QuanticaAlgebra IOpen in the courseHere the equations are yours to write. A story states its facts in plain English, and the two letters, what each one stands for, and one equation for each fact all have to be written down first. After that the system is an ordinary one and 5.2 or 5.3 finishes it. Try the next two stories before any of this has a name.
Problem
A ferry charges one fixed fare for a motorcycle and another for a car. The fares from 4 motorcycles and 3 cars come to 64 dollars, and the fares from 6 motorcycles and 7 cars come to 126 dollars. Enter the fare for one motorcycle, in dollars.
Show a hint
- Two fares are unknown here, one for a motorcycle and one for a car. Name both, then write one equation for each of the two totals.
- With \(m\) dollars for a motorcycle and \(c\) dollars for a car, \(4m+3c=64\) and \(6m+7c=126\). No letter has coefficient \(1\), so isolating one brings fractions. Scale both equations and eliminate the 5.3 way instead.
Show the full solution
With \(m\) dollars for a motorcycle and \(c\) for a car, \(4m+3c=64\) and \(6m+7c=126\). Tripling the first and doubling the second gives \(12m+9c=192\) and \(12m+14c=252\), and subtracting leaves \(5c=60\), so \(c=12\), then \(4m=64-36=28\) and \(m=\boxed{7}\). Check, \(6(7)+7(12)=42+84=126\). Neither total is made of one fare alone, so no single total gives a fare by itself, and the story needs two letters.
Problem
A photo lab made 143 prints on Tuesday, some small and some large. The number of small prints was 5 more than twice the number of large prints. Enter how many large prints the lab made that day.
Show a hint
- Name both counts before anything else, \(L\) for the large prints and \(S\) for the small prints. Two separate facts are stated here, the total of 143 and the description of \(S\) in terms of \(L\), so two equations come out of them.
- With \(L\) large prints and \(S\) small prints the total is \(L+S=143\). Twice the large count is \(2L\), so the second sentence reads \(S=2L+5\). Replace \(S\) in the total.
Show the full solution
\(L+S=143\) and \(S=2L+5\), so \(L+(2L+5)=143\), \(3L=138\), and \(L=\boxed{46}\), with \(S=97\). Check, \(2(46)+5=97\) and \(46+97=143\). Five more than twice \(L\) is \(2L+5\) and not \(2(L+5)\), since only the large count is doubled.
Problem
A greenhouse bench holds shallow trays of 12 cells and deep trays of 20 cells, 37 trays and 588 cells in all. Enter how many of the trays are deep.
Show a hint
- Name the two tray counts. The bench is counted twice, once in trays and once in cells, so there are two facts to translate.
- With \(s\) shallow trays and \(d\) deep trays, \(s+d=37\) and \(12s+20d=588\). Substitute \(s=37-d\).
Show the full solution
\(12(37-d)+20d=588\), so \(444+8d=588\), then \(8d=144\) and \(d=\boxed{18}\). The other 19 trays are shallow, and \(12(19)+20(18)=588\). Counting trays and counting cells are separate measurements, so the story needs two equations, not one.
Problem
A school stationery order was 52 items, all pens and markers. Pens cost 3 dollars each, markers cost 7 dollars each, and the whole order cost 264 dollars. Enter how many of the items were pens.
Show a hint
- Write one equation for the 52 items and a second for the 264 dollars.
- With \(p\) pens and \(m\) markers, \(p+m=52\) and \(3p+7m=264\). Rewrite the first as \(p=52-m\) and put that into the second.
Show the full solution
With \(p\) pens and \(m\) markers, \(p+m=52\) and \(3p+7m=264\). Put \(p=52-m\) into the cost equation to get \(156+4m=264\), so \(4m=108\) and \(m=27\). Then \(p=52-27=\boxed{25}\). The count and the cost are two separate facts about the same order, so each one gets its own equation.
Problem
A hardware bin holds 118 fasteners, each one a bolt or a screw. The number of bolts is 14 less than three times the number of screws. Enter how many bolts are in the bin.
