Revision notes for Edexcel IGCSE Maths Substitution. Open the guide for explanations and worked examples. Written against the Edexcel IGCSE Maths (4MA1) specification, so the content matches what's examinable rather than general Maths background.
Revision notes for Edexcel IGCSE Maths Substitution. Open the guide for explanations and worked examples. Written against the Edexcel IGCSE Maths (4MA1) specification, so the content matches what's examinable rather than general Maths background.
In algebra, letters usually stand for numbers. Substitution means putting the given number in place of the letter, then calculating the answer.
Key words
Replace first, calculate second
Substitution is a two-stage process: first replace each letter with its given number, then use the normal order of operations to calculate.
Substituting into a simple expression
Suppose f=6f = 6f=6 and g=4g = 4g=4. Find the value of 3f+2g3f + 2g3f+2g.
Replace fff with 6 and ggg with 4. Remember that 3f3f3f means 3×f3 \times f3×f.
3f+2g=3(6)+2(4)3f + 2g = 3(6) + 2(4)3f+2g=3(6)+2(4)Multiply before adding.
3(6)+2(4)=18+83(6) + 2(4) = 18 + 83(6)+2(4)=18+8Add the two parts.
18+8=2618 + 8 = 2618+8=26The value of the expression is 26.
In algebra, multiplication signs are often left out.
For example:
You must still follow BIDMAS, the order of operations:
Do not join the numbers
If d=7d = 7d=7, then 4d4d4d means 4×74 \times 74×7, not 47. The missing operation is multiplication.
Substituting into a formula
A formula is c=5d−8c = 5d - 8c=5d−8. Find ccc when d=7d = 7d=7.
Replace ddd with 7.
c=5(7)−8c = 5(7) - 8c=5(7)−8Multiply first.
c=35−8c = 35 - 8c=35−8Subtract.
c=27c = 27c=27So the value of ccc is 27.
Negative numbers need extra care. The safest habit is to put the negative number in brackets when you substitute it.
For example, if r=−3r = -3r=−3, write 4(−3)4(-3)4(−3) rather than 4−34-34−3. This keeps the multiplication clear.
Bracket negative substitutions
When a variable has a negative value, put it in brackets straight away. This avoids most sign errors.
One positive and one negative value
A formula is Q=6p+4rQ = 6p + 4rQ=6p+4r. Find QQQ when p=5p = 5p=5 and r=−3r = -3r=−3.
Substitute p=5p = 5p=5 and r=−3r = -3r=−3.
Q=6(5)+4(−3)Q = 6(5) + 4(-3)Q=6(5)+4(−3)Do the multiplications.
Q=30+(−12)Q = 30 + (-12)Q=30+(−12)Adding a negative is the same as subtracting.
Q=30−12Q = 30 - 12Q=30−12Calculate the final value.
Q=18Q = 18Q=18Subtracting a negative value
Find AAA when A=p−2qA = p - 2qA=p−2q, p=−3p = -3p=−3 and q=−8q = -8q=−8.
Substitute both values using brackets.
A=(−3)−2(−8)A = (-3) - 2(-8)A=(−3)−2(−8)Work out the multiplication part first.
A=(−3)−(−16)A = (-3) - (-16)A=(−3)−(−16)Subtracting a negative becomes adding.
A=−3+16A = -3 + 16A=−3+16Finish the calculation.
A=13A = 13A=13Some formulas contain brackets already. Brackets tell you to calculate that part first.
If the formula is L=m(n−4)L = m(n - 4)L=m(n−4), this means:
A formula with brackets
Find LLL when L=m(n−4)L = m(n - 4)L=m(n−4), m=8m = 8m=8 and n=9n = 9n=9.
Substitute the values into the formula.
L=8(9−4)L = 8(9 - 4)L=8(9−4)Work out the bracket.
L=8(5)L = 8(5)L=8(5)Multiply.
L=40L = 40L=40Sometimes letters are next to each other, such as bcbcbc. This also means multiplication.
Multiplying two negative values
Find aaa when a=6bca = 6bca=6bc, b=−2b = -2b=−2 and c=−5c = -5c=−5.
Substitute both negative values using brackets.
a=6(−2)(−5)a = 6(-2)(-5)a=6(−2)(−5)Multiply the two negative numbers first.
