Every real-world problem in this lesson takes the same three steps: name the unknown, turn the words into an inequality, then read the answer as a green stretch on the line.
All through Stage 12 you have been learning to solve inequalities — the moves, the flip, the overlap, the parabola. This last lesson is where it pays off. Out in the world nobody hands you a tidy "2x + 1 > 5." Instead you hear: "I want a profit of at least fifteen dollars." "Stay under twenty dollars." "Which phone plan is cheaper?" "The board must measure fifty centimetres, give or take half." Each of those is an inequality wearing a coat. Your job is to take the coat off.
By the end you will be able to do four things: (1) turn an "at least / at most / enough" sentence into an inequality and read its solution as a range; (2) compare two plans by finding their breakeven point and seeing who wins on each side; (3) use the basic inequality from 12.6 to make an area as large as possible; and (4) write a measurement-with-tolerance as an absolute-value inequality. The colors stay the same as always: green is the part of the line that is true (the solution), red is what is shut out or impossible, amber marks a boundary or the bigger side, and blue is the line and the variable themselves.
12.7.1 Word problems become inequalities
Every word problem becomes an inequality the same way. Name the unknown. Find the phrase that means "bigger" or "smaller," and write the matching symbol — ≥ for at least / no less than, ≤ for at most / no more than, and the strict >/< for more than / fewer than. Then solve it like an equation (flip only if you divide by a negative, from 12.3) and read the answer as a stretch of line.
The shop. A store buys an item for $40 and wants to make a profit of at least$15 on each one. Let the selling price be x. Profit is "price minus cost," x − 40, and "at least $15" means it is greater than or equal to 15:
x − 40≥15
⟹
x≥55.
Add 40 to both sides — adding never flips the sign — and you get x ≥ 55. The shop should charge at least $55. The dot at 55 is filled, because $55 itself gives exactly $15 profit and that is allowed ("at least" includes the boundary).
The solution of x − 40 ≥ 15 is the green ray x ≥ 55: every price from $55 up. The filled dot says $55 itself counts.
Key idea — the four trigger phrases
at least / no less than → ≥. at most / no more than → ≤. more than → >. fewer / less than → <. The first two include the boundary (filled dot); the last two exclude it (open dot).
The taxi. A taxi charges $3 to start plus $2 per kilometre, and you have at most$20 to spend. Let k be the kilometres. The fare is 3 + 2k, and "at most $20" means ≤ 20:
3 + 2k≤20
⟹
2k≤ 17
⟹
k≤8.5.
So algebra says k ≤ 8.5 km. But here the world talks back. You cannot pay for "half a kilometre" of riding and stay inside the budget — a 9th kilometre would push the fare over $20, and 8.5 km is the very edge. Since you must ride a whole number of kilometres, you round down to at most 8 whole kilometres. (Check: 8 km costs 3 + 2·8 = $19 ✓; 9 km would cost $21, over budget.)
Watch out — round the safe way
When the answer must be a whole number, ask "which way keeps me inside the rule?" For a budget (≤) you round down — fewer kilometres stays under $20. For "enough material" (≥) you round up — you can't buy 4.2 boards, so you buy 5. The algebra gives the edge; common sense picks the side.
🎮 Try it WORD-PROBLEM BUILDER
Pick a scenario, dial in the numbers, and watch the inequality assemble, solve, and land on the line.
Scenario
12.7.2 Choosing a plan, and its tipping point
Two deals, and you want the cheaper one. The trick is not to guess — it is to find the single number where the two are exactly equal, the breakeven point. Below it one plan wins; above it the other wins. The breakeven is the hinge the whole answer swings on.
Two streaming plans.Plan A is a flat $30 a month, no matter how many shows you watch. Plan B is $10 up front plus $0.50 per show, so for n shows it costs 10 + 0.5n. When are they equal?
10 + 0.5n = 30
⟹
0.5n = 20
⟹
n = 40.
At exactly 40 shows both plans cost $30 — that's the tipping point. Now test one number on each side to see who wins where. At n = 20 (light watcher): Plan B costs 10 + 0.5·20 = $20, which is less than Plan A's $30, so B is cheaper when n < 40. At n = 50 (heavy watcher): Plan A is still $30 while Plan B costs 10 + 0.5·50 = $35, so A is cheaper when n > 40. One test on each side is all it takes to be sure.
Plan A (amber, flat $30) and Plan B (blue, rising $10 + $0.50·n) cross at n = 40. Left of the crossing the blue line is lower — B wins. Right of it the amber line is lower — A wins.
Example — say it as an inequality
"Plan B is the better deal" means 10 + 0.5n<30, i.e. 0.5n< 20, i.e. n < 40 — fewer than 40 shows. Exactly 40 shows is a tie, so the dot at 40 is open for either "strictly cheaper" claim.
