Solving Linear Equations: Practice

The equation $x = 7$ is satisfied by exactly one number, namely $7$, and by no other.

$\text{solution set} = \{7\}$

The braces denote the set whose only member is $7$.

An identity holds for every value of $x$. Combining like terms on the left of $x + x$ gives exactly the right side.

$x + x = 2x \text{ for every real } x$

The others are true for a single value or for none.

A contradiction is satisfied by no value of $x$. A number can never equal one more than itself.

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

Since $0 = 1$ is false, the solution set is $\varnothing$.

The symbol $\varnothing$ names the empty set: the set containing nothing at all.

$\varnothing = \{\,\}$

It is different from $\{0\}$, which is a set that does contain one member, the number $0$.

The left side is $0$ for every value of $x$, and $0$ never equals $5$.

$0 \cdot x = 0 \neq 5$

No value works, so there are no solutions and the solution set is $\varnothing$.

A fraction is undefined where its denominator is zero, so set the denominator equal to zero.

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

So $x = 3$ is excluded before any solving begins.

Subtract $4$ from both sides, then divide by $3$.

$3x = 15, \qquad x = 5$

The equation is conditional with the single solution $5$, so the solution set is $\{5\}$.

A safe move is reversible for every $x$. Adding $6$ to both sides is undone by subtracting $6$, and both are defined for every value of $x$.

$L = R \;\Longleftrightarrow\; L + 6 = R + 6$

Multiplying or dividing by $x$ can fail where $x = 0$, and multiplying by $0$ erases all information.

The two sides are the same expression written in a different order, so the equation holds for every value of $x$.

$x + 4 = 4 + x \text{ for all } x$

It is an identity, and its solution set is all of $\mathbb{R}$.

The denominator is $x + 5$. Set it equal to zero to find the excluded value.

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

So $x = -5$ cannot be a solution, no matter what the algebra later produces.

A conditional equation is satisfied by some values and not others. Only $2x - 1 = 5$ singles out particular values.

$2x - 1 = 5 \;\Rightarrow\; x = 3$

It is true at $x = 3$ and false elsewhere. The second and fourth are identities, and the third is a contradiction.

Equivalent equations are defined by their solutions, not their appearance.

$x = 4 \quad\text{and}\quad 3x + 5 = 17 \quad\text{are equivalent}$

Both have solution set $\{4\}$, which is why solving is a chain of equivalent equations ending at one you can read off.