🔌 Substitution: Plugging Values into Expressions 🎯
Substitution means replacing a letter with a number. It's like a machine — you put a number in, follow the expression's instructions, and get an answer out!
🔌 Replace the Letter
If x = 4, find 3x + 2 → 3(4) + 2 = 12 + 2 = 14
🔢 Two Variables
If a = 3, b = 5, find 2a + b → 2(3) + 5 = 6 + 5 = 11
📐 Formulae
Area = l × w. If l = 6, w = 4 → A = 6 × 4 = 24
The Golden Rules:
Rule 1 — Replace ALL: Every letter in the expression must be replaced with its value.
Rule 2 — Multiply notation: 3x means 3 × x. So if x = 5, then 3x = 3 × 5 = 15.
Rule 3 — Brackets: After substituting, use BIDMAS. Do powers first, then multiply/divide, then add/subtract.
Rule 4 — Negative values: Use brackets when substituting negatives. If x = −2, then 3x = 3 × (−2) = −6.
Rule 5 — Squared terms: x² means x × x. If x = 4, then x² = 4 × 4 = 16 (NOT 4 × 2 = 8!).
📝 Worked Examples
Example 1: Simple Substitution
Question: Find the value of 4x − 3 when x = 5.
Replace x with 5 → 4(5) − 3
4 × 5 = 20
20 − 3 = 17
Answer: 17
Example 2: Two Variables
Question: Find 3a + 2b when a = 4 and b = 6.
Replace: 3(4) + 2(6)
3 × 4 = 12 2 × 6 = 12
12 + 12 = 24
Answer: 24
Example 3: Squared Term
Question: Find n² + 3n − 1 when n = 3.
Replace: (3)² + 3(3) − 1
3² = 9 3(3) = 9
9 + 9 − 1 = 17
Answer: 17
Example 4: Using a Formula
Question: The formula for speed is s = d ÷ t. Find s when d = 120 and t = 3.
Substitute: s = 120 ÷ 3
s = 40
Answer: s = 40
🔬 Substitution Calculator
Evaluate ax² + bx + c for any values of a, b, c and x!
🧩 Drag 1: Simple Substitution
If x = 3, find 2x + 1. Drag the correct steps.
6
7
5
9
2 × 3 =
6
+ 1 =
🧩 Drag 2: Two Variables
If a = 5 and b = 2, find 3a − 4b. Drag the correct values.
15
8
7
10
23
3 × 5 = 4 × 2 =
15 − 8 =
🧩 Drag 3: Squared Terms
If n = 4, find n² + 2. Drag the steps.
16
18
8
10
n² = 4 × 4 =
16 + 2 =
🧩 Drag 4: Negative Value
If x = −3, find 2x + 10. Drag the steps.
−6
4
6
16
2 × (−3) =
−6 + 10 =
🧩 Drag 5: Using a Formula
The formula for area of a triangle is A = ½ × b × h. Find A when b = 8 and h = 5.
40
20
80
13
b × h = 8 × 5 =
A = ½ × 40 =
🧩 Drag 6: Full Substitution
If p = 3 and q = 2, find 4p² − 3q + 1. Complete each step.
9
36
6
31
12
43
p² = 3² =
4 × 9 =
3 × q = 3 × 2 =
36 − 6 + 1 =
📝 Practice Questions
Show all substitution steps.
Find 3x + 4 when x = 2.
Find 5n − 7 when n = 3.
Find 2a + 3b when a = 4 and b = 1.
Find 4p − 2q when p = 5 and q = 3.
Find x² when x = 7.
Find n² + n when n = 5.
Find 2x² − 3 when x = 4.
Find 3a² + 2a − 1 when a = 2.
Find 10 − 4x when x = 2.
Find 3x + 2y − z when x = 4, y = 3, z = 5.
The formula for perimeter of a rectangle is P = 2l + 2w. Find P when l = 7 and w = 4.
The formula C = 5n + 3 gives the cost in pence of n items. Find C when n = 6.
Find 4x − 3 when x = −2.
Find x² − 2x + 1 when x = 3.
Two expressions: A = 3n + 5 and B = n² − 1. Find both when n = 4. Which is greater?
The formula v = u + at. Find v when u = 5, a = 3, t = 4.
Find the value of 6ab when a = 2 and b = 5.
If y = 3x − 7, find y when x = 4. Then find x when y = 5.
Find 2(x + 3) when x = 7.
The formula A = πr² (use π = 3.14). Find A when r = 5.
3(2)+4 = 10
5(3)−7 = 8
2(4)+3(1) = 8+3 = 11
4(5)−2(3) = 20−6 = 14
7² = 49
25+5 = 30
2(16)−3 = 32−3 = 29
3(4)+2(2)−1 = 12+4−1 = 15
10−8 = 2
12+6−5 = 13
2(7)+2(4) = 14+8 = 22
5(6)+3 = 33p
4(−2)−3 = −8−3 = −11
9−6+1 = 4
A = 3(4)+5 = 17. B = 16−1 = 15. A is greater
5+3(4) = 5+12 = 17
6(2)(5) = 60
y = 3(4)−7 = 5. If y=5: 5=3x−7 → 3x=12 → x=4
2(7+3) = 2(10) = 20
3.14 × 25 = 78.5
🔥 Challenge: Substitution Problems
The formula for the area of a trapezium is A = ½(a + b)h. Find A when a = 5, b = 9, h = 4.
A taxi costs £(3 + 2n) where n is the number of miles. How much for a 7-mile journey? A 15-mile journey?
Find the value of 5x² − 2x + 3 when x = 3 and when x = −1.
Two number machines: Machine A gives 3n + 1 and Machine B gives n² − 4. For which value of n do both machines give the same output?
The formula for the nth term of a sequence is 4n − 1. Find the 1st, 5th, and 10th terms.
If P = 2(l + w), find l when P = 36 and w = 7.
Evaluate a³ + b² − ab when a = 2 and b = 3.
The stopping distance of a car (in metres) is given by d = v²/20 + v/2, where v is speed in m/s. Find d when v = 10.
Mr Josh has n students. He gives each student (2n + 3) stickers. Write an expression for the total stickers, then find the total when n = 5.
The formula T = 2a + 3b − c. Find T when a = 4, b = 2, c = 5. Then find a new value of c that makes T = 10 when a and b stay the same.
½(5+9)(4) = ½(14)(4) = 28
7 miles: 3+14 = £17. 15 miles: 3+30 = £33
x=3: 45−6+3 = 42. x=−1: 5+2+3 = 10
3n+1 = n²−4 → n²−3n−5=0. Try n=4: 13 vs 12 (no). n=5: 16 vs 21 (no). Try n=−1: −2 vs −3 (no). Actually solve by trial: n=4 gives 13 and 12, n=5 gives 16 and 21. No integer solution — accept checking n=4 and n=5.