Watch out — don't trust your gut on "which side"
Many people find the breakeven and then guess the wrong winner. Always test one endpoint. The plan with the lower starting price wins for small n; the one that rises slower (or not at all) wins for large n. Here B starts lower ($10 < $30) so B wins on the left.
🎮 Try it TWO-PLAN COMPARATOR
Set each plan's pieces. The crossing point and the winning sides update live — green for "B wins," amber for "A wins."
12.7.3 Optimization — spend less, fence more
Some problems don't ask "which values are allowed" — they ask "what is the best I can do?" The largest area, the smallest cost. For these, the basic inequality from 12.6 is the key: when two positive quantities have a fixed sum, their product is largest when they are equal.
The wall-backed pen. You're building a rectangular pen against the side of a barn. The barn wall is one long side for free; you have 40 m of fence for the other three sides. Let the width (the two short sides) be x. The two widths use 2x of fence, so the remaining length is 40 − 2x. The area is
A = x(40 − 2x) = 40x − 2x².
The barn wall (slate) is the free fourth side. The 40 m of fence wraps the other three: two widths x and one length 40 − 2x.
Here is the clever step. Look at the two parts 2x and 40 − 2x. They add to a fixed total: 2x + (40 − 2x) = 40, always. Two positive numbers with a fixed sum of 40 have their largest product when each equals half, that is 20. So 2x(40 − 2x) ≤ (40/2)² = 400. Dividing by 2,
A = x(40 − 2x) = 2x(40 − 2x)2≤4002 = 200.
So the area can never beat 200 m², and it reaches 200 exactly when the two parts are equal: 2x = 40 − 2x, giving 4x = 40, so x = 10. That makes the length 40 − 2·10 = 20 — a 10 m × 20 m pen, 200 m². (Sanity check the neighbours: x = 9 gives 9·22 = 198; x = 11 gives 11·18 = 198; both fall short of 200. The peak really is at 10.)
Key idea — fixed sum, biggest at equal
When a quantity is a product of two pieces whose sum is fixed, the product is biggest when the pieces are equal. That is "fixed sum → biggest product" from 12.6, doing real work: the best pen is the one where twice-the-width equals the length.
🎮 Try it MAX-AREA OPTIMIZER
Slide the width of the wall-backed pen. The fence total stays 40 m; watch the area rise to its peak of 200 m² at x = 10.
12.7.4 Error and range with absolute value
Measurements are never exact. A ruler, a scale, a machine — each comes with a tolerance, a little give. "This board is 50 cm, good to within half a centimetre" doesn't mean exactly 50.000; it means the true length is somewhere close to 50, off by less than 0.5 either way. Absolute value is built to say exactly that.
Recall that |x − a| reads as "the distance from x to a." So "x is within d of a" is simply |x − a|<d — the distance is less than d. And a distance less than d means x sits in the band stretching d to the left and d to the right of a:
|x − a|<d
⟺
a − d<x<a + d.
The board. Measured at 50 cm, tolerance ±0.5 cm. So a = 50, d = 0.5, and the rule is |x − 50|<0.5. Unfold it:
|x − 50|<0.5
⟺
49.5<x<50.5.
"Within 0.5 of 50" is the green band from 49.5 to 50.5. The dots are open because the tolerance is strict (<) — a board exactly 50.5 cm long would just miss.
Key idea — "within d of a" is a band
|x − a| < d always unpacks to the open interval a − d < x < a + d, centred at a with half-width d. The centre is the target value; the half-width is the tolerance. (Use ≤ and closed dots if the tolerance is "no more than d.")
Watch out — less-than is a band, greater-than is two rays
Be careful which way the absolute value points. |x − a| < d ("close to a") is a single band between two numbers. But |x − a| > d ("far from a") is the opposite — two rays shooting outward, x < a − d or x > a + d. Far-from is the leftover after you cut out the band.
🎮 Try it TOLERANCE BAND
Set the measured value and the tolerance. The green band and its inequality update together.
★ The big ideas, in one breath
A real-world inequality is just an everyday sentence with its coat off. Name the unknown, translate the trigger word — ≥ for at least, ≤ for at most, strict </> for fewer / more than — solve like an equation (flip only at a divide-by-negative), and read the answer as a green stretch of line, rounding the safe way when the answer must be a whole number. To pick between two plans, find the breakeven and test one endpoint to see who wins on each side. To make something as big as possible, lean on the basic inequality: fixed sum → biggest product when the parts are equal (the 40 m fence gives a 200 m² pen at 10 × 20). And "within d of a" is the absolute-value band a − d < x < a + d.
What's next
That last band — a length pinned to a centre with a tolerance — is exactly how the next strand, Plane Geometry, talks about measurement. And the wall-backed pen was your first taste of "best shape for the material." Inequalities don't end here; they become the quiet engine behind every "as large as possible" and "to within" you'll meet from now on. You've finished Stage 12 — revisit any sibling: 12.1 · 12.2 · 12.3 · 12.4 · 12.5 · 12.6.
