QuanticaAlgebra IOpen in the course4.4 closed with a promise, equations with no algebra written down at all, just plain English sentences. Here it is kept. A word problem states a fact about an unknown number, and chapter 4 already holds every move needed to find that number once the sentence becomes an equation. Try one cold.
Problem
A trail camera takes the same number of photos each night. After 6 nights, plus the 55 photos from the afternoon it was mounted, its card holds 553 photos. How many photos does it take each night?
Show a hint
- Give the unknown a name in words first. Let \(n\) be the number of photos taken each night, then reread the sentence with that name in it.
- Six nights of \(n\) photos plus the \(55\) from the first afternoon makes \(553\), so \(6n+55=553\).
Show the full solution
The six nights give \(6n\) photos, so \(6n+55=553\). Subtract \(55\), \(6n=498\), and divide by \(6\), \(n=\boxed{83}\). Check against the story, \(6\) nights of \(83\) photos is \(498\), and the first afternoon's \(55\) brings the card to \(553\).
Look at what you just did. You named the unknown, photos per night. You turned the sentence into \(6n+55=553\), solved it with 4.2's moves, answered the question that was asked, and checked the value against the story. That is the entire method of this lesson, and you ran it before it had a name.
The vocabulary translates a piece at a time. The sum of two quantities adds them, their difference subtracts them in the order named, and more than adds. Less than carries a trap. Seven less than \(n\) means \(n\) loses seven, so it is \(n-7\), not \(7-n\). English names the subtracted amount first, and algebra writes it last.
Problem
Translate this sentence into an equation before solving anything. Eleven less than four times a number is \(109\). Enter the number.
Show a hint
- Decide which quantity gets subtracted before writing anything down. Eleven less than a thing means the thing loses eleven.
- The equation is \(4n-11=109\), not \(11-4n=109\).
Show the full solution
Four times the number, less eleven, gives \(4n-11=109\). Add \(11\), \(4n=120\), so \(n=\boxed{30}\). The phrase names the eleven first, but the eleven is the amount taken away, so it is written last.
Problem
A novel and its sequel together run 584 pages, and the sequel is 68 pages longer than the novel. How many pages long is the sequel?
Show a hint
- One letter can carry both books. Let it name the shorter one, and write the sequel's page count from it.
- \(n+(n+68)=584\) gives the novel's pages. Reread the question before you type.
Show the full solution
Let \(n\) be the novel's pages, so the sequel has \(n+68\) and \(n+(n+68)=584\). Then \(2n+68=584\), \(2n=516\), \(n=258\), and the sequel runs \(258+68=\boxed{326}\) pages. The solved letter answers only its own definition, the novel, so the last move is rereading the ask.
Problem
A library cart holds 76 books, hardcovers and paperbacks, with three times as many paperbacks as hardcovers. How many hardcovers are on the cart?
Show a hint
- Name the smaller count. The paperbacks are then written from it, not given a second letter.
- Hardcovers plus three times hardcovers is \(76\), so \(4h=76\).
Show the full solution
Let \(h\) be the hardcover count, so there are \(3h\) paperbacks and \(h+3h=76\). Then \(4h=76\) and \(h=\boxed{19}\). Check, \(19\) hardcovers and \(57\) paperbacks make \(76\) books, and \(57\) is three times \(19\).
Runs of neighboring numbers translate through one variable too. Consecutive integers step by one, so three of them are \(x\), \(x+1\), \(x+2\). Consecutive even numbers step by two, skipping the odd number between, and consecutive odd numbers do the same, so either run is \(x\), \(x+2\), \(x+4\).
Problem
Three lockers in a row carry consecutive numbers, and the three numbers add to 222. Enter the largest of the locker numbers.
Show a hint
- Let \(x\) be the smallest locker number and write the other two from it.
- The equation collects to \(3x+3=222\), and the answer is not \(x\).
Show the full solution
Let \(x\) be the smallest, so \(x+(x+1)+(x+2)=222\). Then \(3x+3=222\), \(3x=219\), \(x=73\), and the largest is \(73+2=\boxed{75}\). Check, \(73+74+75=222\).
Problem
House numbers on one side of a lane run through the odd numbers. Three neighbors' numbers add to 303. Enter the largest of the three.
Show a hint
- Consecutive odd numbers are \(2\) apart, not \(1\) apart.
- With \(x\) the smallest, the equation collects to \(3x+6=303\), and the ask is the largest.
Show the full solution
Neighboring odd numbers step by \(2\), so the three are \(x\), \(x+2\), \(x+4\), and \(3x+6=303\). Then \(3x=297\), \(x=99\), and the largest is \(\boxed{103}\). One side of a street runs through every other number, which is why the step is \(2\).
Some stories state facts at two different moments. An age problem describes today, then jumps ahead, and a jump of six years adds six to every age in the story, both people, not just the one being asked about. Push every age forward, write the fact about the later moment, and the equation that falls out has the variable on both sides, 4.3's shape.
Problem
Mara is 39 years older than her daughter. In 7 years she will be exactly twice her daughter's age. How old is the daughter now, in years?
