Chapter 4.4 — Exercise 4.3 — Reducible Equations

The Substitution Trick, Demonstrated

For 5/(x−1) − 4/(y+1) = −1 and 2/(x−1) − 1/(y+1) = 5, let a = 1/(x−1) and b = 1/(y+1):

This is now an ordinary linear pair. Solving (rearrange the second to b = 2a − 5, substitute into the first): 5a − 4(2a − 5) = −1 → 5a − 8a + 20 = −1 → −3a = −21 → a = 7, then b = 2(7) − 5 = 9. Converting back to x and y:

• a = 7 → 1/(x−1) = 7 → x − 1 = 1/7 → x = 8/7
• b = 9 → 1/(y+1) = 9 → y + 1 = 1/9 → y = −8/9

Every problem in this exercise follows this exact three-stage shape: substitute to linearise, solve the linear pair for a and b, then convert a and b back into x and y. It's worth double-checking the final answer against the very first equation, not just the linear pair in a and b — a correct value of a and b guarantees a correct linear solution, but a small slip while inverting back to x or y (forgetting to flip a fraction, say) can still creep in at the very last step.

Eight Pairs, Each Reduced a Different Way

The substitution itself changes shape depending on what's being divided — sometimes 1/x and 1/y directly, sometimes 1/(x+y) and 1/(x−y), and in one case 1/√x and 1/√y.

The third row is worth a closer look: the denominators are √x and √y, not x and y directly, so converting back at the end means squaring rather than inverting — a = 1/√x = 1/2 gives √x = 2, so x = 4, not 1/2. Missing that one difference between "reciprocal" and "reciprocal of a square root" is the most common slip in this exercise.

The fourth row hides a division step before the substitution can even begin: 6x + 3y = 6xy isn't written in terms of fractions at all until both sides are divided by xy, turning it into 6/y + 3/x = 6 — only then does the usual a = 1/y, b = 1/x substitution apply. Recognising that an equation is secretly reducible, before ever choosing a substitution, is genuinely the harder half of this exercise — the algebra afterward is routine once the right a and b have been named.

The second row is unusual for a different reason: it starts as (x+y)/xy and (x−y)/xy rather than two separate fraction terms, but splitting each numerator over the shared denominator xy turns it into x/xy + y/xy and x/xy − y/xy, which simplify to 1/y + 1/x and 1/y − 1/x — the same a = 1/y, b = 1/x pattern as row six, just arrived at through a different opening move.

A Boat Against the Current

A boat travels 30 km upstream and 44 km downstream in 10 hours; 40 km upstream and 55 km downstream in 13 hours. With x as the boat's still-water speed and y as the stream's speed, upstream speed is x − y and downstream speed is x + y, so time = distance ÷ speed gives:

Substituting a = 1/(x−y) and b = 1/(x+y) and solving gives a = 1/5, b = 1/11 — so x − y = 5 and x + y = 11, giving x = 8 km/h (boat in still water) and y = 3 km/h (the stream). Adding the two resulting equations together (x−y=5 and x+y=11) cancels y directly and leaves 2x=16, a clean finish that avoids a second substitution entirely — the same shortcut works whenever the last step lands on an x+y/x−y pair like this one.

It's worth noticing why upstream and downstream speeds work out to x−y and x+y in the first place: travelling upstream, the current works against the boat, so its effective speed is reduced by the stream's speed; travelling downstream, the current pushes along with the boat, so its effective speed is increased by the same amount. That single physical fact is what turns a two-unknown speed problem into exactly the reciprocal-equation shape this exercise is built around.

Two More Rate Problems

The same reciprocal-of-a-rate idea solves two further problems built around combined speeds and combined work:

• Train and car: Rahim covers 600 km total. 120 km by train plus the rest by car takes 8 hours; 200 km by train plus the rest by car takes 8 hours 20 minutes. Setting up 120/x + 480/y = 8 and 200/x + 400/y = 25/3, then substituting a = 1/x, b = 1/y, gives train speed 60 km/h and car speed 80 km/h.
• Women and men working together: 2 women and 5 men finish an embroidery job in 4 days; 3 women and 6 men finish it in 3 days. With x and y as the number of days one woman and one man would each take alone, 2/x + 5/y = 1/4 and 3/x + 6/y = 1/3 reduce, after substitution, to 18 days for one woman alone and 36 days for one man alone.

In both cases, the substitution variable represents a rate — distance per unit time, or work per day — which is exactly why the reciprocal of the rate (time, or days) is what the original equation was really asking about all along. The work-rate problem in particular relies on a habit worth remembering beyond this one exercise: "1 woman completes 1/x of the job per day" is the natural way to express a rate when the only given information is a total number of days for the whole job, and adding several people's daily rates together is what makes the combined-work equation possible in the first place.

The Chapter, Complete

This exercise closes Pair of Linear Equations by showing that "linear" is often a matter of how an equation is written, not what it fundamentally describes — a boat's speed problem looks nonlinear until upstream and downstream times are separated into their own reciprocal terms, at which point Exercise 4.2's substitution and elimination methods take over unchanged. For the ratio-based classification underlying every pair solved across this chapter, see the chapter introduction, where the whole classification system was established from scratch. The next chapter, Quadratic Equations, moves on from equations that are only linear in disguise to equations that genuinely aren't linear at all, however they're written.