Solutions and Solution Sets

Every technique you’ll learn in this chapter — isolating a variable, factoring, using the quadratic formula, testing intervals — exists to answer one question: which values make the equation or inequality true? This lesson defines that question precisely before any solving method is introduced. Once you know exactly what a solution is, what a solution set is, and how to write one correctly, every later lesson in this chapter is just a faster way of finding it.

None of this requires new algebra. Every check in this lesson is ordinary substitution and arithmetic — the goal is to get the vocabulary and the notation right, because a correct value written in the wrong form (or the wrong number of values reported) is graded as an incomplete answer just as often as an arithmetic mistake is.

What Does It Mean to Solve an Equation or Inequality?

To solve an equation or inequality is to find every value of the variable that makes it a true statement. Each individual value that works is called a solution (or a root, for equations). The complete collection of every solution — no more, no fewer — is called the solution set.

$$ \text{Solve } 3x - 5 = 16 \;\Longrightarrow\; \text{find every } x \text{ for which } 3x - 5 = 16 \text{ is a true statement} $$

Notice the ask is every value. If a problem has three solutions and you report two, you have not fully solved it, even if both of the two you found are correct. This is one of the most common reasons a correct-looking answer loses points.

Checking Whether a Value Is a Solution

Before finding solutions, it helps to be able to check a candidate one. Substitute the candidate value for the variable everywhere it appears, then evaluate both sides independently. If the two sides come out equal (for an equation) or the inequality holds (for an inequality), the candidate is a solution. If not, it isn’t — no matter how it was obtained.

The last row is worth pausing on: \(x=3\) makes both sides equal, and the inequality is strict (\(>\), not \(\ge\)), so equality does not satisfy it. Swapping in a non-strict symbol would change the answer — always check which symbol the problem actually uses.

What Is a Solution Set?

A solution set can contain one value, several values, infinitely many values, or none at all — the equation or inequality itself determines which. Four cases already appear above and below:

• One solution. \(3x - 5 = 16\) has exactly one value that works: \(x = 7\).
• Several solutions. \(x^{2} = 9\) has exactly two: \(x = -3\) and \(x = 3\). Squaring loses the sign, so both are genuinely needed.
• Infinitely many solutions. \(2x + 1 > 7\) is satisfied by every number greater than \(3\) — an infinite interval, not a finite list.
• No solutions. Some equations and inequalities are never true for any real number (covered directly below).

Because a solution set can be infinite, listing every element by hand isn’t always possible. That’s exactly why set-builder notation and interval notation exist — they describe a solution set completely without writing out its members one at a time.

Writing Solution Sets: Set-Builder Notation

Set-builder notation names the set by describing the rule every member must satisfy, rather than listing members:

$$ \{x \mid x > 3\} $$

Read this as “the set of all \(x\) such that \(x\) is greater than \(3\).” The vertical bar \(\mid\) is read “such that”; some textbooks use a colon instead, with the same meaning.

For finite sets, set-builder notation can also just list the members inside braces, separated by commas:

$$ \{-3, 3\} $$

is the full solution set of \(x^{2} = 9\) — braces around the finished list, no rule needed once there are only finitely many values.

Writing Solution Sets: Interval Notation

Interval notation describes an infinite (or continuous) solution set using its two endpoints instead of a rule. The bracket style at each end tells you whether that endpoint itself is included:

Two habits prevent almost every notation mistake. First, the smaller number always goes on the left, regardless of which direction the original inequality symbol pointed: \(x < 5\) becomes \((-\infty, 5)\), not \((5, -\infty)\). Second, infinity always gets a parenthesis, never a bracket — there is no number equal to infinity to “include.”

Three Kinds of Equations: Conditional, Identity, and Contradiction

Simplifying an equation all the way down sometimes eliminates the variable entirely. What happens next depends on what statement is left behind:

$$ 2(x+3) = 2x + 6 \;\Longrightarrow\; 2x+6 = 2x+6 \;\Longrightarrow\; 6 = 6 \quad \text{(always true — an identity)} $$

$$ x + 5 = x + 9 \;\Longrightarrow\; 5 = 9 \quad \text{(never true — a contradiction)} $$

Most equations you’ll solve are conditional — that’s the normal case. But whenever every variable term cancels during simplification, stop and read the remaining constant statement: if it’s true, the solution set is all real numbers; if it’s false, the solution set is empty. Neither case means “no answer” — \(\varnothing\) and \(\mathbb{R}\) are both complete, correct solution sets in their own right.

