Study Companion: Programme F.5 — Linear Equations
F.3 solved an equation by undoing it, one operation at a time, doing the same thing to both sides so the balance stayed level. This programme meets equations that undoing alone cannot finish: the unknown on both sides, fractions with the unknown underneath, and two or three unknowns tied together by as many equations. Every method in it — collecting terms, multiplying through by the LCM, substitution, elimination — is a legal move, and the one idea that makes them legal is a picture. An equation is a statement about where lines meet. One unknown asks where two lines meet; two unknowns, where two lines cross; three unknowns, where three planes share a point. A legal step redraws the picture without moving the meeting point, and each method is a choice of steps that makes the meeting point readable.
The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in five units instead: one unknown, fractions, two unknowns, three unknowns, and the tidying that turns a messy system into one of those shapes.
How to use this companion
Each unit is an argument: a real situation and a question, the attempt a sensible person would make first, the picture that settles it, a check with numbers, and only then the rule. Read it once front to back. The Narration spine at the end of each unit summarises what the video shows.
Unit 01: A Simple Equation Asks Where Two Lines Meet
The problem
Two phone tariffs. Tariff A costs £10 a month plus 5p a minute; tariff B costs £4 a month plus 8p a minute. For someone who hardly calls, B is cheaper; for someone who calls all day, A is. After m minutes the monthly bills, in pounds, are
At how many minutes do the two tariffs cost the same? That is, for which m is 10+0.05m=4+0.08m?
First attempt
F.3 said: move terms across to collect the unknown on one side. Collecting the numbers on the left and the m terms on the right, a hurried first try gives
Check it: at 46.2 minutes, A costs 10+2.31=£12.31 and B costs 4+3.69=£7.69. Not equal. The 0.05m was moved across without changing its sign. Trial and error does no better: at 100 minutes A costs £15 and B £12; at 300 minutes A costs £25 and B £28. The answer is somewhere between, and every closer guess costs another pair of sums.
The picture
Draw both bills against minutes. Each is a straight line: A starts at £10 and climbs 5p a minute, B starts at £4 and climbs 8p a minute.
At m=0 the gap between them is 10-4=£6. Every minute B climbs 3p more than A, so the gap closes by 3p a minute. The question is a chase: a £6 head start, closed at 3p a minute.
Now do what F.3 called balancing: take the same thing, B's bill 4+0.08m, away from both sides. The right side becomes 0 for every m — B's line is flattened onto the axis. The left side becomes the gap, 6-0.03m, and A's line becomes the gap line:
Both lines were changed by exactly the same amount at every m, so wherever they were equal they are still equal, and nowhere else. That is why "the same to both sides" is legal: it redraws the picture without moving the meeting point. The first attempt was illegal for the same reason — moving 0.05m without its sign subtracts it from one side and adds it to the other, which changes the two lines by different amounts and moves the crossing.
Check it with numbers
The gap is 6-0.03m, and it reaches zero when
Both bills at 200 minutes: A is 10+0.05\times200=£20 and B is 4+0.08\times200=£20. Below 200 minutes B is cheaper; above it A is.
The rule it gives
A linear equation in one unknown involves no power of the unknown higher than the first; the book also calls it a simple equation. To solve one, simplify each side until it has the form
Take cx and b from both sides: ax-cx=d-b, so
Read against the picture: d-b is the difference of the two starting values (the head start, with its sign), and a-c is how fast the gap closes. For the tariffs, a=0.05, b=10, c=0.08, d=4, and x=\frac{4-10}{0.05-0.08}=\frac{-6}{-0.03}=200.
When a=c. Then the two lines are parallel and the gap never changes. If b\ne d they never meet and there is no solution: a tariff of £10 plus 5p a minute never costs the same as £4 plus 5p a minute. If b=d as well, the two sides are the same expression, true for every x — an identity, written with \equiv.
Equations that simplify to simple equations. Some equations do not look simple. Take
Expanding, the left side is (4x^2+13x+3)-(x^2+2x-15)=3x^2+11x+18 and the right side is 3x^2-11x-4. Each side is a parabola, but both have the same x^2 term, 3x^2. Subtract 3x^2 from both sides — the same to both sides again — and what is left is the simple equation
In the picture, two parabolas with the same x^2 coefficient differ by a straight line (22x+22), so they meet exactly once, where that line is zero.