Show a hint
- Name both counts, \(s\) for the screws and \(b\) for the bolts. The total of 118 is one fact and the sentence comparing the two counts is the other.
- Three times the screw count is \(3s\), and 14 less than that is \(3s-14\), so \(b=3s-14\). Put that into \(s+b=118\).
Show the full solution
With \(s\) screws and \(b\) bolts, \(s+b=118\) and \(b=3s-14\). Substituting gives \(s+(3s-14)=118\), so \(4s=132\) and \(s=33\), which makes \(b=3(33)-14=\boxed{85}\). Check, \(33+85=118\). Fourteen less than \(3s\) is \(3s-14\) and not \(14-3s\), and the reversed version gives a negative bolt count.
Solving a system produces two numbers, and the question often asks for a third, the difference between them, the larger of the two, or a cost built from both. The work is the same either way. What changes is which number gets typed, and stopping at a letter's value is the most common way to work a story correctly and still answer it wrong.
Problem
A campsite charges one price per tent night and one price per cabin night, unchanged all season. Three tent nights and two cabin nights cost 214 dollars. Five tent nights and four cabin nights cost 396 dollars. Enter how much more one cabin night costs than one tent night, in dollars.
Show a hint
- Name the two nightly prices and write one equation for each booking. Neither equation has a letter with coefficient 1, so elimination is the cheaper route.
- With \(t\) and \(c\) in dollars per night, \(3t+2c=214\) and \(5t+4c=396\). Doubling the first equation matches the \(c\)-columns at \(4c\). The closing sentence asks for a difference, not for either nightly price.
Show the full solution
Doubling \(3t+2c=214\) gives \(6t+4c=428\), and subtracting \(5t+4c=396\) leaves \(t=32\). Then \(96+2c=214\) gives \(c=59\), and the ask is \(c-t=59-32=\boxed{27}\). Check, \(5(32)+4(59)=396\). Typing \(59\) answers a question the problem did not ask, since \(59\) is the cabin price and the story asked for the gap between the two prices.
Problem
A bakery sold 141 loaves, sourdough and rye, with 27 fewer rye loaves than sourdough. Let \(s\) be the number of sourdough loaves and \(r\) the number of rye loaves. A student writes \(s+r=141\) and \(s=r-27\), solves that pair correctly, and both of the student's equations check against the pair that comes out. Fix the equation that is a wrong translation and enter the correct number of rye loaves.
Show a hint
- The student's arithmetic is sound, so the error is in the translation. Solve the student's pair, then read the result against the sentence about 27 fewer rye loaves than sourdough.
- 84 rye against 57 sourdough makes rye the larger count, so the minus sits on the wrong letter. The story says \(r=s-27\).
Show the full solution
The story makes rye the smaller count, so \(r=s-27\). Then \(s+(s-27)=141\), giving \(2s=168\) and \(s=84\), so \(r=\boxed{57}\). The wrong pair and the right pair are the same two numbers swapped, and no algebraic check tells them apart. Only the English says which count is the larger one.
Prealgebra 7.3 gives distance as rate times time, \(d=rt\). A boat moving with a current covers ground at its own still-water speed plus the current's speed, and moving against it at its own speed minus the current's speed. Both speeds are unknown, so name them \(b\) and \(c\). Each trip is one fact, so a round trip is a two-fact story.
Problem
A river tour boat runs 90 km downstream in 3 hours, then the same 90 km back upstream in 5 hours. The current holds one steady speed all day, and the boat's speed in still water is the same on both legs. Enter that still-water speed, in km per hour.
Show a hint
- Each leg has an actual speed you can work out right away, distance divided by time. That is \(90\div 3\) one way and \(90\div 5\) the other.
- Call the boat's still-water speed \(b\) and the current's speed \(c\). The current is with the boat downstream and against it upstream, so \(b+c=30\) and \(b-c=18\), and the \(c\)-terms are already opposite.
