Pair of Linear Equations in Two VariablesIntroduction

It is mela day in Akhila's village, and she makes straight for her two favourite things: the towering Giant Wheel, and the Hoopla stall, where you toss a ring and win whatever prize it lands around completely.

A ride on the Giant Wheel costs ₹3 and a game of Hoopla costs ₹4. By evening Akhila has spent exactly ₹20, and she remembers one more thing: she played Hoopla half as many times as she rode the wheel. How many rides did she take, and how many games did she play?

Let's test a guess. Suppose she took 6 rides. Then she played Hoopla times, and her bill would be ₹. That is more than she spent, so 6 rides is too many.

Crack the Fair Puzzle

Your turn. Change the number of rides and Hoopla games until both clues light up. Every combination you try leaves a dot on the little grid.

Guessing worked here because the answer is two small whole numbers. Now look at the grid: the blue line runs through every combination that passes clue 1, and the green line through every combination that passes clue 2. Your answer sits at the one spot where the two lines cross. Hold on to that picture. It is the big idea of this whole chapter.

From Clues to Equations

Guessing gets slow once the numbers get big, so let's give the two unknowns names. Let be the number of rides Akhila took, and the number of Hoopla games she played. Each clue now becomes an equation.

Clue 1, Hoopla games are half the rides:     (1)

Clue 2, the bill comes to ₹20:     (2)

In both equations, and appear on their own: no squares, no , nothing under a root or in a denominator. That is what makes them linear equations in two variables. In , the highest power of each variable is 120.

Two linear equations in the same two unknowns, taken together, make a pair of linear equations in two variables. Every such pair can be written in the general form

where the s, s and s are real numbers, with and (so that neither equation loses both of its variables). Akhila's Equation (1) becomes , and her Equation (2) becomes , so , and .

So what have we gained? Two clues have become two tidy equations, and the answer you found by guessing, rides and games, makes both of them true: Equation (1) gives ✓ and Equation (2) gives ✓. Equations like these are the tool for the days when guessing is hopeless, and the next few steps show how they lead straight to the answer — no guessing needed.

Linear or Not?

A quick warm-up before we go on. An equation is linear in and when each variable appears only to the first power: never squared, never multiplied by the other, never under a root or in a denominator. A coefficient like is fine, as long as the variable itself stays plain.

Speed round: get five right to move on.

Every Solution Is a Point

Here is the idea that turns equations into pictures. Take Equation (1), , and try the point (6, 3): put and into and you get — exactly what the equation wants. So (6, 3) is a solution of Equation (1). Now try (2, 2): , not 0, so (2, 2) is not a solution. Every point of the grid either passes or fails each equation — and a pair of numbers is exactly what names a point on a graph.

Drag the purple point around the grid, or tap the grid to make it jump. The two meters under the grid work out and for the point you are holding, and light up when the value is right: 0 for Equation (1), 20 for Equation (2). Find two points that pass Equation (1) and two that pass Equation (2) — the grid joins each pair into a line.

Two points are all a straight line needs — and every point on the blue line passes Equation (1) (try a third one and see). Likewise every point on the green line passes Equation (2). Now find the single point that passes both.

It is (4, 2): 4 rides and 2 games of Hoopla, the same answer you found by guessing, found this time by drawing. A pair of numbers that satisfies both equations is called a solution of the pair, and on a graph it is the point where the two lines cross. In this chapter you will learn to find that point in three ways: by drawing (the graphical method), and by two algebraic methods, substitution and elimination, which work even when the crossing point is far too awkward to read off a graph.