Solving Linear Equations: Additional Practice (Set 2)

An identity requires the two sides to agree for every $x$, so the coefficients of $x$ must be equal.

$k = 5$

With $k = 5$ the equation is $5x + 3 = 5x + 3$, true for all $x$. Any other $k$ gives a single solution.

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

$2x + 1 = 2(x + 3) \;\Rightarrow\; 2x + 1 = 2x + 6 \;\Rightarrow\; 1 = 6$

The variable cancels and leaves a false statement, so there is no solution and the solution set is $\varnothing$.

Exclude $x = 1$. The fractions on the left share a denominator, so combine them, or multiply through by $x - 1$.

$3 - 1 = x \;\Rightarrow\; x = 2$

The candidate $x = 2$ is allowed and checks in the original, so the solution set is $\{2\}$.

The excluded value is $x = 1$, and the candidate $x = 2$ is not excluded. Check it in the original.

$\frac{2^2 - 1}{2 - 1} = \frac{3}{1} = 3$

Both sides equal $3$, so $x = 2$ is valid. The cancellation assumed $x \neq 1$, which does not affect this answer.

Equivalent to $x = 2$ means the solution set is exactly $\{2\}$. Squaring introduces a second root.

$x^2 = 4 \;\Rightarrow\; x = 2 \text{ or } x = -2$

So $x^2 = 4$ has solution set $\{-2, 2\}$ and is not equivalent. The other three each have solution set $\{2\}$.

Exclude $x = 3$. Multiplying by $x - 3$ gives $x + c = 2$, so the candidate is $x = 2 - c$. It fails only when it equals the excluded value.

$2 - c = 3 \;\Rightarrow\; c = -1$

Then the sole candidate is extraneous and the solution set is $\varnothing$.

Infinitely many solutions means an identity, so the constant terms must match once the $2x$ cancels.

$k = 7$

Then the equation is $2x + 7 = 2x + 7$, true for all $x$. For any other $k$ the equation is a contradiction with no solution.

Solution set $\mathbb{R}$ is the identity case, where the equation reads $0 = 0$.

$a = 0 \text{ and } b = 0 \;\Rightarrow\; 0 \cdot x = 0$

This is true for every $x$. If $a \neq 0$ there is one solution, and if $a = 0$ with $b \neq 0$ there is none.

Multiplying is reversible only where you can divide back, which fails wherever the factor is zero.

$x - 5 = 0 \;\Rightarrow\; x = 5$

At $x = 5$ the step cannot be undone, so that value can appear as an extraneous root. The nonzero constant $7$ is never zero, so multiplying by it is reversible everywhere.

The denominators are constants, so multiply both sides by the least common denominator $12$.

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

No values are excluded, so the solution set is $\{12\}$.

The original product is zero at $x = 0$ and $x = 3$. Dividing by $x$ assumes $x \neq 0$.

$x(x - 3) = 0 \;\Rightarrow\; x = 0 \text{ or } x = 3$

The step discards $x = 0$, leaving only $x = 3$, so the solution $x = 0$ is lost.

For every $x$ except $5$, the fraction $\frac{x - 5}{x - 5}$ equals $1$, so the equation holds.

$x \neq 5 \;\Rightarrow\; \frac{x - 5}{x - 5} = 1$

Only $x = 5$ is excluded, so infinitely many values (every real number except $5$) are solutions.