E2.6 - Solving linear inequalities (Extended)

E2.6 - Solving linear inequalities (Extended)

A club has 52availableandpays52 available and pays 10 to book a room plus $6 per person. What is the greatest number who can attend? E2.6 turns limits into linear inequalities, solves them with balanced operations and interprets the resulting range.

One inequality

An equation finds where two quantities are equal; an inequality finds which values make one quantity smaller or larger than another. Define the unknown before translating the limit. In the opening problem, if nn is the number attending, the total cost is 10+6n10+6n, and “has 52available”requires52 available” requires 10+6n\le52$.

Preserving an inequality

Adding or subtracting the same quantity on both sides preserves the inequality. Multiplying or dividing both sides by the same positive number also preserves it.

Adding the same number shifts both positions equally on the number line. Multiplication by a positive number stretches their separation without reversing which is greater. This explains why the familiar equation-solving operations work here under these conditions.

Worked example 1: solving a strict inequality

Question. Solve 3x+5<203x+5<20.

Plan. Undo the addition, then divide by the positive coefficient 3.

3x+5<203x<15subtract 5 from both sidesx<5divide both sides by positive 3\begin{aligned} 3x+5&<20\\ 3x&<15 &&\text{subtract 5 from both sides}\\ x&<5 &&\text{divide both sides by positive 3} \end{aligned}

Answer. x<5x<5.

Check. At x=4x=4, the original left side is 17, below 20. At the boundary x=5x=5, it is 20, so the boundary must be excluded.

The next problem replaces a strict limit with a budget that includes equality. Its unknown is also a whole-number count.

Worked example 2: constructing a budget inequality

Question. A room costs 10tobookand10 to book and 6 for each person. A club can spend at most $52. Find the greatest number of people it can afford.

Plan. Form an inclusive cost bound, solve it, then interpret the count.

Let nn be the non-negative integer number of people.

10+6n≤52cost must not exceed the budget6n≤42subtract the fixed chargen≤7divide by 6\begin{aligned} 10+6n&\le52 &&\text{cost must not exceed the budget}\\ 6n&\le42 &&\text{subtract the fixed charge}\\ n&\le7 &&\text{divide by 6} \end{aligned}

Answer. At most 7 people can attend.

Check. Seven people cost 10+6(7)=5210+6(7)=52 dollars, allowed. Eight cost 10+6(8)=5810+6(8)=58 dollars, too much.

Unknowns on both sides can be collected just as in equations: from 5x+1≤2x+135x+1\le2x+13, subtract 2x2x and 1 to obtain 3x≤123x\le12, hence x≤4x\le4. Keep the inequality symbol throughout: the answer is a range, not just its boundary.

Watch out

Solving the corresponding equality finds a boundary. You must still give the correct side of that boundary and decide whether equality is allowed.

Use the maximum spending condition to choose the symbol.

Negative multipliers

Multiplying by a negative number reflects order across zero. For example, 2<52<5, but after multiplying by −1-1, −2>−5-2>-5. The number that was farther right becomes farther left.

Reversing the inequality

When multiplying or dividing both sides by a negative number, reverse the inequality: <↔><\leftrightarrow> and ≤↔≥\le\leftrightarrow\ge.

Adding or subtracting a negative number still preserves direction. It is negative multiplication or division that reverses it. Avoid multiplying by an unknown-sign expression: you cannot choose the direction without knowing its sign.

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