Show the full solution
Let \(b\) be the boat's speed in still water and \(c\) the current's speed. Each leg's actual speed is distance over time, \(90\div 3=30\) downstream and \(90\div 5=18\) upstream, so \(b+c=30\) and \(b-c=18\). Adding gives \(2b=48\), so \(b=\boxed{24}\), with \(c=6\). The current is with the boat one way and against it the other, so the two \(c\)-terms are opposite and one addition finishes the system.
Problem
A small plane covers 1400 km with a steady wind in 4 hours, then covers the same 1400 km back against that wind in 5 hours. The wind holds one steady speed and the plane's speed in still air is the same on both legs. Enter the wind's speed, in km per hour.
Show a hint
- Two speeds are unknown here, the plane's speed in still air and the wind's speed, so give each one a letter. With the wind those two speeds add, and against the wind they subtract.
- Each leg gives a ground speed, \(1400\div 4=350\) with the wind and \(1400\div 5=280\) against it, so \(p+w=350\) and \(p-w=280\). Subtracting the second equation from the first removes \(p\).
Show the full solution
The ground speeds are \(1400\div 4=350\) and \(1400\div 5=280\), so \(p+w=350\) and \(p-w=280\). Subtracting the second equation from the first gives \(2w=70\), so \(w=\boxed{35}\), with \(p=315\). Check, \(1400\div 350=4\) and \(1400\div 280=5\). The ask is the wind's speed, so \(315\) is not the value to enter.
Counts are not the only facts a story can state. A measured length is a fact too, and it translates the same way. Prealgebra 11.5 gives a rectangle's perimeter as \(2L+2W\), so a stated perimeter is one equation, and any second length measured across the same rectangle is the other. Two measurements pin down both dimensions.
Problem
A rectangular garden bed is edged with 100 m of timber all the way around. A soaker hose runs along one full length and both widths, and it measures 64 m. Enter the length of the bed, in meters.
Show a hint
- Name the length \(L\) and the width \(W\), both in meters, and write one equation for the timber and one for the hose. The timber goes all the way around, so it uses each dimension twice.
- The system is \(2L+2W=100\) and \(L+2W=64\). Both equations carry \(2W\), so subtracting one from the other clears \(W\) in a single move.
Show the full solution
The timber gives \(2L+2W=100\) and the hose gives \(L+2W=64\). Both carry \(2W\), so subtracting the second from the first leaves \(L=\boxed{36}\). Then \(36+2W=64\) gives \(W=14\). Check, \(2(36)+2(14)=100\) and \(36+28=64\). Each measured length in the story is its own equation, and two of them are enough to give both dimensions.
A two-digit number is not the sum of its digits and not their product. With tens digit \(t\) and units digit \(u\) the number is worth \(10t+u\), since a digit in the tens place is worth ten times its face value. Reversing the two digits gives \(10u+t\). Both are ordinary expressions in two letters, so a fact about either one is an equation.
Problem
The digits of a two-digit number add to 11, and reversing the digits gives a number 27 greater than the original. Enter the original number.
Show a hint
- The two letters are the digits, not the number. Write the original as \(10t+u\) and the reversed number as \(10u+t\).
- \((10u+t)-(10t+u)\) collects to \(9u-9t\), so the second fact is \(u-t=3\). Pair that with \(t+u=11\).
Show the full solution
The reversal fact is \((10u+t)-(10t+u)=27\), which collects to \(9u-9t=27\), so \(u-t=3\). With \(t+u=11\), adding gives \(2u=14\), so \(u=7\) and \(t=4\), and the number is \(\boxed{47}\). Solving gives the two digits, and the number still has to be rebuilt as \(10t+u\). Typing \(4\) or \(7\) answers a question the story did not ask.
Problem
A repair counter charges one fixed price to reseal a window and one to rehang a door. One week it resealed 12 windows and rehung 7 doors for 438 dollars. The next week it resealed 9 windows and rehung 11 doors for 432 dollars. Enter the total cost of one reseal and one rehang, in dollars.
Show a hint
- The counts of jobs are given, so the two unknowns are the price of a reseal and the price of a rehang. Name those two prices and write one equation for each week's takings.
