Linear inequalities: a stepwise method

Inequalities describe solution sets

Consider x + 4 < 9. Subtracting 4 from both sides gives x < 5. This does not name one value: every number below 5 works. On a number line, place an open circle at 5 because 5 itself makes 5 + 4 < 9 false, then shade to the left.

For x + 4 ≤ 9, the algebra gives x ≤ 5. The closed circle now includes the boundary because 5 + 4 ≤ 9 is true. Words, symbols, and the graph should agree: less than points left, greater than points right, and an equality bar includes the endpoint.

Solve multi-step inequalities

Solve 3(2x − 1) ≤ 5x + 7. Distribute first: 6x − 3 ≤ 5x + 7. Subtract 5x from both sides to obtain x − 3 ≤ 7, then add 3, giving x ≤ 10. Each move applies the same addition or subtraction to both sides, so the comparison keeps its direction.

Check both the boundary and a nearby value. At x = 10, the original sides are 3(19) = 57 and 5(10) + 7 = 57, so the included endpoint works. At x = 11, the sides are 63 and 62, so 63 ≤ 62 is false. Those tests support both the endpoint and the shaded direction.

Keep the original inequality visible. A boundary test checks inclusion; a point from the shaded side checks direction.

Reverse the sign after a negative multiplication or division

Solve −4x + 6 > 18. Subtract 6 to get −4x > 12. Dividing both sides by −4 gives x < −3, and the sign reverses. If it stayed as x > −3, x = 0 would appear valid even though 6 > 18 is false.

The reversal is about order, not a memorized decoration. Since 2 < 5, multiplying both numbers by −1 places them at −2 and −5, where −2 > −5. Addition and subtraction preserve order; multiplication or division by a positive number preserves it; multiplication or division by a negative number reverses it.

Handle compound inequalities and intervals

In −2 < 3x + 4 ≤ 13, x must satisfy both comparisons. Subtract 4 throughout to get −6 < 3x ≤ 9, then divide all three parts by positive 3: −2 < x ≤ 3. The graph has an open endpoint at −2, a closed endpoint at 3, and shading between them. Interval notation is (−2, 3].

An ‘or’ inequality joins separate regions. For x ≤ −1 or x > 4, shade left from a closed point at −1 and right from an open point at 4. Do not shade the gap. Before calculating, identify whether the wording requires an intersection (‘and’) or a union (‘or’).

• Open circle for < or >
• Closed circle for ≤ or ≥
• Shade between endpoints for a bounded ‘and’ statement
• Shade separate rays for an ‘or’ statement

Clear fractions without changing the solution

Solve (2x − 3)/5 ≥ (x + 1)/2. Multiply both sides by 10, a positive common denominator: 2(2x − 3) ≥ 5(x + 1). This becomes 4x − 6 ≥ 5x + 5, then −11 ≥ x, or x ≤ −11. At x = −11 both original sides equal −5, so the boundary belongs.

A fixed positive denominator can be cleared without reversing the sign. A fixed negative multiplier requires a reversal. Never multiply an inequality by a variable expression unless you know its sign; if that expression may be positive, negative, or zero, the problem needs cases and may also have excluded values.

Model a constraint from a word problem

A study center charges a 12-unit registration fee plus 4.5 units per session, and your budget is at most 48 units. If n is the number of sessions, write 12 + 4.5n ≤ 48. Subtract 12 and divide by 4.5 to obtain n ≤ 8. Because sessions are counted, the practical solutions are whole numbers from 0 through 8.

The phrase ‘at most’ includes the limit, while ‘fewer than’ does not. Define the variable and its unit, translate the constraint, solve, then return to context. An algebraic set such as n ≤ 8 also contains negative numbers, but the real situation excludes negative session counts.

Find common mistakes with targeted checks

Frequent errors include reversing the sign after subtraction, failing to reverse it after division by a negative, distributing to only one term, and drawing an endpoint that contradicts the symbol. Another mistake is reporting only the boundary, as if x ≤ 10 meant x = 10.

Use three checks: redo the algebra with reasons beside each move, substitute the boundary when it is allowed, and test one value from each relevant region. A graph gives a visual check, but it cannot repair an inequality typed incorrectly or prove that every symbolic step was valid.

• Did the same operation affect both sides?
• Was the sign reversed only for multiplication or division by a negative?
• Does the endpoint match the strict or inclusive symbol?
• Do sample values agree with the shaded region?

Practice independently and use Eqora for one stuck step

Try three problems without a solution in view: 5x − 7 > 18; 4 − 2x ≥ 10; and −5 < 2x + 1 ≤ 9. Predict the direction first, solve, draw each number line, and test a value. The answers are x > 5, x ≤ −3, and −3 < x ≤ 4; compare only after finishing.

If a result disagrees, show Eqora the original inequality and your first uncertain line. Ask which operation changes the solution set or request a new example with the same trap. Then close the explanation and solve a fresh inequality alone. AI can misread a symbol or calculate incorrectly, so keep your own substitution and number-line check as the final evidence.

Put it into practice now

Choose one problem from your current homework or review set. Attempt it before opening Eqora, then use the app only at the point where your own reasoning stops.

• State what the problem is asking before you solve it
• Identify the first step you cannot justify
• Ask Eqora one focused follow-up about that step
• Finish with a similar problem and no solution in view

The session is complete when the method is clearer, not simply when the worksheet has one more answer.