PrepShorts · Study sheet · Class 9 Mathematics · Chapter 2, Introduction to Linear Polynomials
Chapter 2 · Introduction to Linear Polynomials
Recovering y = ax + b from two observations
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Two bills arrive and no price list came with them. Hidden behind them are a rate and a fixed fee, and two observations are exactly enough to recover both.
The idea
Two measurements are enough to recover a linear relationship completely, and the reason is a counting argument rather than a clever technique: y = ax + b hides exactly two unknown numbers, and each observation you make supplies exactly one equation about them. Two for two, and the relationship is pinned. That framing also tells you precisely when the method breaks — two readings taken at the same x give you the same equation twice and pin nothing, and if the situation was never linear to begin with the arithmetic still hands you an a and a b, which will fit your two points and be wrong everywhere else.
What you should be able to do
- Explain why a relationship of the form
y = ax + bhas exactly two unknowns - Turn each of two observations into an equation in
aandb - Solve the resulting pair by expressing one unknown in terms of the other and substituting
- Interpret the recovered
aandbin the language of the original situation - Obtain
adirectly as the change inydivided by the change inx, and explain why that shortcut works - Check a recovered relationship by feeding both original observations back through it
- State the two circumstances in which two observations are not enough
- Set up the relationship when the roles of the two quantities are assigned the other way round, as the temperature exercise does
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| linear relationship | a pairing of two quantities expressible as y = ax + b | printed on pp. 22 and 26; §2.5 on p. 26 is titled with it |
| linear equation | an equation of degree 1, here one of the two formed from the observations | printed on p. 20 |
| coefficient | the multiplier a, which is the change in y per unit of x | printed on pp. 16 and 18 |
| constant | the number b, the value y takes when x is zero | printed on p. 16 |
| substitute | to replace a letter with a known value or with another expression | printed on pp. 20 and 26 |
| fixed monthly fee | the part of a bill charged regardless of usage — the situation's name for b | printed on p. 26 in Example 11 |
| slope | the name the chapter later gives to a | printed on p. 31, four pages after this topic's material |
| y-intercept | the name the chapter later gives to b | printed on p. 35 |
| simultaneous conditions | two requirements that one pair of numbers must satisfy at once | an added phrasing; not printed in this chapter, which solves the pair without naming the situation |
Standard Hindi vocabulary.
Where people slip up
- "One reading is enough — divide the bill by the usage." ₹350 over 10 GB gives ₹35 a GB, which contradicts the second reading. That division silently assumes the fixed fee is zero. Show it failing at 20 GB.
- "
aandbare found by trial." They are the solution of two equations that the two observations write down for you. Nothing is guessed. - "The two observations can be any two." They must sit at different values of
x. Two bills for the same 10 GB give one equation twice, and no amount of algebra will separateafromb. - "If I get numbers out, the model was right." The arithmetic always produces an
aand ab. It is the third observation that tests whether the relationship was linear at all. This is why checking a further point matters — and it is what the next topic does with a graph. - "
bis the bigger number,athe smaller." In item 2,a = 60andb = 200; in item 1,a = 25andb = 150; in item 3,ais a fraction under one andbis negative. There is no rule of size, only of role. - "A negative
bmeans something has gone wrong." In item 3 it is correct and meaningful: at 0 °F the Celsius reading is below zero. - "In
°C = a °F + b,amust be 9/5 because that is the formula I know." The exercise has assigned the roles the other way round, soais5/9. Read which quantity the equation solves for before reaching for a remembered constant. - "20 and 150 are just answers." They are the price list. The Think and Reflect box on p. 27 exists to force this reading.
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Worked answers to this chapter’s exercises · this video explains Exercise Set 2.5 Q1, Exercise Set 2.5 Q2, Exercise Set 2.5 Q3, End-of-Chapter Exercises Q10, End-of-Chapter Exercises Q11
Transcript1,436 words
Two bills from the same data plan, and no price list anywhere. The first month: ten gigabytes used, and the bill came to three hundred and fifty. The second month: twenty gigabytes, and the bill was five hundred and fifty. That is everything you have. Nothing but two pairs of numbers. And it is enough. From those two lines you can recover the entire price list, and then answer questions about months that have not happened yet.
Here is why it is enough, and it is a counting argument rather than a trick. Assume the bill depends on the usage in the simplest way there is: y equals a x plus b. x is the gigabytes used. y is the bill. Now count what you do not know. There is a, and there is b. Two numbers, and nothing else. Each observation you make will turn into one equation about those two numbers.
Two unknowns, two equations. That is the whole plan, and it also tells you in advance exactly when the plan will fail. Take the first observation and feed it in. x is ten, y is three hundred and fifty. Three hundred and fifty equals ten a plus b. One equation, two unknowns. That is not a solution. It is a condition. And plenty of price lists satisfy it. Twenty a gigabyte with a fixed fee of a hundred and fifty: ten twenties is two hundred, plus a hundred and fifty, three hundred and fifty. It fits.
Thirty-five a gigabyte with no fixed fee at all: ten thirty-fives is three hundred and fifty. That fits too. Nothing per gigabyte and a flat fee of three hundred and fifty. Also fits. Three completely different price lists, and the first bill cannot tell them apart. So ask what they say about a month you have not used yet. Twenty gigabytes. The first predicts five hundred and fifty. The second predicts seven hundred. The third predicts three hundred and fifty, because under that plan the usage costs nothing.