- The system is \(12w+7d=438\) and \(9w+11d=432\). Neither letter has coefficient \(1\), so eliminate one by scaling. The reseal coefficients \(12\) and \(9\) both divide \(36\).
Show the full solution
With \(w\) the price of a reseal and \(d\) the price of a rehang, the two facts are \(12w+7d=438\) and \(9w+11d=432\). Scaling the first by \(3\) and the second by \(4\) gives \(36w+21d=1314\) and \(36w+44d=1728\), and subtracting leaves \(23d=414\), so \(d=18\). Then \(12w+126=438\) gives \(w=26\), so \(w+d=\boxed{44}\). Neither price on its own is the answer, since the question asks for the sum.
Every story here named two quantities and stated two facts, and each system was written one sentence at a time. Some stories name three, a triangle's three angles or three prices on one receipt. Three quantities, three facts. Lesson 5.6, Three or More Variables, raises the count, and elimination still applies, clearing one letter at a time.
Practice these ideas
Practice
A bike shop charges one fixed price to true a wheel and one fixed price to replace a brake pad. Two wheel truings and three pad replacements cost 61 dollars, and two truings and one pad replacement cost 39 dollars. Enter the price of one wheel truing, in dollars.
Show the solution
With \(t\) the price of one truing and \(p\) the price of one pad replacement, \(2t+3p=61\) and \(2t+p=39\). Subtracting the second from the first leaves \(2p=22\), so \(p=11\), and then \(2t+11=39\) gives \(2t=28\) and \(t=\boxed{14}\). Both visits hold the same two truings, so the 22 dollar gap is the cost of two pad replacements.
Practice
A jar holds 45 marbles, each one red or blue. Each red marble weighs 6 grams, each blue marble weighs 4 grams, and the marbles weigh 232 grams in all. Enter how many of the marbles are red.
Show the solution
With \(r\) red marbles and \(b\) blue, \(r+b=45\) and \(6r+4b=232\). Put \(45-r\) in for \(b\) to get \(6r+4(45-r)=232\), so \(2r+180=232\) and \(r=\boxed{26}\). The other 19 marbles are blue, and \(26\cdot 6+19\cdot 4=232\) grams.
Practice
A vending machine holds 30 bars, all of them granola bars or protein bars. Granola bars sell for 2 dollars each and protein bars sell for 3 dollars each, and the 30 bars are worth 79 dollars in all. Enter how many of the bars are protein bars.
Show the solution
With \(g\) granola bars and \(p\) protein bars, \(g+p=30\) and \(2g+3p=79\). Substitute \(g=30-p\) into the second equation. Then \(2(30-p)+3p=79\), so \(60+p=79\) and \(p=\boxed{19}\). The other 11 bars are granola, and \(2\cdot 11+3\cdot 19=79\).
Practice
A workshop packs hinges into small crates and large crates, each crate size holding a fixed number of hinges. One shipment of 4 small crates and 3 large crates holds 117 hinges. Another shipment of 2 small crates and 5 large crates holds 139 hinges. Enter how many hinges one large crate holds.
Show the solution
Let \(x\) be the hinges in a small crate and \(y\) the hinges in a large crate. The first shipment gives \(4x+3y=117\) and the second gives \(2x+5y=139\). Doubling the second gives \(4x+10y=278\), and subtracting the first leaves \(7y=161\), so \(y=\boxed{23}\). A small crate holds \(12\) hinges, and \(4(12)+3(23)=117\) matches the first shipment.
Practice
A print shop charges the same price for every poster and the same price for every sticker. One order of 4 posters and 3 stickers cost 41 dollars. Another order of 2 posters and 7 stickers cost 37 dollars. Enter the price of one poster, in dollars.
Show the solution
With \(p\) dollars for a poster and \(s\) dollars for a sticker, the two orders read \(4p+3s=41\) and \(2p+7s=37\). Doubling the second gives \(4p+14s=74\), and subtracting the first leaves \(11s=33\), so \(s=3\) and \(4p=41-9=32\), which makes \(p=\boxed{8}\). The prices fit the story, 4 posters and 3 stickers cost \(32+9=41\) dollars and 2 posters and 7 stickers cost \(16+21=37\) dollars.