(−2)(−5)=10(-2)(-5) = 10(−2)(−5)=10Multiply by 6.
a=6(10)a = 6(10)a=6(10)The value of aaa is 60.
An index is a power. For example, x2x^2x2 means x×xx \times xx×x, so it is read as “xxx squared”.
When substituting into a square, substitute the value first, then square it.
Squaring a negative number
If x=−3x = -3x=−3, then x2=(−3)2=9x^2 = (-3)^2 = 9x2=(−3)2=9. The brackets matter because the whole value of xxx is being squared.
A formula with a square
Find www when w=4x2+7w = 4x^2 + 7w=4x2+7 and x=−3x = -3x=−3.
Substitute x=−3x = -3x=−3 using brackets.
w=4(−3)2+7w = 4(-3)^2 + 7w=4(−3)2+7Square the negative number first.
w=4(9)+7w = 4(9) + 7w=4(9)+7Multiply before adding.
w=36+7w = 36 + 7w=36+7Finish the calculation.
w=43w = 43w=43Some expressions mix powers, multiplication and subtraction.
A mixed expression with several letters
Find the value of b2−4acb^2 - 4acb2−4ac when a=2a = 2a=2, b=−5b = -5b=−5 and c=3c = 3c=3.
Substitute the values carefully.
b2−4ac=(−5)2−4(2)(3)b^2 - 4ac = (-5)^2 - 4(2)(3)b2−4ac=(−5)2−4(2)(3)Square first.
(−5)2=25(-5)^2 = 25(−5)2=25Work out the multiplication part.
4(2)(3)=244(2)(3) = 244(2)(3)=24Subtract.
25−24=125 - 24 = 125−24=1Some formulas include fractions. The line in a fraction means “divide by”.
Parts of a fraction
For formulas like A=12bhA = \frac{1}{2}bhA=21bh, multiplying by 12\frac{1}{2}21 means finding half of the product bhbhbh.
A formula with one half
Find AAA when A=12bhA = \frac{1}{2}bhA=21bh, b=10b = 10b=10 and h=7h = 7h=7.
Substitute the values.
A=12(10)(7)A = \frac{1}{2}(10)(7)A=21(10)(7)Multiply the base and height.
A=12(70)A = \frac{1}{2}(70)A=21(70)Find half of 70.
A=35A = 35A=35For a formula with a larger fraction, work out the numerator and denominator separately before dividing.
A formula with a fraction bar
Find sss when s=v2−u22as = \frac{v^2 - u^2}{2a}s=2av2−u2, v=8v = 8v=8, u=4u = 4u=4 and a=2a = 2a=2.
Substitute all the values using brackets where helpful.
s=82−422(2)s = \frac{8^2 - 4^2}{2(2)}s=2(2)82−42Work out the powers in the numerator.
s=64−162(2)s = \frac{64 - 16}{2(2)}s=2(2)64−16Finish the numerator and denominator.
s=484s = \frac{48}{4}s=448Divide.
s=12s = 12s=12Longer formulas are not harder if you keep the working tidy. Substitute first, then deal with each part using BIDMAS.
A common example is s=ut+12at2s = ut + \frac{1}{2}at^2s=ut+21at2. This has multiplication, a fraction and a square, so order matters.
A longer substitution
Find sss when s=ut+12at2s = ut + \frac{1}{2}at^2s=ut+21at2, u=−4u = -4u=−4, a=6a = 6a=6 and t=3t = 3t=3.
Substitute the values.
s=(−4)(3)+12(6)(3)2s = (-4)(3) + \frac{1}{2}(6)(3)^2s=(−4)(3)+21(6)(3)2Work out the square first.
s=(−4)(3)+12(6)(9)s = (-4)(3) + \frac{1}{2}(6)(9)s=(−4)(3)+21(6)(9)Work out the multiplication parts.
s=−12+27s = -12 + 27s=−12+27Add the two terms.
s=15s = 15s=15In the exam
Write the substituted expression before calculating, because this can earn method marks even if the final arithmetic goes wrong.
Put brackets around negative values as soon as you substitute them.
Use BIDMAS carefully: brackets, powers, multiplication or division, then addition or subtraction.
Check whether your answer should be positive or negative, especially when the question includes negative numbers.
Check yourself
Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.
Test yourself on this topic, or move on to the next guide.
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