Expressions Using Letter-Numbers

Shabnam is always 3 years older than Aftab. When Aftab is 10, Shabnam is 13; when Aftab is 18, Shabnam is 21 — the relationship 'Shabnam's age = Aftab's age + 3' stays true no matter what Aftab's actual age is. Writing out this sentence every single time is tiresome, so mathematicians use a letter to stand in for the changing quantity. Let a represent Aftab's age; then Shabnam's age is given by the expression a + 3. A letter used this way to represent a number is called a letter-number, and an expression containing one, like a + 3, is called an algebraic expression.

The real power of this shorthand is that it works for every value of a at once. If Aftab is 23 (a = 23), substitute 23 in place of a: Shabnam's age = 23 + 3 = 26. Try a completely different value, say a = 40: Shabnam's age = 40 + 3 = 43. The same expression a + 3 answers the question correctly every single time, for any age Aftab might be — that's what makes it a formula.

Formulas built from letter-numbers show up constantly in geometry too. The perimeter of a square with side length q is 4 times the side, giving the formula 4q (or, written with the multiplication sign shown, 4 x q). If a particular square has a side length of 7 cm, substituting q = 7 into the formula gives a perimeter of 4 × 7 = 28 cm.

Look at the sequence 4, 8, 12, 16, 20, ... — the multiples of 4. The 3rd term is 4 × 3, and the 29th term is 4 × 29. In general, if n stands for the position of a term in this sequence, the nth term is 4 x n. As a standard convention in algebra, the multiplication sign between a number and a letter-number is dropped entirely, and the number is written first: 4 x n is shortened to 4n. The number placed before the letter-number (here, 4) is called the coefficient.

This shorthand makes algebraic expressions quicker to read once you're used to it, but evaluating them still means substituting a number for the letter and multiplying. To find the value of 7k when k = 4, treat 7k as 7 x k and substitute: 7 × 4 = 28. Similarly, to evaluate 5m + 3 when m = 2: since 5m means 5 x m, substitute m = 2 to get 5 × 2 + 3 = 10 + 3 = 13.

A common slip is forgetting that 5m means 5 times m, not '5 and m written together' or '5 placed next to m with no particular meaning.' Whenever you see a number immediately followed by a letter-number with no symbol between them — like 3a, 12x, or 7k — read it as that number multiplying the letter-number, every time.

Suppose a shop sells 5c + 3c + 10c worth of pencils over three days, where c is the price of one pencil and the numbers are how many pencils were sold each day. Since every term involves the same letter-number c, these are called like terms, and they can be combined using the distributive property: 5c + 3c + 10c = (5 + 3 + 10)c = 18c. If a pencil costs ₹50 (c = 50), the total earned is 18 × 50 = ₹900 — and you can check this matches evaluating the original, unsimplified expression at c = 50 too: 5(50) + 3(50) + 10(50) = 250 + 150 + 500 = 900. Both the original and the simplified expression always give the same value.

Now suppose the same shop also earns 11d from selling erasers, where d is the eraser's price. The combined earnings expression is 18c + 11d. Since c and d are different letter-numbers, 18c and 11d are unlike terms — there is no way to combine them into a single term, so 18c + 11d is already in its simplest form.

A longer expression like 3a + 9b - 6 + 8a - 4b - 7a + 16 is simplified by grouping its like terms together: the a-terms (3a, 8a, -7a) combine to (3 + 8 - 7)a = 4a; the b-terms (9b, -4b) combine to (9 - 4)b = 5b; and the plain numbers (-6, 16) combine to 10. So the fully simplified expression is 4a + 5b + 10. Checking this at a = 2, b = 1: the simplified form gives 4(2) + 5(1) + 10 = 8 + 5 + 10 = 23, and the original expression gives the same value when every term is substituted and added up.

Algebraic expressions follow exactly the same bracket rules you've already learned with plain numbers. Take 2(2x - 3) + 8x + 12. Using the distributive property to remove the bracket: 2(2x - 3) = 2(2x) - 2(3) = 4x - 6. So the full expression becomes 4x - 6 + 8x + 12, and combining like terms (4x + 8x = 12x, and -6 + 12 = 6) gives the simplified form 12x + 6. Checking at x = 5: the original expression gives 2(10 - 3) + 40 + 12 = 2(7) + 52 = 14 + 52 = 66, and the simplified form gives 12(5) + 6 = 60 + 6 = 66 — both match.

Now consider 8h - (5 + 7h) + 9, where the bracket is preceded by a '-' sign. Every term inside the bracket flips sign when it's removed: 8h - (5 + 7h) + 9 = 8h - 5 - 7h + 9. Grouping like terms (8h - 7h = h, and -5 + 9 = 4) gives the simplified expression h + 4. Checking at h = 3: the original gives 24 - (5 + 21) + 9 = 24 - 26 + 9 = 7, and the simplified form gives 3 + 4 = 7 — they match.

This also makes it easy to add or subtract two algebraic expressions. To subtract (9a - 6b + 14) from (6a + 9b - 18), write it as (6a + 9b - 18) - (9a - 6b + 14), then flip every sign inside the second bracket: 6a + 9b - 18 - 9a + 6b - 14. Grouping like terms (6a - 9a = -3a, and 9b + 6b = 15b, and -18 - 14 = -32) gives -3a + 15b - 32.

Picture a row of adjoining squares made with matchsticks, where each new square after the first shares one side (one matchstick) with the square before it. The first square alone needs 4 matchsticks. The second square adds 3 more sticks (since one side is shared), making 7. The third adds 3 more again, making 10. The pattern of totals is 4, 7, 10, 13, ... — each new row-length needs 3 more matchsticks than the one before.

Describing this with algebra: if n is the number of squares in the row, the formula for the total matchsticks needed is 3n + 1. Check it against what we already know: for n = 1, 3(1) + 1 = 4 (correct); for n = 2, 3(2) + 1 = 7 (correct); for n = 3, 3(3) + 1 = 10 (correct). Now the formula lets us jump straight to a large value without drawing anything: for a row of 15 squares, the number of matchsticks needed is 3(15) + 1 = 45 + 1 = 46, found instantly instead of by counting.

A formula can also be used in reverse — to find which position produces a given value. If someone has exactly 100 matchsticks, how many squares could they make in a row? Setting 3n + 1 equal to 100 and solving: 3n = 99, so n = 33 — exactly 33 adjoining squares, with no matchsticks left over. This is the real strength of writing a pattern algebraically: a single formula answers both 'what comes next' and 'which position gives this value,' for every possible case.