Check by substitution. It is always wise to put the answer back into the original equation, working out each side separately. For x=-1: the left side is (-3)(2)-(4)(-4)=-6+16=10, and the right side is (-2)(-5)=10. Equal, so x=-1 is right.
Worked example
Solve 5(x-1)+3(2x+9)-2=4(3x-1)+2(4x+3).
This is a=11, b=20, c=20, d=2, so x=\frac{2-20}{11-20}=\frac{-18}{-9}=2. Check: left side 5(1)+3(13)-2=42; right side 4(5)+2(11)=42.
Where this shows up
A Celsius reading C is F=1.8C+32 in Fahrenheit. Is there a temperature at which both scales show the same number? Set C=1.8C+32: here a=1, b=0, c=1.8, d=32, and
-40\ ^\circC is -40\ ^\circF. Look at any thermometer with both scales, such as an outdoor or a freezer thermometer: the two -40 marks sit side by side, the one place where the two lines meet. The tariffs are the same calculation in pounds: at 200 minutes a month the bills are equal at £20, and the rule tells a heavy caller exactly where A starts to pay off.
Narration spine. Draw the two tariff lines, A starting higher and B climbing faster. Try the hurried transposition, m=46, and show the two unequal bills there. Sweep the minutes and watch the gap between the lines shrink. Subtract B from both sides: B's line flattens onto the axis, A's becomes the gap line, and the meeting point stays at 200. Build x=\frac{d-b}{a-c} from the head start and the closing rate, then turn B's slope down to A's and watch the crossing run away to infinity.
Unit 02: Clearing Fractions Changes the Unit, Not the Answer
The problem
A water tank has two inlet taps. Tap A alone fills it in 3 hours; tap B alone fills it in 6 hours. With both taps open, how long does it take?
Call the time x hours. In one hour both taps together fill \frac1x of the tank, A fills \frac13 of it and B fills \frac16, so
First attempt
Average the times: \frac{3+6}2=4.5 hours. Or flip each fraction and add: x=3+6=9 hours. Both are impossible. Tap A on its own fills the tank in 3 hours, and opening B as well can only help, so the answer must be less than 3 hours. The equation has the unknown underneath a fraction bar, and neither guess treats it as a fraction.
The picture
Draw the tank as a bar cut into sixths: 6 is the lowest common multiple (LCM) of 3 and 6, the smallest piece both fractions are made of.
A fills 2 strips an hour, B fills 1 strip an hour: 3 strips an hour, 6 strips to fill \to 2 hours.
In strips, nothing is a fraction any more. A third of a tank is 2 strips, a sixth is 1 strip, and a whole tank is 6 strips. Counting strips over x hours:
That is exactly what multiplying every term of the equation by 6x produces: 6x\times\frac1x=6, 6x\times\frac13=2x and 6x\times\frac16=x. Multiplying through by the LCM changes the unit — from tanks to strips — and so every term has to be multiplied: each one must be recounted in the new unit. Leave one term out, and it is still counted in tanks while the others are in strips.
The x in the multiplier is there because the unknown is itself in a denominator. Multiplying by x is safe only if x\ne0; but x=0 was never allowed, since \frac1x has no meaning there.
Check it with numbers
6=3x, so x=2 hours: less than 3, as it must be. In 2 hours tap A fills \frac23 of the tank and tap B fills \frac26=\frac13, and \frac23+\frac13=1 tank.
The rule it gives
Algebraic fractions. When a simple equation contains fractions, first clear them: multiply every term on both sides by the LCM of the denominators, cancel, and solve the simple equation that is left. F.1 found LCMs of numbers (Unit 05) and F.2 of algebraic denominators (Unit 06).
A numeric LCM. Solve \frac{x+2}2-\frac{x+5}3=\frac{2x-5}4+\frac{x+3}6. The LCM of 2, 3, 4, 6 is 12:
The minus sign in front of \frac{x+5}3 belongs to the whole numerator, so it becomes -4(x+5)=-4x-20. Keep each numerator in brackets until it is multiplied out.
An algebraic LCM. When the denominators contain x, the LCM is the product of the different factors, such as x(x-3)(x-5). Multiplying out usually leaves x^2 terms on both sides, and — as in Unit 01 — they cancel to leave a simple equation.
