Solving Linear Equations: Practice

Subtract $3x$ from both sides.

$5 = -2$

The reduced form is $0 \cdot x = -7$, which is false for every $x$, so the equation is a contradiction with solution set $\varnothing$.

Multiplying by $x - 1$ is the move that can add a solution, because $x - 1$ is zero at $x = 1$.

$x = 1 \;\Rightarrow\; x - 1 = 0$

At that value the cleared equation is automatically true, so $x = 1$ may appear as an extraneous root and must be checked against the domain.

Do not divide by $x$, since it can be zero. Move everything to one side and factor.

$x^2 - 9x = 0 \;\Rightarrow\; x(x - 9) = 0$

The product is zero when $x = 0$ or $x = 9$, so the solution set is $\{0, 9\}$. Dividing by $x$ lost the solution $x = 0$.

Subtract $2x$ from both sides, then add $7$, then divide by $2$.

$2x - 7 = 5, \qquad 2x = 12, \qquad x = 6$

The equation is conditional, so the solution set is $\{6\}$.

Equivalent equations must have identical solution sets. Dividing $3x = 12$ by the nonzero constant $3$ is reversible.

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

Both have solution set $\{4\}$. The other pairs differ: $x^2 = x$ has $\{0, 1\}$, the rational equation has $\varnothing$, and $x^2 = 25$ has $\{-5, 5\}$.

A reduced statement with the variable gone and a true numerical result means the original held for every value of $x$.

$7 = 7 \text{ is always true}$

The equation is an identity, so the solution set is $\mathbb{R}$.

Distribute the left side.

$5x + 5 = 5x + 1 \;\Rightarrow\; 5 = 1$

Subtracting $5x$ leaves the false statement $5 = 1$, so the equation is a contradiction with solution set $\varnothing$.

A solution set of $\varnothing$ means no value works. Subtracting $x$ from $x + 3 = x - 3$ gives a false statement.

$x + 3 = x - 3 \;\Rightarrow\; 3 = -3$

That is impossible, so the solution set is $\varnothing$. The first has $\{3\}$, the third has $\{0\}$, and the fourth is an identity.

A solution set of $\mathbb{R}$ means every value works. Distributing $4(x + 2)$ gives exactly the right side.

$4(x + 2) = 4x + 8$

The two sides are identical, so it is an identity. The first has $\{1\}$, the second has $\{0\}$, and the fourth is a contradiction.

Distribute the left side.

$6x - 2 = 6x - 2$

The two sides are identical, so every real number is a solution. The equation is an identity.

A safe addition uses an expression defined for every value of $x$, so it can be subtracted straight back.

$2x - 5 \text{ is defined for all real } x$

The other three are undefined where a denominator is zero (at $x = 0$, $x = 1$, and $x = -3$), so adding them is not safe everywhere.

The denominator $x - 2$ is zero at $x = 2$, so that value is excluded before solving.

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

Multiplying by $x - 2$ introduced $x = 2$ as an extraneous root. Discarding it leaves nothing, so the solution set is $\varnothing$.