Algebra homework
Linear inequalities homework: signs and boundaries
Keep the order rule visible, check the endpoint and use targeted hints to correct your own inequality working.
Linear inequalities homework asks for a set of values, not merely the value where two expressions become equal. The most common mistake is to treat every step like an equation and forget what multiplying by a negative number does to order. Another is to stop at the boundary value without deciding whether it belongs to the solution. Both errors can survive a quick glance at an otherwise tidy calculation.
This guide works through original one-variable examples with paper, a number line and independent substitutions. It also shows a targeted Lirno workflow when your school permits AI support: capture the complete assignment, verify the scan, ask for a hint, attempt the algebra yourself and check the result. The examples concern linear expressions with known numerical coefficients. Rational, quadratic and absolute-value inequalities need additional methods; do not apply this routine to every inequality you meet.
1. Read the relation before rearranging
The symbols < and > mean strictly less than and strictly greater than. The symbols ≤ and ≥ include equality. In 2x + 3 ≤ 11, the goal is every allowed x whose substituted expression is at most eleven. The corresponding equation 2x + 3 = 11 identifies a boundary, but the equation alone cannot decide which side of that boundary satisfies the original statement.
Write down the domain as well. If x represents a real number, fractions and decimals may be included. If it represents the number of identical notebooks purchased, only nonnegative integers make sense. A mathematical result such as x ≤ 4.5 must then be interpreted within that domain; it does not authorize buying half a notebook. The algebra and the interpretation are separate parts of a complete answer.
Before calculating, copy the comparison symbol carefully and predict what increasing x does to the left-hand expression. For a positive coefficient, increasing x increases that expression; for a negative coefficient, it decreases it. OpenStax’s introduction to solving linear inequalities reviews the underlying rules. Our numerical exercises are original. Keep the assigned notation beside your working so a copied symbol does not silently change the problem.
2. Capture the whole task and inspect the scan
When using Lirno, capture the complete inequality together with the instruction, brackets, domain and any requested diagram. A cropped photograph that omits “integers only” changes the meaning of the answer. Put the sheet flat, avoid shadows and leave space around the line. If your own attempt is relevant, include it clearly or supply it separately rather than covering part of the printed question with handwritten algebra.
Read the captured expression back against the original before asking for help. Check < versus ≤, minus signs, fractions and parentheses. A small horizontal mark can turn strict inequality into an inclusive one; a missing minus can change the entire solution direction. Lirno’s homework-scanning workflow describes capture and review. A successful capture is useful input, not proof that the mathematics or recognition is correct.
Choose a narrow first request, such as asking which operation isolates the variable in 7 − 3x < 16. Do not ask for a full answer simply because a tool can produce one. Write your intended next step before reading the hint. If the scanned question differs from the paper, correct the input first. An explanation for a different inequality cannot validate your homework, even when the reasoning sounds fluent.
3. Addition and subtraction preserve the direction
Start with 7 − 3x < 16. Subtract seven from both sides to obtain −3x < 9. The comparison direction remains unchanged. Subtracting a positive number, adding a negative number or moving a term through an explicitly equal operation does not by itself reverse the relation. The important question is what operation you apply to both complete sides, not whether a minus sign appears somewhere in the line.
Now compare 2x + 3 ≤ 11. Subtract three from both sides: 2x ≤ 8. Divide by positive two: x ≤ 4. These steps preserve order. Test x = 3 in the original: 2(3) + 3 = 9, and 9 ≤ 11 is true. Test x = 5: the left side is thirteen, so it fails. The boundary x = 4 gives eleven and is included because the original symbol allows equality.
Label each operation in your margin. This makes a suspected mistake easier to locate than a chain of unexplained lines. If your homework contains variables on both sides, subtraction still works: 4x + 1 < 2x + 9 becomes 2x + 1 < 9, then 2x < 8 and x < 4. The coefficient after collecting terms determines which division is needed; the visual location of x does not.
4. A negative multiplier reverses order
Return to −3x < 9. Dividing both sides by negative three reverses the relation, producing x > −3. You can see why with an ordinary comparison: 2 < 5, but multiplying both numbers by −1 gives −2 > −5. Their positions reverse on the number line. This is an order rule, not a special instruction triggered by a negative answer.
Check the original inequality at x = 0: 7 − 3(0) = 7, and seven is less than sixteen. At x = −4, the expression becomes nineteen, which is not less than sixteen. At the boundary x = −3, it becomes sixteen; strict < excludes that value. These checks support the direction and endpoint decision. They supplement the equivalent algebraic steps rather than replacing the need to justify those steps.