Beyond the book: excluded values. Before multiplying, note every value of x that makes a denominator zero; those values can never be solutions. If the answer turns out to be one of them, the equation has no solution. Solve \frac{x}{x-2}=\frac2{x-2}+3: multiplying by x-2 gives x=2+3(x-2), so x=2 — but x=2 makes both denominators zero. Multiplying by x-2 multiplied by zero there, and manufactured an equality the original never had. There is no solution. (None of the book's examples produces an excluded value, but its exercises can.)
Worked example
Solve \frac4{x-3}+\frac2x=\frac6{x-5}. Exclude x=0, 3 and 5. Multiply every term by the LCM x(x-3)(x-5):
The 6x^2 terms cancel: 30=18x, so x=\frac{30}{18}=\frac53, which is not excluded. Check: \frac4{5/3-3}=\frac4{-4/3}=-3 and \frac2{5/3}=\frac65=1.2, so the left side is -1.8; the right side is \frac6{5/3-5}=\frac6{-10/3}=-1.8.
Where this shows up
A magnifying glass of focal length f=10 cm obeys the lens formula \frac1f=\frac1u+\frac1v, where u is the distance to the object and v the distance to the sharp image. Hold the glass u=30 cm from a desk lamp: where should the paper go? From \frac1{10}=\frac1{30}+\frac1v, multiply every term by the LCM 30v:
Try it: 15 cm behind the glass, a small sharp image of the lamp appears on the paper, upside down. The tank is just as easy to test with a sink and two taps of different flow, and the rule is the same one: together, the rates add, \frac13+\frac16=\frac12 of a tank an hour.
Narration spine. Show the tank and the two taps. Offer the average, 4.5 hours, and strike it: A alone takes only 3. Cut the tank into six strips. Run the clock: each hour A fills two strips and B one, and the tank is full at 2 hours. Then multiply the equation by 6x and watch each term turn into strips, 6=2x+x, and x=2.
Unit 03: Elimination Swings a Line About the Crossing
The problem
On a kitchen scale, 3 mangoes and 2 apples weigh 1.6 kg. On a pan balance, 4 mangoes balance 3 apples and a 1 kg weight. Assuming the mangoes all weigh the same, and so do the apples, how much does each weigh?
Measure in hundreds of grams, and let a mango weigh x and an apple y:
The second comes from the balance, 4x=3y+10, with 3y taken from both sides.
First attempt
Use the kitchen scale alone. A 400 g mango and a 200 g apple fit (1). So do a 200 g mango and a 500 g apple, and so does "no mango, 800 g apples". One equation in two unknowns has infinitely many solutions: choose any x, and (1) gives a y. F.4 drew exactly this: the graph of an equation is every one of its solutions at once (F.4 Unit 01), and here it is a straight line.
So use both. Adding them gives 7x-y=26; subtracting gives -x+5y=6. Both still contain two unknowns. Combining the equations as they stand does not remove a letter.
The picture
Plot both equations. Each is a line, the line of all its solutions; the pair of weights must lie on both, so it is the point where the lines cross, (4,2).
Two facts about this picture turn it into a method.
Scaling an equation does not move its line. 3(3x+2y)=3\times16, that is 9x+6y=48, is true at exactly the same points as 3x+2y=16. The numbers change; the line does not.
Adding two equations gives a line through the crossing. At the crossing both equations are true, so their sum is true there too. Any combination "3\times(1) plus s\times(2)",
is a line through (4,2). As s grows from 0, the line swings about the crossing like a door on its hinge. At s=2 the y-coefficient 6-3s is zero, and the line is standing upright:
A vertical line through the crossing says x=4 outright. Elimination is choosing the multiples that swing a line upright (or flat), so that one coordinate of the crossing can be read off.
Check it with numbers
The book's rule chooses those multiples at once: multiply each equation by the coefficient of y in the other one. The y-coefficient in (2) is 3 and in (1) it is 2:
The y terms are now the same size with opposite signs, so add: 17x=68, x=4. Put x=4 in (1): 12+2y=16, y=2. Check in the other equation, (2): 16-6=10. A mango weighs 400 g and an apple 200 g.
