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Algebra fluency: indices, algebraic fractions and formulae
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Logarithms, the binomial series and calculus all build on index laws and confident handling of algebraic fractions. Securing these now makes Further Pure much smoother.
Laws of indices
am Γ an = am + n am Γ· an = am β n (am)n = amn
a0 = 1 aβn = 1an a1n = βΏβa amn = (βΏβa)m
2723 = (β27)2 = 32 = 9
Same base trick: 4x = 8 means 22x = 23, so x = 32.
a0 = 1 aβn = 1an a1n = βΏβa amn = (βΏβa)m
2723 = (β27)2 = 32 = 9
Same base trick: 4x = 8 means 22x = 23, so x = 32.
Algebraic fractions
Simplify: factorise top and bottom, then cancel common factors.
x2 β 9x2 + x β 6 = (x + 3)(x β 3)(x + 3)(x β 2) = x β 3x β 2
Add or subtract: use a common denominator.
2x + 3x + 1 = 2(x + 1) + 3xx(x + 1) = 5x + 2x(x + 1)
x2 β 9x2 + x β 6 = (x + 3)(x β 3)(x + 3)(x β 2) = x β 3x β 2
Add or subtract: use a common denominator.
2x + 3x + 1 = 2(x + 1) + 3xx(x + 1) = 5x + 2x(x + 1)
Rearranging formulae
Do the same to both sides until the new subject is alone.
If the subject appears twice, collect those terms on one side and factorise.
y = x + 2x β 3
y(x β 3) = x + 2
xy β 3y = x + 2
xy β x = 3y + 2
x(y β 1) = 3y + 2, so x = 3y + 2y β 1
If the subject appears twice, collect those terms on one side and factorise.
y = x + 2x β 3
y(x β 3) = x + 2
xy β 3y = x + 2
xy β x = 3y + 2
x(y β 1) = 3y + 2, so x = 3y + 2y β 1
The volume of a sphere is V = 43Οr3. Find r when V = 500 cm3.
- Rearrange: r3 = 3V4Ο
- r3 = 15004Ο = 119.36...
- r = β119.36... = 4.923...
Answer: r = 4.92 cm (3 s.f.)
Know all six index laws, including negative and fractional powers. Factorise before cancelling algebraic fractions. When the subject appears twice, collect it on one side and factorise.
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