Show a hint
- Write both ages now from one letter, then push both of them forward \(7\) years.
- The equation is \(d+46=2(d+7)\), the 4.3 shape. Collect the variable terms.
Show the full solution
Let \(d\) be the daughter's age now, so Mara is \(d+39\), and in \(7\) years \(d+39+7=2(d+7)\). Then \(d+46=2d+14\), so \(d=\boxed{32}\). Check against the story, in \(7\) years they are \(39\) and \(78\), and \(78\) is twice \(39\).
Problem
A cafe till holds 28 coins, every one a quarter or a dime, and together they are worth 5.05 dollars. How many of the coins are quarters?
Show a hint
- One letter carries both counts. If \(q\) of the coins are quarters, the other \(28-q\) are dimes.
- Multiply the whole equation by \(100\) the 4.4 way, so the hundredths become whole numbers.
Show the full solution
With \(q\) quarters and \(28-q\) dimes, \(0.25q+0.10(28-q)=5.05\). Multiply by \(100\), \(25q+280-10q=505\), so \(15q=225\) and \(q=\boxed{15}\). Check in the story, \(15\) quarters are \(3.75\) dollars and \(13\) dimes are \(1.30\), which is \(5.05\) in all.
Problem
A rectangular parade banner is 5 cm longer than it is wide, and its perimeter is 350 cm. How long is the banner, in centimeters?
Show a hint
- Let \(w\) be the width and write the length in terms of \(w\). The perimeter uses each of them twice.
- \(4w+10=350\) gives the width. That is not what was asked.
Show the full solution
Let \(w\) be the width, so the length is \(w+5\) and \(2(w+w+5)=350\). Then \(4w+10=350\), \(4w=340\), \(w=85\), and the length is \(85+5=\boxed{90}\) cm. Solving gives the width, but the question asks for the length, which is why rereading the question is part of the routine.
A wrong translation solved correctly passes every check against its own equation, because substituting a solution back into the equation it solves always works. The only check that reaches the translation is substituting into the English itself. That is 4.3's check-the-original rule pushed one step earlier, since here the original is the sentence.
Problem
A puzzle card reads, twelve less than six times a number is 138. Dev writes \(12-6n=138\), solves his own equation correctly, and gets \(n=-21\), and a check in that equation passes. Check the story instead. Six times \(-21\), then twelve less, gives \(-138\), not \(138\), so the translation is backwards. Fix it and enter the correct number.
Show a hint
- Twelve less than six times \(n\) starts from \(6n\) and takes twelve away.
- The correct equation is \(6n-12=138\).
Show the full solution
The correct translation is \(6n-12=138\), so \(6n=150\) and \(n=\boxed{25}\). Dev's check passed because any correctly solved equation confirms itself. Only the story can catch the flip.
Problem
A carpenter saws off half of a board, then trims off another three quarters of a meter, and 4.5 meters remain. How long was the board, in meters?
Show a hint
- Let \(x\) be the board's full length in meters and write what each cut takes away from it.
- Multiply \(x-\frac{x}{2}-\frac{3}{4}=4.5\) through by \(4\), the 4.4 move.
Show the full solution
Let \(x\) be the full length, so \(x-\frac{x}{2}-\frac{3}{4}=4.5\). Multiply by \(4\), \(4x-2x-3=18\), so \(2x=21\) and \(x=\boxed{10.5}\) meters. Check in the story, half of \(10.5\) leaves \(5.25\), and trimming \(0.75\) more leaves \(4.5\).
An equation is a claim (4.1), balance moves solve it (4.2), on both sides (4.3), cleanup reaches the disguised ones (4.4). Translation feeds it anything plain English can say. Some stories name two quantities with no fact writing one from the other, so one letter stops being enough. Chapter 5 begins there, with 5.1 Two Variables, Two Equations.
Practice these ideas
Practice
The sum of a number and 97 is 189. Enter the number.
Show the solution
The sentence translates to \(n+97=189\). Subtract \(97\) from both sides, so \(n=\boxed{92}\). Check, \(92+97=189\).
Practice
The difference of a number and 26 is 91. Enter the number.
Show the solution
\(n-26=91\). Add \(26\) to both sides, so \(n=\boxed{117}\). Check, \(117-26=91\). The difference of \(a\) and \(b\) is \(a-b\), in the order named.
Practice
Eight more than seven times a number is 162. Enter the number.
Show the solution
\(7n+8=162\). Subtract \(8\), so \(7n=154\), and divide by \(7\), so \(n=\boxed{22}\). Check, \(154+8=162\).
Practice
Nine less than six times a number is 219. Enter the number.
Show the solution
\(6n-9=219\). Add \(9\), so \(6n=228\), and \(n=\boxed{38}\). Check, \(228-9=219\). English names the nine first, algebra writes it last.
Practice
A florist puts the same number of stems in each of 8 bouquets and has 17 stems left over, from 201 stems in all. How many stems are in each bouquet?
Show the solution
Let \(s\) be the stems in each bouquet. Then \(8s+17=201\), so \(8s=184\) and \(s=\boxed{23}\). Check, \(8\cdot 23=184\), plus the \(17\) left over is \(201\).
