A2a - Substituting numerical values and expressions into formulae ... - Free Printable

We are given:
> x = 2 and y = –5

We need to evaluate each expression from (a) to (r), and then spot the links — meaning, notice how later expressions relate to earlier ones (e.g., same value, built from previous, algebraically equivalent, etc.).

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Let’s compute each one step by step.

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🔢 First, compute basic values:

- x = 2
- y = –5
- x² = 4
- y² = 25
- xy = 2 × (–5) = –10
- x²y = 4 × (–5) = –20
- xy² = 2 × 25 = 50
- x²y² = 4 × 25 = 100
- (xy)² = (–10)² = 100

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## ✔ Now compute each expression:

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(a) x + x + x = 3x = 3×2 = 6

(b) 3x = 3×2 = 6 → Same as (a)

→ *Link: (a) and (b) are algebraically identical.*

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(c) 3x + 7 = 6 + 7 = 13

(d) 3x + 17 = 6 + 17 = 23

(e) 3x + 17 + y = 23 + (–5) = 18

(f) 3x + 17 – y = 23 – (–5) = 23 + 5 = 28

(g) 3x + 17 – 2y = 23 – 2×(–5) = 23 + 10 = 33

(h) –2y + 3x + 17 = same as (g) → 33

→ *Link: (g) and (h) are commutative rearrangements — same terms, same value.*

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(i) –2y = –2×(–5) = 10

(j) xy = 2×(–5) = –10

(k) x²y = 4×(–5) = –20

(l) x²y² = 4×25 = 100

(m) (xy)² = (–10)² = 100 → same as (l)

→ *Link: (l) and (m) are equal — since (xy)² = x²y²*

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(n) xy² = 2×25 = 50

(o) x² + 3x = 4 + 6 = 10 → same as (i)

→ *Link: (o) = (i) → x² + 3x = 4 + 6 = 10; –2y = 10*

(p) x(x + 3) = 2×(2 + 3) = 2×5 = 10 → same as (i) and (o)

→ *Link: (p) expands to x² + 3x → same as (o)*

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(q) (x – 2)(x + 3) = (2 – 2)(2 + 3) = 0 × 5 = 0

(r) x² + x – 6 = 4 + 2 – 6 = 0 → same as (q)

→ *Link: (q) expands to x² + x – 6 → so (q) and (r) are algebraically identical.*

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## 🧩 Let’s summarize all answers:

| Expression | Value |
|------------|-------|
| (a) x+x+x | 6 |
| (b) 3x | 6 |
| (c) 3x+7 | 13 |
| (d) 3x+17 | 23 |
| (e) 3x+17+y | 18 |
| (f) 3x+17–y | 28 |
| (g) 3x+17–2y | 33 |
| (h) –2y+3x+17 | 33 |
| (i) –2y | 10 |
| (j) xy | –10 |
| (k) x²y | –20 |
| (l) x²y² | 100 |
| (m) (xy)² | 100 |
| (n) xy² | 50 |
| (o) x²+3x | 10 |
| (p) x(x+3) | 10 |
| (q) (x–2)(x+3) | 0 |
| (r) x²+x–6 | 0 |

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## 🔗 Spotting the Links:

Here are the key relationships:

1. (a) = (b) — both are 3x → identical expressions.
2. (g) = (h) — same terms, reordered → commutative property.
3. (l) = (m) — because (xy)² = x²y² → exponent rules.
4. (i) = (o) = (p) — all equal 10.
- (i): –2y = 10
- (o): x² + 3x = 4 + 6 = 10
- (p): x(x+3) = 2×5 = 10 → which expands to x² + 3x → so (p) = (o)
5. (q) = (r) — (x–2)(x+3) expands to x² + x – 6 → so they are algebraically identical.
6. (e) = (d) + y → 23 + (–5) = 18
7. (f) = (d) – y → 23 – (–5) = 28
8. (g) = (d) – 2y → 23 – 2×(–5) = 33
9. (n) = x × y² = 2 × 25 = 50
10. (k) = x² × y = 4 × (–5) = –20

Also note:

- The sequence from (a) to (f) builds up: starting with 3x, adding constants, then adding/subtracting y or 2y.
- Expressions (o), (p), (i) all converge to 10 — interesting coincidence? Actually, it’s because:
> x² + 3x = 4 + 6 = 10
> –2y = –2×(–5) = 10
So numerically equal, but algebraically different.

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## ✔ Final Answer Summary:

All computed values:

- (a) 6
- (b) 6
- (c) 13
- (d) 23
- (e) 18
- (f) 28
- (g) 33
- (h) 33
- (i) 10
- (j) –10
- (k) –20
- (l) 100
- (m) 100
- (n) 50
- (o) 10
- (p) 10
- (q) 0
- (r) 0

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## 🔗 Key Links Identified:

- Identical expressions: (a)=(b), (g)=(h), (l)=(m), (q)=(r)
- Same numerical value, different form: (i)=(o)=(p)=10
- Building blocks: (d) is used in (e), (f), (g)
- Algebraic expansion: (p) expands to (o); (q) expands to (r)
- Exponent identity: (l) and (m) use (xy)² = x²y²

This exercise helps reinforce substitution, algebraic manipulation, and recognizing equivalent expressions — even when they look different!

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✔ Done!