They agreed at ten and they scatter at twenty. That scattering is what the second bill is for. The second bill was five hundred and fifty. So five hundred and fifty equals twenty a plus b. One of those three survives. The other two are eliminated by a single line of evidence. And notice what the second observation had to do. It had to sit at a different usage. Two equations. Ten a plus b is three hundred and fifty. Twenty a plus b is five hundred and fifty.
Take the first and make b the subject. b is three hundred and fifty minus ten a. Now put that into the second, in place of b. Five hundred and fifty equals twenty a, plus the quantity three hundred and fifty minus ten a. Twenty a minus ten a is ten a. So five hundred and fifty is ten a plus three hundred and fifty. Ten a is two hundred. So a is twenty.
Now go back for b. Three hundred and fifty minus two hundred is a hundred and fifty. The rule is y equals twenty x plus a hundred and fifty. And here is the step it would be easy to skip. Twenty and a hundred and fifty are not answers. They are the price list. The twenty is what one gigabyte costs. Every extra gigabyte adds twenty to the bill. The hundred and fifty is what you pay before you use anything at all. Set x to zero and the whole first term vanishes; a hundred and fifty is what is left.
It is the monthly fee, and it was never written on either bill. Now the payoff. What would fifteen gigabytes cost? Twenty fifteens is three hundred, plus a hundred and fifty, is four hundred and fifty. A number that appears on neither bill, and you can be confident in it. That was the careful route. There is a faster one, and it is worth seeing why it works. Look at the two equations again. Both end in plus b. The same b.
So subtract one from the other and the b disappears without any rearranging. Five hundred and fifty minus three hundred and fifty is two hundred. Twenty a minus ten a is ten a. Two hundred equals ten a, so a is twenty. One line. Then put a back into either equation for b, exactly as before. Written as a single fraction it is the difference in the bills over the difference in the usage. Two hundred over ten.
And that fraction should look familiar. You have met it before. The difference in the output, divided by the difference in the input. Take a pattern of tiles growing one, three, five, seven. Its rule is y equals two x minus one. The gap between consecutive terms is two. The coefficient is also two. That was not a coincidence and it is not one here either. For any rule of this shape, the change in y divided by the change in x is the coefficient, and the constant never appears in the calculation at all.
It cancels, every time, because both readings carry the same one. So the rate per gigabyte and the gap in a sequence are the same quantity, met in two different rooms. Before trusting the rule, feed both observations back through it. Both, not one. Twenty times ten is two hundred, plus a hundred and fifty, is three hundred and fifty. The first bill. Twenty times twenty is four hundred, plus a hundred and fifty, is five hundred and fifty. The second bill.
Both match, so the rule is consistent with everything you know. And you can see why one check is not a check. Thirty-five a gigabyte with no fee also gives three hundred and fifty at ten gigabytes. It passes the first test perfectly and fails the second by a hundred and fifty. Checking one observation only tells you that you did the arithmetic right. Now the failures the counting argument predicted, and there are two of them.
The first. Suppose both your bills are for ten gigabytes. If they agree, the second equation is the first one written twice. It rules nothing out. Every price list that survived the first bill still survives. If they disagree, it is worse. Nothing survives at all. No pair of numbers can make the same usage produce two different bills. So the two readings do not merely have to be two. They have to sit at different values of x.
That is a fact about the observations, not about the algebra. The second failure is more dangerous, because nothing goes wrong visibly. Suppose the machine you are measuring squares its input rather than scaling it. At ten it returns a hundred. At twenty it returns four hundred. Two clean readings. Run them through the method. The difference is three hundred over ten, so a is thirty, and b comes out as minus two hundred.
y equals thirty x minus two hundred. It is a perfectly good rule and it passes both checks. But feed it fifteen and it says two hundred and fifty, while the machine says two hundred and twenty-five. In fact the recovered rule agrees with the machine at exactly the two inputs you measured, and disagrees everywhere else. The arithmetic will always hand you an a and a b. Whether the relationship was ever linear is a separate question, and it takes a third observation to ask it.
One last case, where the setup itself is the trap. Two temperature scales. Ice melts at zero on one and thirty-two on the other. Water boils at a hundred on the first and two hundred and twelve on the second. Suppose you are asked for the rule in this direction: the first scale equals a, times the second scale, plus b. Read that carefully. The second scale is now the input.
So the difference in the outputs is a hundred, and the difference in the inputs is a hundred and eighty. A hundred over a hundred and eighty is five ninths. Five ninths, not nine fifths. The famous number is the one you get going the other way, and which one you get depends entirely on which quantity the rule solves for. And b comes out negative, which is not an error. It says that when the second scale reads zero, the first reads below zero, and it does.
Two unknowns, two readings. Count them first, and the method, its shortcut, and both of its failures all follow from the count.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Linear growth and linear decayClass 9 · Ch 2, Introduction to Linear Polynomials
- A constant difference is the signature of a linear patternClass 9 · Ch 2, Introduction to Linear Polynomials
- What makes a polynomial linear, and the equation you get by fixing its valueClass 9 · Ch 2, Introduction to Linear Polynomials
Comes up again in
- Why two points are enough to draw the lineClass 9 · Ch 2, Introduction to Linear Polynomials
- What a and b do to the line: slope, y-intercept, and parallel familiesClass 9 · Ch 2, Introduction to Linear Polynomials