3D - Solving linear inequalities

3D - Solving linear inequalities

An inequality can have infinitely many solutions, so finding one successful value is only a start. You will rearrange linear inequalities, describe their full solution sets and combine conditions while keeping track of which endpoints are allowed.

Equivalent inequalities

For which values of xx is 7−2x7-2x at least as large as x−5x-5? Values such as 00 and 33 work, but 55 does not. We need a rule describing every value that works.

Solution set

The solution set of an inequality contains all values that make it true. Equivalent inequalities have the same solution set.

The signs << and >> are strict: equality is excluded. The signs ≤\le and ≥\ge include equality.

Adding or subtracting the same quantity on both sides preserves the order. Multiplying or dividing both sides by a positive number does too. For example, 2<52<5 gives 6<156<15 after multiplication by 33.

A negative multiplier reverses the order: 2<52<5, but −6>−15-6>-15. On a number line, multiplication by −1-1 reflects both numbers in zero, so the number that was on the left moves to the right. This is why division by a negative number also reverses the inequality sign. Division by zero is never allowed.

Worked example 1: unknowns on both sides

Question. Find the set of values of xx for which 5x−7<2x+85x-7<2x+8.

Plan. Collect the xx-terms on the left and the constants on the right.

5x−7<2x+83x−7<8subtract 2x3x<15add 7x<5divide by positive 3\begin{aligned} 5x-7&<2x+8\\ 3x-7&<8 &&\text{subtract }2x\\ 3x&<15 &&\text{add }7\\ x&<5 &&\text{divide by positive }3 \end{aligned}

Answer. x<5x<5.

Check. At x=5x=5, both original sides equal 1818, so the boundary is excluded. Subtracting the original right side from the left gives 3(x−5)3(x-5), which is negative exactly when x<5x<5.

Now choose an inequality whose rearrangement gives a negative coefficient of xx. The final division must reverse the sign.

Worked example 2: division by a negative number

Question. Find the set of values of xx for which 7−2x≥x−57-2x\ge x-5.

Plan. Collect the variable terms, then divide by their coefficient.

7−2x≥x−5−3x≥−12subtract x and 7x≤4divide by −3 and reverse the sign\begin{aligned} 7-2x&\ge x-5\\ -3x&\ge-12 &&\text{subtract }x\text{ and }7\\ x&\le4 &&\text{divide by }-3\text{ and reverse the sign} \end{aligned}

Answer. x≤4x\le4.

Check. The original difference is (7−2x)−(x−5)=3(4−x)(7-2x)-(x-5)=3(4-x). This is nonnegative exactly when x≤4x\le4. At x=4x=4, both original sides equal −1-1, so equality is allowed.

You could instead obtain 12≥3x12\ge3x and divide by positive 33. Both routes describe the same set.

Watch out

The line −3x≥−12⇒x≥4-3x\ge-12\Rightarrow x\ge4 is wrong: division by −3-3 reverses the sign. The correct result is x≤4x\le4. Adding or subtracting a negative number does not itself reverse a sign.

Keep the inequality sign throughout the next solution.

Representing solution sets

A number line shows a whole solution set at once, including its endpoints.

Use an open circle for an excluded endpoint and a filled circle for an included endpoint. Extend the solution to the left for smaller values and to the right for larger values; an arrow shows that it continues without a final value.

[DIAGRAM: asset_name: Strict and inclusive endpoints on a number line; asset_slug: linear-endpoints; caption: The sets x less than minus 2 and x less than or equal to minus 2 differ only at the endpoint; description: Two number lines share the boundary minus 2. The upper solution extends left from an open circle at minus 2 and is labelled x less than minus 2. The lower solution extends left from a filled circle at minus 2 and is labelled x less than or equal to minus 2. Their interval forms are shown alongside.]
The sets x less than minus 2 and x less than or equal to minus 2 differ only at the endpoint

Interval and set notation

An interval describes a continuous range. Round brackets exclude a finite endpoint; square brackets include it. For example, (−2,3](-2,3] means −2<x≤3-2<x\le3.

The symbol R\mathbb R means the real numbers. In set-builder notation, {x∈R:x<5}\{x\in\mathbb R:x<5\} means “the real numbers xx for which x<5x<5”.

The symbols −∞-\infty and ∞\infty describe continuation without a bound. They are not real endpoints to include, so always use round brackets next to them.

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