E2.6 - Solving linear inequalities

E2.6 - Solving linear inequalities

Solving an inequality finds a whole range of values rather than one required value. You will preserve or reverse its direction correctly, solve compound inequalities, and interpret the result when a quantity must be a whole-number count.

Balancing a linear inequality

A game charges three points per move and a two-point starting fee. Your total must be below fourteen. How many moves can you afford? An equation would locate a boundary; an inequality describes the allowed side of it.

For a number xx of moves, the total is 3x+23x+2, so the condition is 3x+2<143x+2<14. Solving means rewriting the condition directly in terms of xx.

Adding or subtracting the same number on both sides preserves order. For example, 2<52<5 gives 6<96<9 after adding four. Both positions move equally far on the number line, so their left-to-right order stays the same. Multiplying or dividing both sides by a positive number also preserves order.

Worked example 1: a strict inequality

Question. Solve 3x+2<143x+2<14.

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

3x<12subtract two from both sidesx<4divide by positive three\begin{aligned} 3x&<12 && \text{subtract two from both sides}\\ x&<4 && \text{divide by positive three} \end{aligned}

Answer. x<4x<4. On a number line, use an open circle at 44 and an arrow left.

Check. x=3x=3 gives 11<1411<14, which is true. The boundary x=4x=4 gives 14<1414<14, which is false; x=5x=5 also fails.

Change only the sign to inclusive. The balancing steps stay the same, while the boundary becomes allowed.

Worked example 2: an inclusive inequality

Question. Solve 3x+2≤143x+2\le14.

Plan. Subtract two and divide by positive three, retaining the inclusive sign.

3x≤12subtract twox≤4divide by positive three\begin{aligned} 3x&\le12 && \text{subtract two}\\ x&\le4 && \text{divide by positive three} \end{aligned}

Answer. x≤4x\le4. Use a filled circle at 44 and an arrow left.

Check. x=4x=4 now gives the true statement 14≤1414\le14. The value x=5x=5 gives 17≤1417\le14, which is false.

If the letter appears on both sides, collect it by adding or subtracting the same expression on both sides. For example, 5x<2x+95x<2x+9 becomes 3x<93x<9, then x<3x<3. Subtracting 2x2x preserves order for every particular value of xx.

Watch out

The answer to 3x+2<143x+2<14 is not x=4x=4. Four is the boundary, and it fails the strict comparison. Keep the inequality sign in the final answer.

Multiplying or dividing by a negative

Multiplication by a negative number reverses order. Start with 2<52<5. Multiplying by −1-1 gives −2>−5-2>-5, because reflecting the positions across zero swaps their left-to-right order. A further positive scale change preserves this new order. Division by a negative also reverses order because it is multiplication by a negative reciprocal.

Reversing an inequality

When multiplying or dividing both sides by the same negative number, reverse the inequality sign: << becomes >>, and ≤\le becomes ≥\ge, or vice versa. Adding or subtracting a negative number does not reverse it.

An inclusive boundary stays inclusive when the sign reverses: equal quantities remain equal after multiplication.

Worked example 3: dividing by a negative

Question. Solve 7−2x>17-2x>1.

Plan. Isolate the negative multiple of xx, then divide by −2-2 and reverse the sign.

−2x>−6subtract sevenx<3divide by negative two\begin{aligned} -2x&>-6 && \text{subtract seven}\\ x&<3 && \text{divide by negative two} \end{aligned}

Answer. x<3x<3.

Check. In the original inequality, x=2x=2 gives 3>13>1, true, while x=4x=4 gives −1>1-1>1, false. The boundary x=3x=3 gives equality and is excluded.

Replace the coefficient −2-2 by −12-\frac12. The negative coefficient still reverses order, but multiplying is now the convenient inverse operation.

Worked example 4: multiplying by a negative

Question. Solve 7−x2>17-\frac{x}{2}>1.

Plan. Isolate −x2-\frac{x}{2}, then multiply by −2-2, reversing the sign.

−x2>−6subtract sevenx<12multiply by negative two\begin{aligned} -\frac{x}{2}&>-6 && \text{subtract seven}\\ x&<12 && \text{multiply by negative two} \end{aligned}

Answer. x<12x<12.

Check. x=10x=10 gives 7−5=2>17-5=2>1, true. x=14x=14 gives 7−7=0>17-7=0>1, false. At 1212, the sides are equal, so the boundary is excluded.

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