FA18, HA22 - Linear inequalities

FA18, HA22 - Linear inequalities

An inequality can allow a whole range of values rather than one answer. Learn to read and solve inequalities in one variable, list integer solutions when requested, and represent solution sets on number lines. This teaches FA18 and the shared one-variable linear work within HA22.

Interpreting inequalities

A craft kit costs £5 plus £3 for each refill. With at most £20 to spend, can you buy exactly five refills? The phrase “at most” decides whether spending the full £20 is allowed.

The symbols << and >> exclude equality; ≤\le and ≥\ge include it. Thus “at most 2020” means ≤20\le20, while “less than 2020” means <20<20. “At least 33” means ≥3\ge3, including 33 itself.

Solution set and bounded interval

The solution set contains every value making an inequality true. A bounded interval has both a lower and an upper boundary. In −2<x≤3-2<x\le3, both −2<x-2<x and x≤3x\le3 must be true at the same time.

Unless the question restricts the variable, values between integers are allowed too. For example, 0.40.4 satisfies −2<x≤3-2<x\le3. If the question says that xx is an integer, list only whole numbers and their negatives, including zero where permitted.

Worked example 1: two excluded boundaries

Question. Write down all the integers satisfying −2<x<3-2<x<3.

Plan. Start just above the lower boundary and stop before the upper one.

The lower condition excludes −2-2, so the first integer is −1-1. The upper condition excludes 33, so the last is 22.

Answer. −1, 0, 1, 2-1,\ 0,\ 1,\ 2.

Check. Each listed integer is greater than −2-2 and less than 33. The neighbouring integers −2-2 and 33 fail one condition each, so none is missing.

Now change only the upper comparison to include equality. One extra integer becomes possible.

Worked example 2: an included upper boundary

Question. Write down all the integers satisfying −2<x≤3-2<x\le3.

Plan. Keep the lower boundary excluded but include the upper one.

The integers strictly above −2-2 begin at −1-1. This time 33 satisfies x≤3x\le3, so include it.

Answer. −1, 0, 1, 2, 3-1,\ 0,\ 1,\ 2,\ 3.

Check. Substituting 33 gives −2<3-2<3 and 3≤33\le3, both true. Substituting −2-2 still gives a false strict lower comparison.

A chain is an “and” condition, not a choice between its two sides.

Watch out

For −2<x≤3-2<x\le3, including −2-2 because “it is an endpoint” is wrong. Test the actual symbol: −2<−2-2<-2 is false. Equality is included only where the extra line is present.

Apply both boundaries in a fresh interval.

Solving inequalities

Solving an inequality means finding an equivalent, simpler description of all its solutions.

Adding or subtracting the same quantity on both sides preserves the order. Multiplying or dividing by a positive number also preserves it. For example, 2<52<5 becomes 4<104<10 when both sides are doubled.

Multiplying by a negative number reverses the order: 2<52<5 becomes −2>−5-2>-5 after multiplication by −1-1. On a number line this reflects both values in zero and swaps which is further right. Dividing by a negative number has the same effect.

Negative multiplication or division

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

Use this rule when the sign of the multiplier or divisor is known. Never divide by zero.

Worked example 3: a positive coefficient

Question. Solve 3x+5≤203x+5\le20.

Plan. Subtract 55, then divide by the positive coefficient 33.

3x≤15subtract five from both sidesx≤5divide by positive three\begin{aligned} 3x&\le15 &&\text{subtract five from both sides}\\ x&\le5 &&\text{divide by positive three} \end{aligned}

Answer. x≤5x\le5.

Check. The boundary x=5x=5 gives 20≤2020\le20, which is true. A smaller value, x=4x=4, gives 17≤2017\le20; a larger value, x=6x=6, gives 23≤2023\le20, which is false.

Change only the sign of the coefficient. The division now reverses the inequality.

Worked example 4: a negative coefficient

Question. Solve −3x+5≤20-3x+5\le20.

Plan. Subtract 55, then divide by −3-3 and reverse the sign.

−3x≤15subtract fivex≥−5divide by negative three and reverse\begin{aligned} -3x&\le15 &&\text{subtract five}\\ x&\ge-5 &&\text{divide by negative three and reverse} \end{aligned}

Answer. x≥−5x\ge-5.

Check. At x=−5x=-5, the original gives 20≤2020\le20. At x=0x=0, it gives 5≤205\le20, true. At x=−6x=-6, it gives 23≤2023\le20, false. The direction must therefore include values above −5-5.

The unknown may appear on both sides, just as in an equation. Collect it using the same operation on both sides.

Worked example 5: unknowns on both sides

Question. Solve 2x+7<5x−22x+7<5x-2.

Plan. Collect the unknown on the right to keep its coefficient positive.

7<3x−2subtract two x9<3xadd two3<xdivide by positive three\begin{aligned} 7&<3x-2 &&\text{subtract two x}\\ 9&<3x &&\text{add two}\\ 3&<x &&\text{divide by positive three} \end{aligned}

Answer. x>3x>3, the same statement as 3<x3<x.

Check. x=4x=4 gives 15<1815<18, true. At x=3x=3, both sides are 1313, so the strict inequality excludes the boundary.

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