5.1 Solving Linear Equations and Inequalities
Key Takeaways
- Multiplying or dividing an inequality by a negative number reverses the inequality symbol, the single rule that separates inequality solving from equation solving.
- An equation that simplifies to a true statement such as 8 = 8 has infinitely many solutions; one that simplifies to a false statement such as 5 = 9 has no solution.
- Clearing fractions by multiplying every term by the least common denominator converts a messy equation into an integer equation in one step.
- Compound inequalities joined by 'and' produce an intersection graphed as a bounded segment; those joined by 'or' produce a union graphed as two rays.
- On a number line, a strict inequality uses an open circle and a non-strict inequality uses a closed circle.
5.1 Solving Linear Equations and Inequalities
Competency 2 skill 4 covers linear and absolute value equations or inequalities with one or two variables, with solutions represented algebraically or graphically. This section handles the linear half; section 5.2 handles absolute value.
Solving linear equations
Every legal move keeps the equation balanced: add, subtract, multiply, or divide both sides by the same quantity (never zero for multiplication or division).
The efficient order is: distribute → combine like terms on each side → collect variables on one side → isolate.
Solve 5(2x − 3) + 4 = 3(x + 8) − 6 Distribute: 10x − 15 + 4 = 3x + 24 − 6 Combine: 10x − 11 = 3x + 18 Collect: 7x = 29 Isolate: x = 29/7
Non-integer answers are common and correct; a fraction is not evidence of an error.
Clearing fractions first is almost always worth it. For (2x)/3 − 1/4 = (x + 5)/6, the LCD of 3, 4, and 6 is 12. Multiply every term: 8x − 3 = 2(x + 5), so 8x − 3 = 2x + 10, giving 6x = 13 and x = 13/6. Every term must be multiplied, including those without fractions.
Special solution cases
These appear on the exam as "how many solutions" items.
| Final form | Meaning | Solution set |
|---|---|---|
| x = 7 | One value works | Exactly one solution |
| 8 = 8 (variables vanish, true) | Always true | Infinitely many solutions |
| 5 = 9 (variables vanish, false) | Never true | No solution |
3(x + 4) = 3x + 12 → 3x + 12 = 3x + 12 → 0 = 0. Infinitely many solutions — an identity. 4x + 7 = 4x − 2 → 7 = −2. No solution — parallel lines with different intercepts.
Structurally: identical coefficients and identical constants give infinitely many solutions; identical coefficients with different constants give none.
Solving linear inequalities
Every equation rule carries over except one.
[!IMPORTANT] Multiplying or dividing both sides by a negative number reverses the inequality symbol. Adding or subtracting a negative does not reverse it.
Solve −4x + 9 > 25 Subtract 9: −4x > 16 (no reversal — subtraction) Divide by −4: x < −4 (reversal — division by a negative)
Verify with a test value: x = −10 gives −4(−10) + 9 = 49 > 25 ✓, while x = 0 gives 9 > 25 ✗. Testing one value from your claimed solution set catches sign-flip errors instantly.
An alternative that avoids the rule entirely is to move the variable to whichever side keeps it positive: from −4x + 9 > 25, add 4x and subtract 25 to get −16 > 4x, then divide by positive 4 to get −4 > x, which reads x < −4.
Graphing on a number line
+---------------------------------------------------------------------------+
| x > 3 open circle at 3, shade right o------------> |
| x >= 3 closed circle at 3, shade right *------------> |
| x < 3 open circle at 3, shade left <------------o |
| x <= 3 closed circle at 3, shade left <------------* |
+---------------------------------------------------------------------------+
Open circle means the endpoint is excluded (strict < or >); closed circle means included (≤ or ≥).
Compound inequalities
"And" (conjunction) requires both conditions simultaneously — an intersection.
−5 ≤ 2x + 1 < 9 Subtract 1 from all three parts: −6 ≤ 2x < 8 Divide all three by 2: −3 ≤ x < 4
The graph is a segment, closed at −3 and open at 4. Whatever you do, do it to all three parts.
"Or" (disjunction) requires at least one condition — a union.
x < −2 or x ≥ 5 graphs as two rays pointing away from each other, with a gap between −2 and 5.
A conjunction whose parts do not overlap, such as x > 6 and x < 1, has no solution. A disjunction that covers everything, such as x < 4 or x > 1, has all real numbers as its solution.
Two-variable linear inequalities
Graph the boundary line, then shade a half-plane.
- Graph the line from the corresponding equation. Use a dashed line for < or >, a solid line for ≤ or ≥.
- Test a point not on the line — (0, 0) whenever the line misses the origin.
- Shade the side containing the point if the test succeeds; shade the other side if it fails.
Graph y < −2x + 6. The boundary y = −2x + 6 is dashed. Testing (0, 0): 0 < 6 is true, so shade the side containing the origin — below and left of the line.
When the inequality is written with the variable term on the right or with a negative coefficient, do not shortcut by "shading below for <." Always test a point; the shortcut fails for forms like −3x + y > 2 unless you first solve for y.
Modeling with inequalities
Translate the constraint words precisely: "at least" is ≥, "at most" is ≤, "more than" is >, "fewer than" is <, "no more than" is ≤, and "no less than" is ≥.
A club has $450 for shirts costing $12 each plus a flat $75 setup fee. How many shirts can it buy? 12s + 75 ≤ 450 → 12s ≤ 375 → s ≤ 31.25. Since shirts are whole objects, the answer is 31 shirts.
That final step — interpreting a fractional bound in a discrete context — is where items separate mechanical solvers from modelers. Rounding 31.25 up to 32 exceeds the budget.
Solve for x: 7 − 3(x + 2) ≥ 4x + 15
How many solutions does the equation 2(6x − 9) = 3(4x − 6) have?
A field trip costs a $220 bus fee plus $9 per student for admission. The budget is $700. Which inequality models the situation, and what is the maximum number of students?