If you instead divide −3x < 9 by positive three, you obtain −x < 3; the direction stays the same because the divisor was positive. A further multiplication by −1 gives x > −3. Both routes agree. Ask Lirno for an explanation of the specific reversal if you need one, using the tutor’s support levels. Then restate the rule in your own words and identify the exact operation where it applies.
5. Keep brackets and fractions under control
Consider 3(2 − x) ≥ 12. Expand accurately to 6 − 3x ≥ 12. Subtract six to get −3x ≥ 6, then divide by −3 and reverse the relation: x ≤ −2. The boundary is included. Substituting x = −2 into the original gives twelve; x = −3 gives fifteen and also works. The nearby value x = −1 gives nine and fails.
A fraction does not automatically require reversing the sign. In (x − 1)/2 > 3, multiplying both sides by positive two gives x − 1 > 6, hence x > 7. In (x − 1)/(−2) > 3, multiplying by negative two gives x − 1 < −6, hence x < −5. The denominator’s known sign controls the operation. Write parentheses around the whole numerator so you do not accidentally divide only one term.
This routine assumes a nonzero numerical divisor. Do not multiply an inequality by an expression containing x when its sign is unknown; that can require cases, and zero may make division invalid. If your task contains x in a denominator, stop and identify the appropriate method with your teacher or course material. Keeping that boundary clear is more useful than extending a familiar shortcut beyond its valid conditions.
6. Show the boundary and solution direction
For x > −3, mark negative three with an open point and indicate all values to its right. For x ≤ −2, use a filled point at negative two and indicate the left side. Open means the endpoint is excluded; filled means it is included. The direction comes from greater or less, not from whether the endpoint itself is positive or negative.
If your class requests interval notation, x > −3 is (−3, ∞) and x ≤ −2 is (−∞, −2]. Infinity is not a numerical endpoint that can be included. Some local conventions use different bracket shapes, so follow your teacher’s notation while keeping the inclusion meaning explicit. Label the boundary on the number line; an unlabelled arrow and dot leave the reader unable to tell which set you mean.
A word problem adds a final filter. Suppose a club has a budget of forty units of money, pays a fixed fee of seven and spends six per notebook. Write 7 + 6n ≤ 40, obtain n ≤ 5.5, and restrict n to nonnegative integers. The feasible counts are zero through five. Check five costs thirty-seven and six costs forty-three. The largest feasible count is five; the algebraic boundary five and a half is not itself an allowed purchase.
7. Handle identities and contradictions honestly
Not every linear inequality leaves a boundary. For 2x + 3 < 2x + 8, subtract 2x to obtain 3 < 8. That statement is always true, so every value in the stated domain satisfies the original. Over the real numbers, the solution is all real numbers. Within a smaller contextual domain, the solution is every value in that domain.
In 2x + 3 > 2x + 8, the same subtraction gives 3 > 8, which is false. No value satisfies it. Do not divide by the vanished coefficient or report x = 0 simply because the variable disappeared. The result is a statement about the whole solution set. Test two different x values if you want to understand why the same contradiction remains after substitution.
To check your own attempt with Lirno, provide both the original and your labelled steps, then request the first unsupported transformation. A suggestion still needs verification: AI can misread or reason incorrectly. Recalculate that line independently and use values on both sides of a proposed boundary. School rules apply, and Lirno does not guarantee correctness, grades, mastery or permission. A fluent explanation is not a substitute for a mathematically justified step.
8. Turn the error into focused practice
Make three fresh exercises that isolate different decisions: 5x − 4 > 11, −2x + 1 ≥ 9, and 4(x + 1) ≤ 12. Their solutions are x > 3, x ≤ −4 and x ≤ 2. Work each from the original, mark the operation that could reverse the sign and test an included value, an excluded value and the endpoint. Cover these results until your own attempt is complete.
Then change one symbol while keeping the numbers. Replacing > with ≥ in the first exercise includes three without changing which side is shaded. Changing −2 to positive two in the second exercise changes the calculation to x ≥ 4. Explain why that alters the solution direction. Such small variations reveal whether you understand the operation or merely remember the appearance of one worked line.
Lirno’s study-practice workflow can support follow-up questions after your corrected attempt. Keep the practice specific to the mistake: sign reversal, parentheses, endpoint or domain. You can also complete every exercise here on paper without an app. Finish by writing one complete conclusion that states the solution set and its domain. That sentence connects the algebra, the diagram and the actual homework question.