Solution Sets for Inequalities

An inequality’s solution set is almost always an interval (or a union of intervals) rather than a short list, because if one value satisfies \(x > 3\), so does every larger value — infinitely many numbers, not just one. Compound inequalities combine two conditions with “and” or “or,” and the two words change the solution set in opposite ways:

• “And” (both conditions must hold) → intersection. \(-1 < x + 2 \le 6\) really means ”\(-1 < x+2\) and \(x + 2 \le 6\)” simultaneously — solve by applying the same operation to every part at once: subtracting \(2\) gives \(-3 < x \le 4\), interval \((-3, 4]\).
• “Or” (either condition is enough) → union. ”\(x < -2\) or \(x > 5\)” has two separate pieces that never overlap, written with the union symbol: \((-\infty, -2) \cup (5, \infty)\).

The union symbol \(\cup\) is doing real work in that second case — there is no single interval that captures “less than \(-2\)” and “greater than \(5\)” at once, because the two pieces don’t touch.

Worked Example A: Checking Whether a Value Is a Solution

Determine whether \(x = -4\) is a solution of \(2(x+3) - 1 \ge 3x + 5\).

Step 1 — substitute \(-4\) for every \(x\):

$$ 2(-4+3) - 1 \; \ge \; 3(-4) + 5 $$

Step 2 — evaluate the left side and the right side independently:

$$ 2(-1) - 1 = -3 \qquad \text{and} \qquad -12 + 5 = -7 $$

Step 3 — check whether the resulting numeric statement is true:

$$ -3 \ge -7 \quad \text{(true)} $$

Since substituting \(-4\) produces a true statement, \(x = -4\) is a solution of the inequality. The two sides are evaluated completely separately before comparing — never simplify across the inequality symbol as if it were an equation.

Worked Example B: Finding and Writing the Solution Set of a Linear Equation

Find the solution set of \(5(x-2) = 3x + 4\).

Step 1 — distribute the left side:

$$ 5x - 10 = 3x + 4 $$

Step 2 — collect variable terms on one side and constants on the other:

$$ 5x - 3x = 4 + 10 \;\Longrightarrow\; 2x = 14 $$

Step 3 — isolate \(x\):

$$ x = 7 $$

The variable did not cancel, so this is a conditional equation with exactly one solution. Write the solution set with braces, even for a single value:

$$ \{7\} $$

Reporting the bare number \(7\) instead of the set \(\{7\}\) is common and usually accepted informally, but the set notation is the technically complete answer — it’s the form that generalizes correctly once an equation has more than one solution.

Worked Example C: Finding and Writing the Solution Set of a Compound Inequality

Solve \(3 \le 2x - 1 < 9\) and express the solution set in interval notation.

Step 1 — add \(1\) to all three parts:

$$ 4 \le 2x < 10 $$

Step 2 — divide all three parts by \(2\) (positive, so the inequality directions don’t flip):

$$ 2 \le x < 5 $$

Step 3 — convert to interval notation, matching each endpoint’s bracket to its symbol:

The left end uses \(\le\) (bracket, included); the right end uses \(<\) (parenthesis, excluded):

$$ [2, 5) $$

Both sides of a compound inequality get every operation at the same time — there’s no way to solve the left half and the right half as separate problems, since \(x\) has to satisfy both conditions from a single expression.

Worked Example D: Classifying an Equation

Determine whether \(4(x+1) - 2x = 2x + 4\) is conditional, an identity, or a contradiction.

Step 1 — distribute and combine like terms on the left:

$$ 4x + 4 - 2x = 2x + 4 \;\Longrightarrow\; 2x + 4 = 2x + 4 $$

Step 2 — try to isolate \(x\):

$$ 2x - 2x = 4 - 4 \;\Longrightarrow\; 0 = 0 $$

Step 3 — read the remaining statement:

The variable cancelled completely, and \(0 = 0\) is always true. This is an identity — every real number is a solution.

$$ \text{Solution set: } \mathbb{R}, \text{ or } (-\infty, \infty) $$

Had step 2 produced something false instead, like \(0 = 4\), the equation would be a contradiction with solution set \(\varnothing\) instead — the arithmetic decides which one it is, not a guess.