The rule it gives
Simultaneous linear equations with two unknowns. One linear equation in x and y has infinitely many solutions, a whole line of them. Two such equations may have just one pair of values that satisfies both simultaneously: the crossing of their lines.
Solution by elimination.
- Multiply each equation by the coefficient of the chosen unknown in the other equation, so that its two terms have the same size.
- If those terms have opposite signs, add the equations; if they have the same sign, subtract one from the other. Either way the unknown vanishes.
- Solve for the remaining unknown, substitute into either original equation to find the other, and check in the remaining original.
Solution by substitution. Solve one equation for one unknown, substitute that expression into the other equation, solve, then substitute back. The expression often contains a fraction. For 5x+2y=14 (1) and 3x-4y=24 (2): from (1), y=7-\frac{5x}2. Into (2):
In (1): 20+2y=14, y=-3. Check: 20-6=14 and 12+12=24.
The two methods find the same crossing. Substitution slides along line (1) — y=7-\frac{5x}2 is that line, one point for each x — until it reaches line (2). Elimination swings a line about the crossing until it points straight at the answer. Substitution is quickest when one unknown already has a coefficient of 1; elimination avoids fractions otherwise.
Beyond the book: when the lines do not cross. The book says the pair may have one solution. If the lines are parallel, they never cross, and elimination removes both unknowns at once, leaving something false: 3x+2y=16 and 6x+4y=20 give, from (1)\times2-(2), 0=12. No solution. If the two equations are one line written twice, such as 3x+2y=16 and 6x+4y=32, elimination leaves 0=0, and every point of the line is a solution.
Worked example
Solve 7x-4y=23 (1) and 4x-3y=11 (2). Eliminate y: multiply (1) by 3 and (2) by 4.
The y terms have the same sign, so subtract: 5x=25, x=5. In (2): 20-3y=11, y=3. Check in (1): 35-12=23.
Where this shows up
A 12 V battery with a 3\ \Omega resistor and a 7 V battery with a 1\ \Omega resistor both drive current into a shared 2\ \Omega resistor. With branch currents I_1 and I_2 (in amperes), the shared resistor carries I_1+I_2, and Kirchhoff's voltage law round each loop gives
Eliminate I_2: (1)\times3 gives 15I_1+6I_2=36 and (2)\times2 gives 4I_1+6I_2=14; same sign, so subtract: 11I_1=22, I_1=2 A. Then 5(2)+2I_2=12, I_2=1 A, and the shared resistor carries 3 A. Check in loop 2: 2(2)+3(1)=7 V. The fruit is the version to try at home: any kitchen scale and two kinds of fruit give an equation like (1), and it takes a second, different weighing to pin down both weights.
Narration spine. Write the scale's equation and slide a point along its line: every point fits. Draw the balance's line; the weights are where the two cross. Scale equation (1) by three and watch its line stay put. Then add growing multiples of (2): the line swings about the crossing until it stands upright at x=4, the y-coefficient reading zero. Finish with y=2 and the check.
Unit 04: Three Unknowns Need the Same Letter Eliminated Twice
The problem
Every food label gives the energy in kilocalories next to the grams of fat, carbohydrate and protein. The energy is not measured by burning each food; it is calculated from a fixed energy per gram of each. Could those three numbers be recovered from three labels?
Take three portions (their numbers made round for the working):
| portion | fat (g) | carbohydrate (g) | protein (g) | energy (kcal) |
|---|---|---|---|---|
| 1 | 2 | 3 | 1 | 34 |
| 2 | 1 | 2 | 3 | 29 |
| 3 | 3 | 1 | 2 | 39 |
With f, c and p the kilocalories per gram of fat, carbohydrate and protein:
First attempt
Use Unit 03. Eliminate p from (1) and (2): (1)\times3-(2) gives
Now take another pair, (1) and (3), and eliminate whatever looks easiest — say c: (3)\times3-(1) gives 7f+5p=83. Two equations, but (4) is about f and c and the new one is about f and p. Between them they still hold all three unknowns, and no combination of the two removes two letters at once. Stalled.
The picture
A linear equation in three unknowns is a plane in three-dimensional space, with axes f, c and p. Two planes meet in a line, and the solution is the one point shared by all three planes.