Practice
Every crate of apples weighs the same and every crate of pears weighs the same. Three crates of apples and two crates of pears weigh 63 kg together, and one crate of apples weighs 6 kg more than one crate of pears. Enter the weight of one crate of pears, in kilograms.
Show the solution
Let \(a\) be the weight of one crate of apples in kilograms and \(p\) the weight of one crate of pears. Then \(3a+2p=63\) and \(a=p+6\), so \(3(p+6)+2p=63\), giving \(5p+18=63\) and \(p=\boxed{9}\). Check, \(a=15\) and \(3(15)+2(9)=63\). The 3 and the 2 are crate counts and the letters are weights, so \(3a\) is the weight of all three apple crates.
Practice
A nursery's fern count is 8 less than three times its palm count, and the two counts add to 140. Enter how many palms the nursery has.
Show the solution
With \(p\) palms and \(f\) ferns, \(f=3p-8\) and \(f+p=140\), so \((3p-8)+p=140\), giving \(4p=148\) and \(p=\boxed{37}\). Check, \(f=103\), \(103+37=140\) and \(3(37)-8=103\). Reversing the order to \(f=8-3p\) gives \(p=-66\), a negative palm count, so the order the sentence names is what the translation has to keep.
Practice
A stall sold 56 cups of cider. Small cups cost 3 dollars each and large cups cost 5 dollars each, and the stall took in 204 dollars from those 56 cups. Enter how many more small cups than large cups it sold.
Show the solution
With \(s\) small cups and \(L\) large, \(s+L=56\) and \(3s+5L=204\). Substituting \(s=56-L\) gives \(3(56-L)+5L=204\), so \(168+2L=204\), \(L=18\), and \(s=38\). The ask is \(s-L=38-18=\boxed{20}\). The question asks for the difference, not for either count, so the subtraction at the end is easy to skip.
Practice
A ranger counted deer and elk on one range. Three times the deer count is 40 more than the elk count, and the elk count is 8 less than twice the deer count. Enter how many elk the ranger counted.
Show the solution
With \(d\) deer and \(e\) elk, \(3d=e+40\) and \(e=2d-8\). Substituting gives \(3d=(2d-8)+40\), so \(d=32\), and then \(e=2(32)-8=\boxed{56}\). Check, \(3(32)=96\) and \(56+40=96\). A story can pin a pair down with two comparisons and no total at all.
Practice
An airport walkway is 96 m long. Walking along it in the direction it moves, a traveler covers the 96 m in 24 seconds. Walking back along it against its direction, at the same walking speed, the traveler covers the same 96 m in 48 seconds. Enter that walking speed in meters per second, the speed the traveler would walk on a floor that does not move.
Show the solution
The two ground speeds are \(96\div 24=4\) and \(96\div 48=2\). With \(w\) the walking speed and \(m\) the walkway's speed, \(w+m=4\) and \(w-m=2\). Adding the two gives \(2w=6\), so \(w=\boxed{3}\). Then \(m=1\), which fits both times. With only one trip you would know just \(w+m\), so the second trip is what separates the two speeds.
Practice
A rectangular courtyard has a perimeter of 84 m, and its length is 4 m more than its width. Enter its area, in square meters.
Show the solution
With \(L\) the length and \(W\) the width, \(2L+2W=84\) and \(L=W+4\). Dividing the first by \(2\) gives \(L+W=42\), and substituting \(L=W+4\) gives \(2W+4=42\), so \(W=19\) and \(L=23\). The area is \(23\times 19=\boxed{437}\) square meters. Neither letter is the answer on its own, since the question asks for the product of the two sides.
