A22 - Linear inequalities and number lines

A22 - Linear inequalities and number lines

An inequality describes a range of possible values. You will solve linear inequalities, explain negative division, show solutions on number lines and choose permitted whole-number answers in a context.

Ranges satisfying a linear inequality

A minibus costs £40 plus £8 per passenger. With £100 to spend, can you afford eight passengers? An inequality will give the whole range of affordable counts, rather than just one trial.

You already know

To solve a linear equation, perform the same operation on both sides. Addition, subtraction and division by a positive number work in the same way for an inequality.

An inequality sign compares values. The difference between strict and inclusive signs matters at the boundary.

Strict and inclusive inequalities

x<4x<4 means xx is strictly less than 4, so 4 is excluded. x≤4x\leq4 includes 4. Similarly, >> excludes equality and ≥\geq includes it. A solution set contains every value making the comparison true.

Unless a domain is stated, do not assume the unknown is an integer. For example, x<4x<4 allows 3.9, zero and negative numbers.

Worked example 1: a strict comparison

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

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

3x+2<143x<12subtract 2 on both sidesx<4divide by positive 3\begin{aligned} 3x+2&<14\\ 3x&<12 &&\text{subtract 2 on both sides}\\ x&<4 &&\text{divide by positive 3} \end{aligned}

Why keep the direction? Both operations preserve numerical order.

Answer. x<4x<4.

Check. x=3x=3 gives 11<1411<14, true. The boundary x=4x=4 gives 14<1414<14, false.

Change only << to ≤\leq. The algebra is unchanged, but the endpoint is now included.

Worked example 2: including the boundary

Question. Solve 3x+2≤143x+2\leq14.

Plan. Use the same two inverse operations.

3x+2≤143x≤12subtract 2x≤4divide by positive 3\begin{aligned} 3x+2&\leq14\\ 3x&\leq12 &&\text{subtract 2}\\ x&\leq4 &&\text{divide by positive 3} \end{aligned}

Why retain the equality bar? The original comparison allows equality.

Answer. x≤4x\leq4.

Check. At x=4x=4, 14≤1414\leq14 is true. At x=5x=5, 17≤1417\leq14 is false.

Variables on both sides use the same collection method. For 4x+1≥2x+94x+1\geq2x+9, subtracting 2x2x and 1 gives 2x≥82x\geq8, hence x≥4x\geq4.

Brackets and fractions also follow the equation methods. Expanding 2(x−1)≤x+32(x-1)\leq x+3 gives 2x−2≤x+32x-2\leq x+3, then x≤5x\leq5. Multiplying (x+1)/3>2(x+1)/3>2 by positive 3 gives x+1>6x+1>6, hence x>5x>5.

Watch out

Writing x=4x=4 instead of x≤4x\leq4 loses the other solutions. A boundary value is not usually the entire solution set.

Change only the right-hand number from Worked example 2.

Negative multiplication reverses order

Multiplying or dividing both sides by a negative number reverses their order.

Start with 2<52<5. Multiplying by −1-1 gives −2>−5-2>-5: the points have reflected across zero. The greater original number becomes the smaller negative number. Every negative multiplier includes this reversal.

For example, −3x<12-3x<12 gives x>−4x>-4 after dividing by −3-3. A permitted input x=0x=0 gives 0<120<12. A wrong-side input x=−5x=-5 gives 15<1215<12, false.

Worked example 3: dividing by a negative

Question. Solve 7−2x≤17-2x\leq1.

Plan. Isolate the negative multiple of xx, then reverse the comparison when dividing.

7−2x≤1−2x≤−6subtract 7x≥3divide by −2 and reverse\begin{aligned} 7-2x&\leq1\\ -2x&\leq-6 &&\text{subtract 7}\\ x&\geq3 &&\text{divide by }-2\text{ and reverse} \end{aligned}

Why reverse only at the last step? Subtracting 7 preserves order; division by −2-2 reverses it.

Answer. x≥3x\geq3.

Check. x=3x=3 gives 1≤11\leq1; x=4x=4 gives −1≤1-1\leq1. Both are allowed.

Change only the comparison to a strict one. The endpoint will be excluded.

Worked example 4: a strict reversed comparison

Question. Solve 7−2x<17-2x<1.

Plan. Subtract 7, then divide by the negative coefficient.

−2x<−6subtract 7x>3divide by −2 and reverse\begin{aligned} -2x&<-6 &&\text{subtract 7}\\ x&>3 &&\text{divide by }-2\text{ and reverse} \end{aligned}

Why is the answer greater than 3? Larger xx makes 7−2x7-2x smaller.

Answer. x>3x>3.

Check. x=3x=3 gives 1<11<1, false; x=4x=4 gives −1<1-1<1, true.

You can avoid negative division by moving the variable instead. From 7−2x≤17-2x\leq1, add 2x2x, subtract 1 and divide by 2:

7≤1+2x⟹6≤2x⟹3≤x.7\leq1+2x\quad\Longrightarrow\quad6\leq2x\quad\Longrightarrow\quad3\leq x.

This is the same range as x≥3x\geq3.

Watch out

A negative sign somewhere in the question does not trigger reversal. From x−5<8x-5<8, add 5 to get x<13x<13. Reverse only when multiplying or dividing both sides by a negative number.

Do not divide by an unknown unless its sign is known. In xy>3xxy>3x, the direction after dividing by xx would depend on whether xx is positive or negative; division is impossible if x=0x=0.

Now change only the right-hand number from Worked example 4.

Open and closed number-line endpoints

A number line shows the allowed values and whether each endpoint belongs to the solution.

[DIAGRAM: asset_name: Strict and inclusive number-line solutions; asset_slug: 1ma1_m32_number_lines; caption: Endpoint style records inclusion, while the shaded direction records the allowed values.; description: Three horizontal number lines show x > 2 with an open circle and a right arrow, x ≥ 2 with a filled circle and a right arrow, and −1 < x ≤ 3 with an open circle at −1, filled circle at 3 and a highlighted segment between them.]
Endpoint style records inclusion, while the shaded direction records the allowed values.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account