x = 1 \text{ or } x = -2$

Multiplying by $x + 2$ added the root $x = -2$, which does not satisfy the original $x - 1 = 0$. So $x = -2$ is extraneous, while $x = 1$ is the genuine solution.

The only excluded value is $x = 0$, where the denominator vanishes, and the candidate $x = 3$ is not excluded.

$\frac{1}{3} = \frac{1}{3}$

Checking confirms it, so $x = 3$ is a valid solution.

Exclude $x = -1$, then multiply every term by $x + 1$.

$x + (x + 1) = 1 \;\Rightarrow\; 2x + 1 = 1 \;\Rightarrow\; x = 0$

The candidate $x = 0$ is allowed and checks in the original, so the solution set is $\{0\}$.

Equivalence is defined solely by solutions.

$\text{equivalent} \;\Longleftrightarrow\; \text{same solution set}$

The form, the number of terms, and the particular steps between them are irrelevant; only the solution sets must match.

Exclude $x = 1$. Multiplying by $x - 1$ gives $x + a = 5$, so the candidate is $x = 5 - a$. It fails only when it equals the excluded value.

$5 - a = 1 \;\Rightarrow\; a = 4$

Then the sole candidate $x = 1$ is extraneous and the solution set is $\varnothing$. For any other $a$ there is one valid solution.

Exactly one solution is the case $ax = b$ with $a \neq 0$.

$3x = 12 \;\Rightarrow\; x = 4$

The form $0 \cdot x = 0$ gives all of $\mathbb{R}$, and $0 \cdot x = 4$ gives $\varnothing$.