Mathematics (Core)Exam code: 0580

Introduction to Algebra

Algebra and Sequences

PractiseFlashcards

Replacing letters with numbers

  • Substitution means putting given numbers in place of the letters, then evaluating
  • Write the expression or formula down first, then replace each letter, then calculate
  • Follow the order of operations exactly as with any numerical calculation

Negative values need brackets

  • A negative value substituted for a letter is written in brackets, which keeps the sign attached through powers and multiplications
    • Where x = −4, x² is (−4)² = 16, not −16
  • Without brackets a calculator applies the power before the sign, giving the wrong result

Worked example

Substituting a negative value

Given a = 5 and b = −3, find the value of (i) 3a − b² (ii) 2b(a + b).

Solution:

  • Substitute a negative value inside brackets every time, so the sign cannot be separated from the number
  • (i) The expression becomes 3 × 5 − (−3)²
  • BIDMAS puts the index before the subtraction, so square first
  • (−3)² = (−3) × (−3) = 9, positive because two negatives multiply to a positive
  • 15 − 9 = 6
  • Without the bracket this reads 3 × 5 − −3², and since the power is applied before the sign, −3² is −9 and the answer comes out as 15 + 9 = 24, which is wrong
  • (ii) The expression becomes 2 × (−3) × (5 + (−3))
  • Brackets first: 5 + (−3) is 5 − 3 = 2
  • Then left to right: 2 × (−3) = −6, and −6 × 2 = −12
  • The answer is negative because exactly one of the factors is negative

Exam tip

Write the substituted line first

Substituting into a formula was set in 8 of the 22 papers. Almost every scheme gives the first mark for the line with the numbers in place, such as 3 × 5 − (−3)² above or 58 = 2v² − (−8) below. So write that line down before working anything out, rather than doing it in your head: if the arithmetic after it slips, the line still keeps one mark.

Worked example

Finding a letter that is squared in a formula

In the formula E = 2v² − u, find the positive value of v when E = 58 and u = −8.

Solution:

  • Substitute the values you know, with the negative in brackets: 58 = 2v² − (−8)
  • Subtracting −8 is the same as adding 8, so 58 = 2v² + 8
  • Take 8 from both sides: 2v² = 50, then divide by 2: v² = 25
  • The formula gives v², so finish with a square root: v = √25 = 5
  • Check: 2 × 5² − (−8) = 50 + 8 = 58, which is the value of E given

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