Lesson 1.2 · Unit 1 · 6-EE.2c

Evaluating an expression without introducing a sign error

The order of operations is a convention, not a law of arithmetic. It exists so that everyone reading \( 3 + 4 \cdot 5 \) gets the same answer, and the only reason it matters is that people agreed on it. What is not conventional is the bracket discipline that keeps negative values from going wrong when you substitute.

The method
  1. Grouping symbols first, working from the innermost outward. Brackets, and also the implicit grouping in a fraction bar and under a radical.
  2. Then exponents, left to right.
  3. Then multiplication and division together, left to right as they appear. Neither outranks the other.
  4. Then addition and subtraction together, left to right. Again neither outranks the other.
  5. A fraction bar groups. \( \frac{a + b}{c} \) means the whole of \( a + b \) is divided by \( c \), so evaluate the top completely, evaluate the bottom completely, then divide.
  6. When substituting, put every value in brackets. Write \( 3(-2)^2 \), not \( 3-2^2 \). This single habit prevents most substitution errors.
  7. \( -3^2 \) and \( (-3)^2 \) are different. The first is \( -(3^2) = -9 \), because the exponent binds tighter than the negative sign. The second is \( 9 \).

Where students lose marks: substituting a negative value without brackets. If \( x = -4 \), then \( x^2 \) is \( (-4)^2 = 16 \), but writing it as \( -4^2 \) gives \( -16 \). The bracket is not optional decoration; it is what tells the exponent what it applies to.

Worked example

The problem. Evaluate \[ \frac{3(x - y)^2 - 4z}{2x + z} \] when \( x = 7 \), \( y = 3 \) and \( z = -1 \).

Step one: rewrite with brackets around every substituted value. Before doing any arithmetic: \[ \frac{3\big((7) - (3)\big)^2 - 4(-1)}{2(7) + (-1)} \] The brackets around \( -1 \) are the ones that matter, and putting them in costs nothing.

Step two: the fraction bar groups, so work top and bottom separately. Do not attempt to combine them until both are single numbers.

Step three: innermost grouping in the numerator. \( (7) - (3) = 4 \), so the numerator becomes \( 3(4)^2 - 4(-1) \).

Step four: exponent next. \( (4)^2 = 16 \), giving \( 3(16) - 4(-1) \). Note that the exponent applies to the 4 only, not to the 3 in front of it.

Step five: multiplication before subtraction. \( 3(16) = 48 \) and \( -4(-1) = +4 \). A negative times a negative is positive, and this is the sign step the problem is really testing. The numerator is \( 48 + 4 = 52 \).

Step six: the denominator. \( 2(7) = 14 \), then \( 14 + (-1) = 13 \).

Step seven: divide. \( \dfrac{52}{13} = 4 \).

Step eight: check the sign step by redoing it wrong on purpose. A student who reads \( -4z \) as \( -4 \) times \( 1 \) rather than times \( -1 \) gets a numerator of \( 48 - 4 = 44 \) and an answer of \( \frac{44}{13} \), which is not an integer. When an answer that should be clean comes out ugly, a sign error is the first thing to check.

Practice · 10 problems

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

  1. Evaluate \( 3 + 4 \cdot 5 \).
    Show the full solution

    Multiplication first: \( 4 \cdot 5 = 20 \), then \( 3 + 20 \). 23

  2. Evaluate \( (3 + 4) \cdot 5 \).
    Show the full solution

    Grouping first: \( 3 + 4 = 7 \), then \( 7 \cdot 5 \). 35

  3. Evaluate \( -5^2 \) and \( (-5)^2 \).
    Show the full solution

    The exponent binds tighter than the negative sign, so the first is \( -(25) \). In the second the brackets make the base \( -5 \). \( -25 \) and \( 25 \)

  4. Evaluate \( 2x^2 - 3x \) when \( x = -2 \).
    Show the full solution

    \( 2(-2)^2 - 3(-2) = 2(4) + 6 = 8 + 6 \). 14

  5. Evaluate \( \dfrac{a + b}{a - b} \) when \( a = 9 \) and \( b = 3 \).
    Show the full solution

    Top: \( 9 + 3 = 12 \). Bottom: \( 9 - 3 = 6 \). Then \( 12 \div 6 \). 2

  6. Evaluate \( 24 \div 4 \cdot 3 \) and explain why the answer is not 2.
    Show the full solution

    Multiplication and division have equal priority and are done left to right, so \( 24 \div 4 = 6 \) first, then \( 6 \cdot 3 = 18 \). Getting 2 comes from doing the multiplication first, treating \( 4 \cdot 3 = 12 \) as though multiplication outranked division. It does not; they are one level worked left to right. 18; division and multiplication have equal priority and go left to right

  7. Evaluate \( -x^2 + 4x - 1 \) when \( x = -3 \).
    Show the full solution

    Substitute with brackets: \( -(-3)^2 + 4(-3) - 1 \). The exponent applies to \( -3 \) inside the bracket, giving 9, and the leading minus then negates it: \( -9 \). Next \( 4(-3) = -12 \). So \( -9 - 12 - 1 \). \( -22 \)

  8. Evaluate \( \dfrac{2(x + 3)^2}{x - 1} \) when \( x = 5 \).
    Show the full solution

    Numerator: \( (5 + 3) = 8 \), then \( 8^2 = 64 \), then \( 2(64) = 128 \). Denominator: \( 5 - 1 = 4 \). Then \( 128 \div 4 \). 32

  9. A student evaluates \( 5 - 2(x - 4) \) at \( x = 1 \) and gets 9. Find their error and the correct value.
    Show the full solution

    Correctly: \( 5 - 2(1 - 4) = 5 - 2(-3) = 5 + 6 = 11 \). Getting 9 comes from subtracting the 2 from the 5 first, computing \( 3(x-4) = 3(-3) = -9 \), and then taking the sign wrong; either way the error is treating the subtraction as higher priority than the multiplication. Multiplication comes first, and the \( -2 \) travels with the bracket. 11; they subtracted before multiplying

  10. Explain why the order of operations is a convention rather than something that could be proved, and what would go wrong without one.
    Show the full solution

    Nothing in arithmetic forces \( 3 + 4 \cdot 5 \) to mean \( 3 + (4 \cdot 5) \) rather than \( (3 + 4) \cdot 5 \). The symbols alone are ambiguous, and mathematicians chose one reading so that written expressions have a single meaning. The choice is not arbitrary in spirit, since giving multiplication priority lets a polynomial like \( 3x^2 + 2x + 1 \) be written without brackets, which is why the convention is convenient. Without an agreed order the same expression would evaluate differently for different readers, so no formula could be communicated reliably, and every expression would need brackets everywhere. The symbols are ambiguous on their own; the convention makes written expressions unambiguous, and without it the same expression would mean different things to different readers

← 1.1 The real numbers and the properties1.3 Quantities, units and conversion →

Keep going in the full Algebra 1 course

All of U1 (Unit 1: Expressions, Quantities and the Properties Behind Algebra) is free to everyone, with no account. The course has 75 lessons, each with worked examples and practice problems that come with full solutions. A free account can also start a free month of the whole course, with no card.

Open this lesson in the courseSee the whole course