Substitution & expansion
AdvancedSubstitution = putting numbers into an expression; expansion = removing brackets with the distributive law. Master both and algebra flows.
After this lesson, you can:
- Substitute: replace variable with number
- Distributive law: a(b+c) = ab + ac
- With negative: watch signs
Before this lesson you should know: Introduction to algebraic expressions
1Summary 重點摘要
Substitution = putting numbers into an expression; expansion = removing brackets with the distributive law. Master both and algebra flows.
2Vocabulary cards Vocabulary
S
Substitution代入
Replace variable with number.
E
Expand展開
Remove brackets.
D
Distributive law分配律
a(b+c) = ab + ac.
B
Bracket括號
( ) grouping.
T
Term項
Parts separated by +/−.
C
Coefficient係數
Number in front of variable.
S
Simplify化簡
Combine like terms.
E
Expression代數式
Variables, constants, operators.
3Step by step Steps
- 1. Substitution: one variableSubstitution means replacing the variable with a concrete value, then calculating by the order of operations. Example: 3x + 2 when x = 4 — replace x with 4, giving 3(4) + 2 = 12 + 2 = 14. Two steps only: replace, then calculate. Note that 3(4) means 3 times 4, with the multiplication sign usually omitted. A common error is reading 3x + 2 as 34 + 2, gluing the digits together; the correct reading is 3 times x, then add 2. Writing out each step keeps substitution marks safe.
- 2. Substitution: two variablesTwo variables substitute the same way — each takes its own value. Example: 2x + 3y when x = 1 and y = 2 — replace x with 1 and y with 2, giving 2(1) + 3(2) = 2 + 6 = 8. The point: every variable takes only its own value, never swapped. A common error is reading 2x as 21 when x = 1, or 3y as 32 when y = 2; the correct reading is always 'coefficient times the value'. Write brackets around each substituted value, e.g. 2(1) + 3(2) — vital when the value is negative.
- 3. Expand: single bracketExpansion uses the distributive law: the number outside the bracket multiplies every term inside. Example: 3(x + 2) = 3 × x + 3 × 2 = 3x + 6. The point: the outside number multiplies every term, not just the first. A life view: each gift costs 3 dollars, take x of them plus 2 more, and the total is 3x + 6 dollars. A common error is multiplying only the first term, writing 3(x + 2) as 3x + 2; a check: after expansion the number of terms should match the inside — two terms in, two terms out.
- 4. Expand: with negativeWhen a negative sits outside the bracket, its sign travels with every term. Example: −2(x − 3) = (−2) × x + (−2) × (−3) = −2x + 6; note (−2) × (−3) gives +6, since two minuses make a plus. Two common errors: writing −2(x − 3) as −2x − 6, forgetting that negative times negative is positive; and multiplying only the first term by −2. A check: substitute a number — at x = 1 the original gives −2(1 − 3) = −2 × (−2) = 4, while the expansion gives −2(1) + 6 = 4; matching values mean the work is right.
- 5. Expand then simplifyWith two brackets, expand each first, then collect like terms. Example: 3(x + 2) + 2(x − 1) = 3x + 6 + 2x − 2. Collecting: x terms go together, 3x + 2x = 5x; constants go together, 6 + (−2) = 4; the answer is 5x + 4. Common errors: expanding 2(x − 1) as 2x − 1, missing the second term; or mishandling the sign in 6 + (−2). A good habit: after expanding, copy the whole expression once, circle the x terms and the constants separately, then collect each group — the order shows itself.
- 6. Simplify first then substituteEvaluation offers two roads. Example: 3(x + 2) when x = 1. Method one, substitute directly: 3(1 + 2) = 3(3) = 9. Method two, expand first then substitute: 3x + 6 = 3(1) + 6 = 9. The answers agree, because the expression is equivalent before and after expansion. When the expression is complicated, simplifying first is usually faster and safer, since the number of multiplications drops. A common error is forgetting to compute inside the bracket first — 1 + 2 comes before multiplying; brackets first is the first rule of the order of operations.
- 7. ApplicationA geometry application: a rectangle's length is (x + 2) cm and its width is x cm — what is the perimeter? Perimeter = 2 × (length + width) = 2[(x + 2) + x] = 2(2x + 2) = 4x + 4 cm. If x = 5, the perimeter is 4(5) + 4 = 24 cm; then the length is 7 and the width 5, and 2(7 + 5) = 24 checks out. A common error is adding length and width once and stopping, forgetting the factor 2; remember, a perimeter is the total of four sides.
4Diagrams Diagrams
5Worked example Example
Example: when x = 2, find the value of 3(x + 4) − 2x.
- Expand first: 3(x + 4) = 3x + 12, so the expression becomes 3x + 12 − 2x.
- Collect like terms: 3x − 2x = x, so the expression simplifies to x + 12.
- Substitute x = 2: x + 12 = 2 + 12 = 14.
- Check with the original: 3(2 + 4) − 2(2) = 3 × 6 − 4 = 18 − 4 = 14, matching the answer.
- Compare the two roads: direct substitution needs 3 × 6 and then minus 4; simplify-then-substitute needs one addition — faster and safer.
- Answer: The answer is 14. Expand 3(x + 4) to 3x + 12, collect like terms to get x + 12, substitute x = 2 to get 14; the original expression checks out: 3(2 + 4) − 2(2) = 14. Simplify first, substitute second — the least work and the safest route.
6Common mistakes Mistakes
✗ 3(x + 2) = 3x + 2 (forget constant).
✓ Distributive: 3x + 6.
✗ −2(x − 3) = −2x − 6 (sign wrong).
✓ −2x + 6 (negative × negative = positive).
✗ After expanding, no combining.
✓ Always combine like terms.
✗ No brackets when substituting.
✓ Use brackets: 3(4) + 2, not 34 + 2.
7Quick check Quick check
Q13x + 2 when x = 4 = ?
Answer: 14.
Did you get it right?
Q22x + 3y when x=1, y=2 = ?
Answer: 8.
Did you get it right?
Q33(x + 2) = ?
Answer: 3x + 6.
Did you get it right?
Q4−2(x − 3) = ?
Answer: −2x + 6.
Did you get it right?
Q53(x + 2) + 2(x − 1) = ?
Answer: 5x + 4.
Did you get it right?
8Recap Recap
- Substitute: replace variable with number
- Distributive law: a(b+c) = ab + ac
- With negative: watch signs
- After expanding, combine like terms
- Simplify first can be faster