Practice
A 102-meter rope is cut into two pieces, one 20 meters longer than the other. How long is the shorter piece, in meters?
Show the solution
Let the shorter piece be \(x\), so the longer is \(x+20\), and \(x+(x+20)=102\). Combine, \(2x+20=102\), so \(2x=82\) and \(x=\boxed{41}\). Check against the story, \(41+61=102\) and \(61\) is \(20\) more than \(41\).
Practice
A drama club has 152 members, with three times as many actors as crew members. How many actors does it have?
Show the solution
Let \(c\) be the crew count, so there are \(3c\) actors. Then \(c+3c=152\), so \(4c=152\) and \(c=38\). The actors number \(3\cdot 38=\boxed{114}\). Check, \(114+38=152\). The letter \(c\) is the crew count, and the question asks for actors, so the final answer is \(3c\), not \(c\).
Practice
Two consecutive integers add to 141. Enter the smaller one.
Show the solution
\(x+(x+1)=141\), so \(2x+1=141\), \(2x=140\), and \(x=\boxed{70}\). Check, \(70+71=141\).
Practice
Three consecutive even integers add to 174. Enter the largest.
Show the solution
\(x+(x+2)+(x+4)=174\), so \(3x+6=174\), \(3x=168\), and \(x=56\). The largest is \(x+4=\boxed{60}\). Check, \(56+58+60=174\).
Practice
A rectangular parking lot is five times as long as it is wide, and its perimeter is 156 meters. How wide is the lot, in meters?
Show the solution
Let \(w\) be the width in meters, so the length is \(5w\). The perimeter is \(2(w+5w)=156\), so \(12w=156\) and \(w=\boxed{13}\). Check, the length is \(65\), and \(2(13+65)=156\).
Practice
A coach is 37 years older than his son. In 10 years he will be exactly twice as old as his son is then. How old is the son now, in years?
Show the solution
Let \(s\) be the son's age now, so the coach is \(s+37\). In 10 years, \(s+47=2(s+10)\), so \(s+47=2s+20\) and \(s=\boxed{27}\). Check, in ten years they are \(37\) and \(74\), and \(74\) is twice \(37\).
Practice
Nine times a number is 136 more than five times the number. Enter the number.
Show the solution
\(9n=5n+136\). Subtract \(5n\), so \(4n=136\) and \(n=\boxed{34}\). Check, \(9\cdot 34=306\) and \(5\cdot 34+136=306\).
Practice
Thirty-six less than half a number is 95. Enter the number.
Show the solution
\(\frac{n}{2}-36=95\). Add \(36\), so \(\frac{n}{2}=131\), and multiply by \(2\), so \(n=\boxed{262}\). Check, half of \(262\) is \(131\), and \(131-36=95\).
Practice
Fair ride tickets cost 2.50 dollars each, plus a flat 2.25 dollars for a wristband. Joon pays 84.75 dollars in all. How many tickets does he buy?
Show the solution
Let \(t\) be the number of tickets. Then \(2.50t+2.25=84.75\). Multiply by \(100\), \(250t+225=8475\), so \(250t=8250\) and \(t=\boxed{33}\). Check, \(33\) tickets cost \(82.50\) dollars, plus \(2.25\) is \(84.75\).
Practice
Aquarium tickets cost 5.50 dollars for adults and 4.25 dollars for students. A group of 35 people pays 185 dollars in all. How many adults are in the group?
Show the solution
Let \(a\) be the adults, so \(35-a\) are students. Then \(5.50a+4.25(35-a)=185\). Multiply by \(100\), \(550a+425(35-a)=18500\), so \(550a+14875-425a=18500\), \(125a=3625\), and \(a=\boxed{29}\). Check, \(29\) adults pay \(159.50\) dollars, \(6\) students pay \(25.50\), and the total is \(185\).
Practice
Three years from now, Ada will be four times as old as she was three years ago. How old is Ada now, in years?
Show the solution
Let \(n\) be Ada's age now. Then \(n+3=4(n-3)\), so \(n+3=4n-12\), \(15=3n\), and \(n=\boxed{5}\). Check, in three years she is \(8\), three years ago she was \(2\), and \(8\) is four times \(2\).
Practice
A worksheet reads, eighteen less than nine times a number is 198. Kai writes \(18-9n=198\), solves it correctly, and gets \(n=-20\), and a check in his own equation passes. The story check fails, since nine times \(-20\), then eighteen less, is \(-198\). Fix the translation and enter the correct number.
Show the solution
\(9n-18=198\). Add \(18\), so \(9n=216\) and \(n=\boxed{24}\). Check, \(9\cdot 24=216\), and eighteen less is \(198\). Kai's check passed because a correctly solved equation always confirms itself, and only the story catches the flip.
Practice
Twice the sum of a number and 71 is 104 less than four times the number. Enter the number.
Show the solution
\(2(n+71)=4n-104\). Distribute, \(2n+142=4n-104\). Subtract \(2n\), then add \(104\), so \(246=2n\) and \(n=\boxed{123}\). Check, twice the sum is \(2\cdot 194=388\), and \(4\cdot 123-104=388\).