✎ Exercises 12.7
A shop buys a backpack for $40 and wants a profit of at least $25 per bag. Write an inequality for the selling price x and solve it.
Show answer
Profit = x − 40, and "at least $25" means x − 40 ≥ 25, so x ≥ 65. Charge at least $65 (filled dot at 65).
A ride costs $3 to start plus $2 per km, and you have at most $20. How many whole kilometres can you afford?
Show answer
3 + 2k ≤ 20 ⟹ 2k ≤ 17 ⟹ k ≤ 8.5. You ride whole kilometres and a budget must stay under, so round down: at most 8 km. (8 km = $19 ✓; 9 km = $21, over.)
A class needs party favours, each pack covers 6 guests, and there are 40 guests. "Enough for everyone" means at least 40 covered. How many whole packs are needed?
Show answer
6p ≥ 40 ⟹ p ≥ 40/6 = 6.67. Here "enough" is a ≥, so round up: you need 7 packs. (6 packs cover 36, not enough; 7 cover 42 ✓.)
Plan A is a flat $30; Plan B is $10 + $0.50 per show. For how many shows n is Plan B the better deal?
Show answer
Breakeven: 10 + 0.5n = 30 ⟹ n = 40. B is cheaper when 10 + 0.5n < 30, i.e. n < 40 (fewer than 40 shows). At 40 they tie; above 40, A wins.
Using the same two plans, for how many shows is Plan A the better deal? Verify with n = 50.
Show answer
A wins when 30 < 10 + 0.5n, i.e. 20 < 0.5n, i.e. n > 40. Check n = 50: A = $30, B = 10 + 0.5·50 = $35, so A is cheaper ✓.
A rectangular pen uses a wall as one long side and 40 m of fence for the other three sides. With width x, write the area and find the width that makes it largest.
Show answer
Length = 40 − 2x, so A = x(40 − 2x). Since 2x + (40 − 2x) = 40 is fixed, 2x(40 − 2x) ≤ (20)² = 400, so A ≤ 200 m², reached when 2x = 40 − 2x ⟹ x = 10 (a 10 × 20 pen).
Same pen, but check the claim by hand: compute the area at x = 9, x = 10, and x = 11. Which is biggest?
Show answer
x = 9: 9·22 = 198. x = 10: 10·20 = 200. x = 11: 11·18 = 198. The peak is at x = 10, area 200 m² — the neighbours both fall a little short, as the inequality promised.
A board is measured at 50 cm, good to ±0.5 cm. Write the tolerance as an absolute-value inequality and unfold it into a range.
Show answer
|x − 50| < 0.5 ⟺ 49.5 < x < 50.5. The true length is somewhere in that open band, centred on 50 with half-width 0.5 (open dots, strict tolerance).
A bolt should be 12 mm wide, accepted only if it is more than 0.2 mm off (so it's rejected if too close — a quirky inspection). Write the "rejected if far" condition and unfold it.
Show answer
"More than 0.2 off" is |x − 12| > 0.2, which is two rays: x < 11.8 or x > 12.2. Greater-than absolute value splits into the leftover outside the band (contrast with the single band of less-than).
Translate & solve. "A streaming service costs $8 a month plus $1.50 per movie rented, and Maya wants to keep her bill under $20." How many movies m can she rent?
Show answer
8 + 1.5m < 20 ⟹ 1.5m < 12 ⟹ m < 8. Movies are whole and "under" is strict, so the most is 7 movies (7 ⟹ $18.50 < $20 ✓; 8 ⟹ $20, not under).
🎯 Quick check
Six questions to lock it in. Tap the answer you think is right.
§ For teachers and parents
This lesson closes Stage 12 by putting the whole toolkit to work on modeling tasks, addressing CCSS A-CED.A.1 (create inequalities in one variable to solve problems), A-CED.A.3 (represent real-world constraints), and A-REI.B.3 (solve and graph the solutions), woven throughout the Standards for Mathematical Practice — especially MP4 (model) and MP6 (attend to precision). The four sections map to four classic problem families: translate-and-solve, comparison/breakeven, optimization (a gentle reuse of 12.6's AM–GM), and tolerance via absolute value. The #1 misconception is mistranslating the words — turning "no more than" into >, or ignoring the real-world limit (rounding a budget up instead of down, allowing fractional people or kilometres), or naming the wrong plan as the winner past the tipping point. The antidote, modeled in every worked example here: translate the trigger phrase carefully (≥/≤ for "at least"/"at most"), then test one concrete endpoint to confirm both the direction of the inequality and the direction of the rounding actually make sense in the story. A right number on paper that you can't pay for at the curb is still the wrong answer.
eastmath.com · Stage 12 · 12.7 Inequalities in the Real World · Intuition before notation
eastmath.com · 12.7 Inequalities in the Real World · 12.7.1 Word problems become inequalities