Common Mistakes to Avoid

• Stopping after finding one solution when there are more. \(x^{2}=9\) has two solutions, \(-3\) and \(3\); reporting only \(3\) is an incomplete solution set, not a partially-correct one.
• Writing \(x = \varnothing\) instead of \(\varnothing\) alone. The empty set is the entire answer to a no-solution problem — it isn’t assigned to \(x\) as if it were a value.
• Flipping which endpoint gets a bracket. A bracket belongs on the side with \(\le\) or \(\ge\); a parenthesis belongs on the side with \(<\) or \(>\), and on every \(\infty\) or \(-\infty\) endpoint without exception.
• Writing the larger number first in an interval. \((5,2)\) is not a valid interval; the smaller value always comes first, regardless of which way the original inequality pointed.
• Confusing “and” with “or.” An “and” compound inequality needs both conditions at once (an intersection, often one connected interval); an “or” compound inequality needs only one (a union, often two separate pieces). Swapping them changes the solution set completely.
• Treating an inequality like an equation when checking a candidate. Evaluate both sides fully, then compare with the original inequality symbol — don’t simplify the two sides together as though there were an equals sign between them.
• Ignoring whether the inequality is strict. A value that makes both sides exactly equal satisfies \(\le\) or \(\ge\) but not \(<\) or \(>\) — always check the actual symbol in the problem, not the one from a similar-looking problem.
• Declaring “no solution” too early. An equation only has an empty solution set if the variable cancels and leaves a false statement. If the variable is still present, keep solving.

Where This Shows Up Later

• Linear equations and inequalities. The isolation techniques used in Worked Examples B and C are the entire method for solving any linear equation or inequality — this lesson supplies the vocabulary those lessons assume.
• Quadratic equations. \(x^{2}=9\) already previewed the pattern: quadratics typically produce a solution set with two elements, found by factoring or the quadratic formula rather than by inspection.
• Absolute value equations and inequalities. These routinely split into two separate conditions joined by “and” or “or,” exactly the compound-inequality machinery introduced above.
• Systems of equations. A solution to a system is a value (or pair of values) that satisfies every equation in the system simultaneously — the same substitution-and-check idea from this lesson, just applied to more than one equation at once.
• Rational and radical equations. Both routinely introduce extraneous values during solving — candidates that satisfy a simplified equation but fail the original one. Checking every candidate by direct substitution, as in Worked Example A, is exactly how those get caught.

Practice Problems

Work each problem yourself before opening the answer.

Problem 1. Determine whether \(x = 5\) is a solution of \(3x - 7 = 2x - 2\).

Problem 2. Determine whether \(x = -2\) is a solution of \(4 - 3x \le 9\).

Problem 3. Find the solution set of \(4x + 9 = 2x - 3\), and write it using set notation.

Problem 4. Find the solution set of \(x^{2} = 25\).

Problem 5. Determine whether \(3(x-1) = 3x - 3\) is conditional, an identity, or a contradiction.

Problem 6. Determine whether \(2x + 7 = 2x - 1\) is conditional, an identity, or a contradiction.

Problem 7. Solve \(3x - 2 > 10\) and write the solution set in interval notation.

Problem 8. Solve \(-2x + 5 \le 13\) and write the solution set in interval notation. (Watch the direction of the inequality.)

Problem 9. Solve \(1 < 2x + 3 \le 11\) and write the solution set in interval notation.

Problem 10. Write the solution set of ”\(x < -1\) or \(x \ge 6\)” using interval notation.

Quick Reference

Once checking a candidate and writing a solution set both feel automatic, the rest of this chapter is just adding solving techniques on top of vocabulary you already have — starting with isolating a variable in a straightforward linear equation. Refresh the algebra those techniques lean on in Polynomials and Rational Expressions if it’s been a while, or browse the rest of the Algebra lessons as new ones publish.