POPStudying

Pearson Edexcel 4MA1 IGCSE  ·  Topic G3.13

G3.13 Algebraic Substitution

Image  ·  914 × 270

01 — Explanation

What does "substitution" mean?

Substitution means replacing each letter in an expression or formula with a given number, then working out the result using BIDMAS. 3x−5y3x-5y with x=−3,y=2x=-3,y=2 means calculate 3×(−3)−5×23\times(-3) - 5\times2, not "33 add −3-3."

02 — Explanation

How do I avoid sign errors when substituting negative numbers?

Always put the substituted value in brackets, every time — even for expressions that look simple.

5z2 with z=−10  ⟶  5×(−10)2=5×100=5005z^2 \text{ with } z=-10 \;\longrightarrow\; 5\times(-10)^2 = 5\times100=500

Common Misconception: Writing 5×−1025\times-10^2 without brackets. Without them, BIDMAS squares the 1010 first and keeps the minus sign outside, giving the wrong answer 5×(−100)=−5005\times(-100)=-500. The bracket is what protects the negative sign through the squaring.

03 — Explanation

How do I substitute into an expression with multiple terms and different operations?

Substitute every letter first, in brackets, before touching BIDMAS — this separates the two skills (translation, then calculation) so you're never trying to do both at once.

Example: Evaluate 4a−b2+c24a-\dfrac{b}{2}+c^2 when a=3a=3, b=−8b=-8, c=−2c=-2.

  1. Substitute: 4(3)−(−8)2+(−2)24(3)-\dfrac{(-8)}{2}+(-2)^2.
  2. Work left to right through BIDMAS: (−2)2=4(-2)^2=4 first (indices), then 4(3)=124(3)=12 and (−8)÷2=−4(-8)\div2=-4.
  3. Combine: 12−(−4)+4=12+4+4=2012-(-4)+4 = 12+4+4=20.

Exam Tip: When a fraction bar covers an expression like b2\dfrac{b}{2}, treat the whole top as already "bracketed" — divide only after any arithmetic above the line is finished. The same logic applies to a fraction with an expression on both top and bottom, such as a+bc\dfrac{a+b}{c}: simplify the numerator fully first, then divide by cc last.

04 — Explanation

What's the standard, mark-securing method?

  1. Record the original expression/formula.
  2. Substitute, bracketing every value.
  3. Prioritize — apply BIDMAS (indices/powers before multiplication, multiplication before addition/subtraction).
  4. Calculate the final number.

Exam Tip: Writing out the substitution line explicitly (step 2) earns you a method mark even if you go on to make an arithmetic slip — never skip straight to the answer.

05 — Explanation

How do I substitute into a formula and then solve for another letter?

Substitute first, then treat the result as a normal equation to solve.

Example: P=2l+2wP=2l+2w. Given P=20P=20 cm, w=4w=4 cm, find ll.

  1. Substitute: 20=2l+2(4)20 = 2l + 2(4).
  2. Simplify: 20=2l+820=2l+8.
  3. Solve: 2l=122l=12, so l=6l=6 cm.

06 — Explanation

How do I substitute into an expression involving a square root?

Work out everything inside the root first (following BIDMAS exactly as normal), and only take the square root as the very last step.

Example: Pythagoras' theorem states c=a2+b2c=\sqrt{a^2+b^2}. Find cc when a=5a=5 and b=12b=12.

  1. Substitute: c=(5)2+(12)2c=\sqrt{(5)^2+(12)^2}.
  2. Square first: (5)2=25(5)^2=25, (12)2=144(12)^2=144.
  3. Add inside the root: 25+144=16925+144=169.
  4. Take the square root last: c=169=13c=\sqrt{169}=13.

Common Misconception: Taking the square root of aa and bb separately before adding (e.g. 25+144=5+12=17\sqrt{25}+\sqrt{144}=5+12=17). The root applies to the whole sum, not to each term individually — always finish everything inside the root first.

07 — Worked example

Worked Example

T=2x2−3yT = 2x^2 - 3y. Find the value of TT when x=−4x=-4 and y=−5y=-5.

  1. Record: T=2x2−3yT=2x^2-3y.
  2. Substitute, bracketing both values: T=2(−4)2−3(−5)T=2(-4)^2-3(-5).
  3. Prioritize BIDMAS — square first: (−4)2=16(-4)^2=16. Then multiply: 2×16=322\times16=32 and 3×(−5)=−153\times(-5)=-15.
  4. Calculate: T=32−(−15)=32+15=47T = 32 - (-15) = 32+15=47.