Eliminating p from (1) and (2) gives (4), an equation with no p in it. Every point on the line where planes (1) and (2) meet satisfies (4), whatever its height p. So (4) is that line seen from above, looking down the p-axis onto the (f,c) floor. Eliminate p again, from a different pair, and the line where two other planes meet lands on the same floor:
Two lines on one floor: that is Unit 03, and they cross. Eliminating c from the second pair instead dropped the other line onto the (f,p) wall — a different view, and a line on the floor cannot be crossed with a line on the wall. That is why the same unknown must be eliminated from both pairs. Each elimination drops one dimension, and both results must land in the same flat picture.
Check it with numbers
(1)\times2-(3): 4f+6c+2p-(3f+c+2p)=68-39, so
Solve (4) and (5) as in Unit 03: (5)\times5-(4) gives 25c-7c=145-73, so 18c=72 and c=4. From (5), f=29-20=9. Climb back up to the third unknown using an original equation: in (1), 18+12+p=34, so p=4.
Check in the two originals not used for that last step: (2): 9+8+12=29; (3): 27+4+8=39. Fat carries 9 kcal per gram, carbohydrate and protein 4 each.
The rule it gives
Simultaneous linear equations with three unknowns — "just an extension of the work with two unknowns":
- Take one pair of the equations and eliminate one unknown, giving (4).
- Take a different pair and eliminate the same unknown, giving (5). An original equation that already lacks that unknown can serve as one of the pair directly.
- Solve (4) and (5) as two equations in two unknowns.
- Substitute both values into one original equation to find the third unknown.
- Check by substituting all three values into the other two original equations.
The book calls the combining step solving by equating coefficients. The working is longer than for two unknowns, but no step is new. As with two unknowns, three planes need not share exactly one point — beyond the book, they can meet along a line or not all together — and elimination then ends in 0=0 or in something false.
Worked example
Solve 3x+2y-z=19 (1), 4x-y+2z=4 (2), 2x+4y-5z=32 (3). Eliminate z twice.
(1)\times2+(2): 6x+4y-2z+4x-y+2z=38+4, so 10x+3y=42 (4).
(1)\times5-(3): 15x+10y-5z-2x-4y+5z=95-32, so 13x+6y=63 (5).
(4)\times2-(5): 20x-13x=84-63, so 7x=21 and x=3. In (4): 30+3y=42, y=4. In (2): 12-4+2z=4, z=-2.
Check in (1): 9+8+2=19. In (3): 6+16+10=32.
Where this shows up
The numbers 9, 4 and 4 kcal per gram are the conversion factors food labels in the UK and EU actually use (fibre and alcohol have factors of their own). Check a label: 10 g of fat, 30 g of carbohydrate and 5 g of protein should be listed at about 90+120+20=230 kcal, give or take rounding and fibre.
The same elimination finds currents. A ladder circuit of three loops side by side, every resistor 2\ \Omega, with an 8 V source in the first loop and a 4 V source in the middle one, has loop currents I_1, I_2, I_3 (amperes) satisfying
Equation (3) already has no I_1, so it is one of the pair. Eliminate I_1 from the other pair: (1)+(2)\times2 gives 10I_2-4I_3=16 (4). Then (4)+(3): 8I_2=16, I_2=2 A; from (3), I_3=1 A; from (1), 4I_1=12, I_1=3 A. Check in (2): -6+12-2=4 V.
Narration spine. Show the three portions and their equations. Eliminate p from the first pair and c from the second, and strike the mismatch: one line on the floor, one on the wall. Redo it, eliminating p twice, and draw both lines on the (f,c) floor: they cross at (9,4). Climb back to p=4 and check in the two unused equations: 9, 4, 4, the numbers on every food label.
Unit 05: Tidying an Equation Never Moves Its Line
The problem
A hill walk goes x km uphill at 3 km/h and then y km on the level at 4 km/h, and takes 2 h 15 min. Coming back the same way — 4 km/h on the level, 6 km/h downhill — takes 1 h 45 min. How long is each part of the route?
Time is distance over speed, so, in hours,
First attempt
Clear the fractions by multiplying by 12: (1) becomes 4x+3y on the left — and, in a hurry, the right side is left alone:
With (2) correctly multiplied by 12, 2x+3y=21, subtracting gives 2x=\frac94-21 and x=-9.4 km. A negative distance uphill. One term was not multiplied, and the answer is nonsense.
