Algebra replaces specific numbers with letters so that a single rule can describe infinitely many situations at once. A formula like $A = \pi r^2$ works for a circle of radius $3\text{ cm}$ or radius $300\text{ m}$ — the letter $r$ becomes a specific number only when you substitute a value into it. Almost every other algebra topic you meet — equations, sequences, functions — depends on getting substitution right, especially when negative numbers, fractions, or powers are involved.

Why does one formula work for every possible case?

A letter in a formula is a placeholder, not a fixed unknown value like in an equation. Because $A=\pi r^2$ was derived from the general geometry of any circle, the relationship between $A$ and $r$ holds true no matter which number you substitute for $r$. This is the core idea behind generalisation: instead of memorising "the area of a circle with radius 3 is $9\pi$" as an isolated fact, you learn one rule that instantly answers the question for every circle that could ever be drawn.

What does it mean to "substitute" into a formula?

Substituting means replacing every occurrence of a letter with a given number, then working out the result using the normal order of operations (BIDMAS — brackets, indices, division/multiplication, addition/subtraction). For $P = 2(l+w)$ with $l=7$ and $w=4$:

How do I substitute negative numbers without making sign errors?

Always put the negative value inside brackets before you substitute it — this stops a power or a following operator from attaching to the wrong part of the expression. Given $v = u+at$ with $u=-8$, $a=5$, $t=3$:

What if a formula has more than one letter, or letters raised to a power?

Substitute every letter at the same time, evaluate any powers first (still using brackets around negative values), and only then simplify the top and bottom of a fraction separately — never cancel across the fraction bar before both parts are fully evaluated. For $A = \dfrac{b^2-c}{2b}$ with $b=-3$, $c=4$:

Worked Example 1

Find the value of $D = \dfrac{2m-n^2}{m+n}$ when $m=-4$ and $n=3$.

1. Substitute both values in brackets: $D = \dfrac{2(-4)-(3)^2}{(-4)+3}$
2. Evaluate the power and the multiplication first: $D = \dfrac{-8-9}{-1}$
3. Simplify the numerator, then divide: $D = \dfrac{-17}{-1} = 17$

Worked Example 2

The kinetic energy of a moving object is given by $E = \dfrac{1}{2}mv^2$. Find $E$ when $m = 6$ and $v = -5$.

1. Substitute the negative velocity in brackets, since it is being squared: $E = \dfrac{1}{2}(6)(-5)^2$
2. Evaluate the power first: $(-5)^2 = 25$, so $E = \dfrac{1}{2}(6)(25)$
3. Multiply through: $E = \dfrac{1}{2}(150) = 75$