Practice
Plum boxes sell for 4 dollars and fig boxes sell for 7 dollars. A stall sold 11 more plum boxes than fig boxes and took in 187 dollars from those boxes. Let \(p\) be the number of plum boxes and \(f\) the number of fig boxes. Nia writes \(f=p+11\) and \(4p+7f=187\), solves that pair correctly, and gets \(p=10\) and \(f=21\). Her values satisfy both equations she wrote. Fix the wrong translation and enter the correct number of fig boxes.
Show the solution
The stall sold 11 more plum boxes than fig boxes, so the first equation should read \(p=f+11\). Then \(4(f+11)+7f=187\) gives \(11f+44=187\), so \(11f=143\) and \(f=\boxed{13}\). Then \(p=24\), and \(4(24)+7(13)=96+91=187\). Nia's two values are whole numbers and they fit both of the equations she wrote, so only a reread of the English sentence catches the error.
Practice
An isosceles triangle has a perimeter of 95 cm, and each of its two equal sides is 5 cm shorter than twice the base. Enter the length of the base, in centimeters.
Show the solution
With \(b\) the base and \(s\) one equal side, \(2s+b=95\) and \(s=2b-5\). Substituting gives \(2(2b-5)+b=95\), so \(5b-10=95\) and \(b=\boxed{21}\). Check against the story, \(s=2(21)-5=37\) and \(2(37)+21=95\). The perimeter counts the equal side twice, so the first equation is \(2s+b\) and not \(s+b\).
Practice
A two-digit number is 7 times the sum of its digits, and reversing its digits gives a number 18 less than the original. Enter the original number.
Show the solution
With \(t\) the tens digit and \(u\) the units digit, the number is \(10t+u\) and its reversal is \(10u+t\). The first fact reads \(10t+u=7(t+u)\), which simplifies to \(3t=6u\), so \(t=2u\). The second reads \(10u+t=10t+u-18\), which simplifies to \(t-u=2\). Then \(2u-u=2\) makes \(u=2\) and \(t=4\), so the number is \(\boxed{42}\). Both facts hold for 42, since \(7(4+2)=42\) and \(24=42-18\).
Practice
A canteen charges the same price for every roll and the same price for every bowl of soup. Three rolls and two bowls of soup cost 37 dollars, and five rolls and three bowls of soup cost 59 dollars. Enter the cost of four rolls and three bowls of soup, in dollars.
Show the solution
With \(r\) the price of a roll and \(s\) the price of a bowl of soup, \(3r+2s=37\) and \(5r+3s=59\). Tripling the first gives \(9r+6s=111\) and doubling the second gives \(10r+6s=118\), so subtracting leaves \(r=7\), and \(21+2s=37\) gives \(s=8\). Four rolls and three bowls of soup cost \(28+24=\boxed{52}\) dollars. Neither price is the answer, and the closing sentence names a third amount built from both.
Practice
Two trains leave stations 468 km apart at the same moment and travel toward each other at steady speeds, meeting 3 hours later. One train is 14 km per hour faster than the other. Enter how far the faster train traveled, in kilometers.
Show the solution
With \(x\) the slower rate and \(y\) the faster, in km per hour, the two distances add to 468, so \(3x+3y=468\) and \(x+y=156\). The second fact is \(y=x+14\), so \(2x+14=156\), \(x=71\), and \(y=85\). The faster train travels \(3(85)=\boxed{255}\) km. The letters stand for rates and the question asks for a distance, so multiply the faster rate by the 3 hours.
Practice
A ticket window sells lawn seats at 18 dollars and pavilion seats at 32 dollars. One night it sold 11 fewer pavilion seats than lawn seats and took in 1898 dollars. Enter how much more money came from pavilion seats than from lawn seats, in dollars.
Show the solution
With \(L\) lawn seats and \(P\) pavilion seats, \(P=L-11\) and \(18L+32P=1898\), so \(18L+32(L-11)=1898\), giving \(50L=2250\) and \(L=45\), with \(P=34\). Pavilion money is \(32(34)=1088\) and lawn money is \(18(45)=810\), so the difference is \(\boxed{278}\) dollars. Neither 45 nor 34 is the answer, since each count still has to be multiplied by its own price before the subtraction.