Q32

5 marksLong AnswerSection D

Solve the following inequation :

$\frac{2x - 1}{12} - \frac{x - 11}{3} < \frac{3x + 1}{4}, x \in R$

Q: Solve the following inequation : $\frac{2x - 1}{12} - \frac{x - 11}{3} < \frac{3x + 1}{4}, x \in R$

Clear the fractions by multiplying throughout by the LCM and simplify. **Step 1 - Multiply by LCM 12 (positive, so inequality sign is unchanged):** - $(2x - 1) - 4(x - 11) 40/11$ **Final answer:** - Solution set: $x > 40/11$, i.e., x belongs to $(40/11, \infty)$. - ($40/11$ is approximately 3.64.)

Linear Inequalities

Solving Linear Inequalities in One Variable

Official Answer

Clear the fractions by multiplying throughout by the LCM and simplify.

Step 1 - Multiply by LCM 12 (positive, so inequality sign is unchanged):

$(2x - 1) - 4(x - 11) < 3(3x + 1)$

Step 2 - Expand:

$2x - 1 - 4x + 44 < 9x + 3$
$-2x + 43 < 9x + 3$

Step 3 - Collect terms:

$43 - 3 < 9x + 2x$
$40 < 11x$

Step 4 - Divide by 11 (positive):

$x > 40/11$

Final answer:

Solution set: $x > 40/11$ , i.e., x belongs to $(40/11, \infty)$ .
( $40/11$ is approximately 3.64.)

linear inequalityLCMinequality signsolution setinterval notationx > 40/11clearing fractionsreal numbers

Marking Scheme

11 mark: identifying LCM = 12 and multiplying every term correctly.
21 mark: correct expansion $(2x - 1) - 4(x - 11) < 3(3x + 1)$ , i.e. handling the $-4(x - 11)$ sign properly.
31 mark: simplifying to $-2x + 43 < 9x + 3$ .
41 mark: reaching $40 < 11x$ by collecting like terms.
51 mark: final solution $x > 40/11$ , i.e. x in $(40/11, \infty)$ , with correct interval notation.

Hint

Multiply every term by the LCM 12 (positive, so the sign stays the same), simplify to $40 < 11x$ , then divide by 11.

Quick Oral Answer

I multiply through by the LCM 12, expand carefully to get $-2x + 43 < 9x + 3$ , collect terms to $40 < 11x$ , and divide by 11 to get $x > 40/11$ ; since I only used positive multipliers the sign never flips.

Analysis & Explanation

Solving a linear inequality follows the same algebra as an equation, with one crucial extra rule about the direction of the inequality sign.

Concept:

To remove fractions we multiply every term by the LCM of the denominators (12, 3, 4), which is 12. Because 12 is positive, the inequality sign stays the same. The problem then reduces to a simple linear inequality in x.

The golden rule:

The inequality sign flips only when you multiply or divide both sides by a negative number. Here every multiplication/division is by a positive number, so the '<' sign never flips, and the answer is $x > 40/11$ .

Handling the -2x term:

Rather than dividing by -2 (which would flip the sign) partway through, it is safer to move the x-terms to the side that keeps their coefficient positive, giving $40 < 11x$ directly.

Exam trap:

Forgetting to distribute the minus sign over $(x - 11)$ , i.e., writing $-x - 11$ instead of $-x + 11$ after the 4( ) expansion, and mishandling the sign when isolating x.

Real-world link:

Linear inequalities describe feasible ranges everywhere in business: budget limits, minimum production targets, and the very constraints that define a linear programming region.

Common Mistakes

1Not distributing the negative sign over $(x - 11)$ , writing $-4x - 44$ instead of $-4x + 44$ .
2Flipping the inequality sign even though multiplication and division are by positive numbers only.
3Leaving the answer as an equation $x = 40/11$ or writing a closed bracket $[40/11, \infty)$ instead of the open interval.

Interesting Facts

Unlike an equation, an inequality usually has infinitely many solutions forming an interval, which is why the answer is a ray on the number line rather than a single point.

The rule 'flip the sign when multiplying by a negative' is a direct consequence of the fact that multiplying by a negative reverses order on the number line (if $a < b$ then $-a > -b$ ).

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Frequently Asked Questions

When does the inequality sign flip while solving?

The sign flips only when you multiply or divide both sides by a negative number. In this problem the LCM 12 and the divisor 11 are both positive, so the '<' sign is preserved throughout and the final answer is $x > 40/11$ .

Why is the solution an interval and not a single value?

An inequality is satisfied by a whole range of x-values, not just one. Here every real number greater than 40/11 satisfies the inequation, so the solution set is the open interval $(40/11, \infty)$ shown as a ray on the number line.