Finding the Unknown

Picture an old-fashioned two-pan weighing scale, perfectly balanced, with some weights on each side. If exactly the same weight is lifted off both pans at once, or exactly the same weight is added to both pans at once, the scale stays balanced. Nothing about this needs proof: it is simply what balance means. This one everyday fact turns out to be the entire foundation for solving equations. An equation is the mathematical version of a balanced scale: two expressions, the Left Hand Side and the Right Hand Side, joined by an equal sign, claiming they weigh exactly the same. Since performing the same operation on both sides of a true equation cannot break the balance, this becomes a completely safe, always-true move: add the same amount to both sides, subtract the same amount from both sides, multiply both sides by the same number, or divide both sides by the same non-zero number. Every method in this chapter is really just this one balance-scale idea, applied again and again.

A matchstick pattern of triangles has 3 sticks in step 1, 5 sticks in step 2, 7 sticks in step 3, and so on: the nth step needs 2n + 1 sticks. Someone wants to build a step in this pattern using exactly 99 sticks. Which step number would that be? The question becomes: for which value of n does 2n + 1 equal 99? Writing this down gives 2n + 1 = 99, a statement that two expressions, 2n + 1 and 99, are equal. A statement like this, claiming two algebraic expressions are equal, is called an equation. In 2n + 1 = 99, the left side, 2n + 1, is called the Left Hand Side (LHS), and the right side, 99, is called the Right Hand Side (RHS). Finding the value or values of the letter-number that make LHS actually equal to RHS is called solving the equation.

One honest way to solve 2n + 1 = 99 is to guess a value for n and check. Try n = 10: the LHS becomes 21, far too low. Try n = 40: the LHS becomes 81, closer. Try n = 49: the LHS becomes 99, exactly right. This trial and error method always works eventually, but it can take a very long time, especially when the numbers involved are large or awkward, or when there is no obvious starting guess. A faster, systematic method is needed, one that works out the answer directly instead of hunting for it.

Before applying the balance idea to an unknown, it helps to see it work on numbers everyone already trusts. Suppose it is known that 200 + 350 - 40 = 510. What is the value of 200 + 350? There is no need to add 200 and 350 directly: since addition and subtraction are inverse operations, subtracting -40 from both sides of the known equation removes it from the left and leaves 200 + 350 = 510 + 40 = 550. Or suppose it is known that 12 x 15 x 6 = 1080. What is 12 x 15? Multiplication and division are inverse operations, so dividing both sides of the known equation by 6 gives 12 x 15 = 1080 divide 6 = 180. In both cases, nothing new or risky is happening: an operation is being undone on one side by applying its inverse to both sides, and the balance is never broken. This is exactly the tool needed to peel operations away from an unknown letter-number until it stands alone.

Solve 6x - 4 = 20. The unknown x has two things done to it: it is multiplied by 6, then 4 is subtracted. To undo this and leave x alone, work backwards: undo the subtraction first, then undo the multiplication. Add 4 to both sides: 6x - 4 + 4 = 20 + 4, which gives 6x = 24. Now divide both sides by 6: 6x divide 6 = 24 divide 6, which gives x = 4. Check by substituting back into the original equation: 6 x 4 - 4 = 24 - 4 = 20, which matches the RHS exactly, so x = 4 is confirmed correct. Now consider an equation with the unknown on both sides: 5y + 8 = 3y + 20. Subtract 3y from both sides first, to gather every y term onto one side: 5y + 8 - 3y = 3y + 20 - 3y, giving 2y + 8 = 20. From here it proceeds exactly as before: subtract 8 from both sides to get 2y = 12, then divide both sides by 2 to get y = 6. Having the unknown on both sides looks harder at first, but it only takes one extra balance-scale step before the method is identical.

A print shop charges a fixed design fee of ₹80 plus ₹15 for every poster printed. A customer's total bill comes to ₹350. How many posters were printed? Let p stand for the number of posters. The total cost is 80 + 15p, and this must equal 350, giving the equation 80 + 15p = 350. Subtracting 80 from both sides gives 15p = 270, and dividing both sides by 15 gives p = 18. So 18 posters were printed, and it is worth checking: 80 + 15 x 18 = 80 + 270 = 350, which matches. A second kind of problem compares two changing amounts. Two savers start with different amounts and save different amounts every month: saver A starts with ₹3000 and saves ₹450 a month, saver B starts with ₹3900 and saves ₹360 a month. After how many months will they have equal savings? Let m be the number of months. Saver A has 3000 + 450m rupees, saver B has 3900 + 360m rupees, and setting them equal gives 3000 + 450m = 3900 + 360m. Subtracting 360m from both sides gives 3000 + 90m = 3900, subtracting 3000 from both sides gives 90m = 900, and dividing both sides by 90 gives m = 10. After 10 months, both savers have the same amount.

A jar holds a total of 84 marbles shared between two children. One child has 20 marbles more than the other. How many marbles does each child have? This problem seems to have two unknowns: the amount each child has. Call the smaller amount y, so the child with more has y + 20. Writing both amounts this way, using only ONE letter-number, means the total-of-84 condition becomes a single equation: y + (y + 20) = 84. This simplifies to 2y + 20 = 84, then 2y = 64 (subtracting 20 from both sides), then y = 32 (dividing both sides by 2). So one child has 32 marbles and the other has 32 + 20 = 52 marbles, and indeed 32 + 52 = 84. The key trick with two unknowns is to describe the second one in terms of the first, so that only one letter-number ever needs to be solved for.

Forming expressions with symbols and solving equations for an unknown was a major achievement of ancient Indian mathematics, called bijaganita: bija means seed, since the answer to a problem lies hidden inside the unknown number the same way a whole tree lies hidden inside a seed, and solving the problem means helping it grow out, step by step. Around 628 CE, the mathematician Brahmagupta explained in his book Brahmasphutasiddhanta how to add, subtract and multiply unknown quantities using symbols, in much the same spirit as x and y are used today, and gave a direct formula for solving any equation of the form Ax + B = Cx + D. Roughly two centuries later, these ideas travelled to the mathematician Al-Khwarizmi, working in present-day Iraq, whose book on the subject, written around 825 CE, was later translated into Latin and carried the term al-jabr into Europe: this eventually became the English word algebra. Long before either of them, the Bakhshali manuscript, one of the oldest surviving Indian mathematical texts (roughly 300 CE), already contains problems that are solved by exactly this kind of equation-based reasoning.

Ask someone to think of any number, double it, add 10, divide the result by 2, then subtract the original number they thought of. Whatever number they started with, the answer always comes out as 5. This can feel like real magic, but algebra removes the mystery completely. Let the starting number be x. Doubling gives 2x. Adding 10 gives 2x + 10. Dividing by 2 gives x + 5, since both 2x and 10 divide by 2 separately. Subtracting the original number x gives (x + 5) - x, which is simply 5, whatever x was. The trick is not magic at all: it is an equation-building machine in disguise, carefully designed so that the x always cancels itself out at the very last step, leaving the same plain number every single time.