The picture
Plot both equations exactly as they stand, fractions and all. Each is still a straight line: every term is a number times x, a number times y, or a plain number.
Multiplying every term of (1) by 12 gives 4x+3y=27, which holds at exactly the same points: the same line (Unit 03). Multiplying only the left side gives 4x+3y=\frac94, a parallel line much nearer the origin — the line has moved, and so has the crossing. That is the whole rule for tidying: expanding brackets, collecting terms and clearing fractions are legal precisely because they leave every line where it was. Only the way the line is written changes, until the system has the plain shape of Unit 03.
Check it with numbers
(1)\times12: 4x+3y=27 (3). (2)\times12: 2x+3y=21 (4). The y terms have the same sign, so subtract: 2x=6, x=3. In (4): 6+3y=21, y=5. So 3 km uphill and 5 km on the level.
Check in the original times: out, \frac33+\frac54=1+1.25=2.25 h, which is 2 h 15 min; back, \frac36+\frac54=0.5+1.25=1.75 h, which is 1 h 45 min.
The rule it gives
Pre-simplification. Sometimes the given equations must be simplified before substitution or elimination can be applied. Expand the brackets and collect terms; clear fractions by multiplying each equation, every term on both sides, by its own LCM. Then solve as before.
Brackets. 2(x+2y)+3(3x-y)=38 and 4(3x+2y)-3(x+5y)=-8 expand to 11x+y=38 (3) and 9x-7y=-8 (4). (3)\times7+(4): 86x=258, so x=3 and y=38-33=5. Check in (4): 27-35=-8.
Answers need not be whole numbers. The crossing of two lines is wherever it is; the examples of Unit 03 were chosen to land on whole numbers, and most real systems do not (see the worked example). Leave the answer as an exact fraction.
Beyond the book: new variables. Some systems are not linear in the unknowns as written, but are linear in something else. \frac2x+\frac3y=2 and \frac4x-\frac3y=1 contain \frac1x and \frac1y, never x or y alone. Write u=\frac1x and v=\frac1y: then 2u+3v=2 and 4u-3v=1, a system of the Unit 03 shape. Adding, 6u=3, u=\frac12, v=\frac13, so x=2 and y=3. The book's Further problems use the same idea with u=\frac1{x+8y}.
Worked example
Solve
(1) has LCM 20: 4(2x-1)+2(x-2y)=5(x+1), so 8x-4+2x-4y=5x+5, and
(2) has LCM 12: 4(3y+2)+6(4x-3y)=3(5x+4), so 12y+8+24x-18y=15x+12, and
(3)\times9: 45x-36y=81. (4)\times5: 45x-30y=20. Subtract: -6y=61, so y=-\frac{61}6. In (3): 5x=9+4y=9-\frac{244}6=-\frac{190}6, so x=-\frac{19}3.
On a graph the crossing is at about (-6.333,-10.167), nowhere near a grid point, which is why it is found by algebra and not read off.
Where this shows up
Two unknown resistors R_1 and R_2 measure 2.4\ \Omega in parallel. With a second copy of R_2 also in parallel, the three measure 1.5\ \Omega. Parallel resistances add as reciprocals, so
Not linear in R_1 and R_2 — but linear in the conductances G_1=\frac1{R_1} and G_2=\frac1{R_2}, in siemens: G_1+G_2=\frac5{12} and G_1+2G_2=\frac23. Subtracting, G_2=\frac23-\frac5{12}=\frac14 S, so R_2=4\ \Omega; then G_1=\frac5{12}-\frac3{12}=\frac16 S, so R_1=6\ \Omega. Check: \frac{6\times4}{6+4}=2.4\ \Omega. Choosing conductance instead of resistance is choosing the variable that makes the lines straight. The walk is checkable with any walking app: it reports distance and time for each part of a route, and two timed trips over the same ground give a system like this one.
Narration spine. Show the route, uphill then level, with its two times. Write the two equations with their fractions and plot them as they stand: two lines crossing. Multiply only the left side of (1) by 12 and watch its line jump towards the origin — struck out. Multiply every term: the line does not move. Subtract the tidy equations to reach x=3, y=5, and mark 3 km uphill and 5 km level on the route.