Expressions using Letter-Numbers · Lesson 6 of 11
Equivalent Expressions and Correcting Mistakes
“Explain when expressions agree for every allowed value and locate the steps that cause algebra errors.”
• Distinguish equivalence from agreement at one chosen value. • Use a counterexample to show two expressions are not equivalent. • Interpret how brackets change the meaning of an expression. • Obtain equivalent expressions by counting a picture in different ways. • Correct incomplete simplifications and invalid transformations.
Equivalent for Every Allowed Value
Two expressions may look different and still describe the same quantity. For example, l + b + l + b and 2l + 2b both give a rectangle’s perimeter. Their agreement is explained by a valid regrouping of the terms, rather than by how similar they look.
Expressions that have the same value for every allowed assignment of their letters. Valid arithmetic transformations can show their equivalence.
Trying numerical values is helpful, but one matching result alone cannot establish equivalence. For instance, x + 2 and 2x both equal 4 when x = 2. At x = 3, however, their values are 5 and 6. A single counterexample—an allowed value at which the expressions disagree—is enough to show they are not equivalent.
Problem
Are x + 2 and 2x equivalent?
- 1.At x = 2, both expressions give 4, so this one case does not separate them.
- 2.Test x = 3: x + 2 gives 5, while 2x gives 6.
- 3.They disagree at an allowed value. Therefore they are not equivalent, despite matching at x = 2.
A Product Is Different from a Sum
The expression 5u means five times u. The expression 5 + u means add five to u. These are different instructions, so we should compare their meanings and values rather than assuming the same symbols make them equivalent.
| u | 5u | 5 + u |
|---|---|---|
| 2 | 10 | 7 |
| 5 | 25 | 10 |
| 8 | 40 | 13 |
| 11 | 55 | 16 |
Problem
Compare 5u and 5 + u when u = 8. What does the result tell you?
- 1.Evaluate the product: 5u = 5 × 8 = 40.
- 2.Evaluate the sum: 5 + u = 5 + 8 = 13.
- 3.Since 40 and 13 differ, these expressions are not equivalent. No further testing is needed to disprove equivalence.
Brackets can change which quantity a multiplier acts on. In 10y − 3, first take ten times y, then remove three. In 10(y − 3), first remove three from y, then multiply the whole difference by ten. The second instruction removes ten groups of three, not just one group of three.
Problem
Are 10y − 3 and 10(y − 3) equivalent?
- 1.Expand the bracketed expression: 10(y − 3) = 10y − 30.
- 2.The first expression removes 3; the second removes 30. At y = 2, they give 17 and −10.
- 3.They are not equivalent. The first is always 27 greater than the second because (10y − 3) − (10y − 30) = 27.
| y | 10y − 3 | 10(y − 3) |
|---|---|---|
| 0 | −3 | −30 |
| 2 | 17 | −10 |
| 7 | 67 | 40 |
| 10 | 97 | 70 |
Different Counting Routes, One Total
Counting a picture in several ways gives a practical explanation of equivalence. The entries in this diagram include eight 3s, two rs, and two ss. Adding by rows, collecting identical entries, or doubling identical halves must produce the same total, provided every entry is counted once.
Problem
Write three expressions for the total in the diagram and simplify them.
- 1.Rows give 12 + (r + s) + (r + s) + 12. Collecting yields 2r + 2s + 24.
- 2.Counting like entries gives 8 × 3 + 2r + 2s, again 24 + 2r + 2s.
- 3.The upper half totals 12 + r + s, and the lower half is identical. Doubling gives 2(12 + r + s) = 24 + 2r + 2s. All three methods count exactly the same entries.
If a different picture contains two 5y entries, two x entries, −6, and 2, its total is 10y + 2x − 4. The numerical entries combine separately from the letter terms. Reordering the counting route does not change signs or turn unlike quantities into like ones.
Mind the Mistake, Mend the Mistake
To correct an algebraic transformation, find the first rule that was broken. Sometimes the equality is false because a sign or multiplier was mishandled. Sometimes the expression is equivalent but has not reached the requested form. Separating these cases makes your correction precise.
| Expression | Claim to check | Correction and reason |
|---|---|---|
| 3a + 2b | 5 | Keep 3a + 2b; the letter parts differ. |
| 3b − 2b − b | 0 | Correct: coefficient total is zero. |
| 6(p + 2) | 6p + 8 | 6p + 12; multiply both bracket terms by 6. |
| (4x + 3y) − (3x + 4y) | x + y | x − y; subtract both terms in the second bracket. |
| 5 − (2 − 6z) | 3 − 6z | 3 + 6z; the second minus changes to plus. |
| 2 + (x + 3) | 2x − 6 | x + 5; addition is not distribution. |
| 2y + (3y − 6) | −y + 6 | 5y − 6; the plus preserves the bracket signs. |
| 7p − p + 5q − 2q | 7p + 3q | 6p + 3q; subtract the hidden coefficient 1 of p. |
| 5(2w + 3x + 4w) | 10w + 15x + 20w | Equivalent but incomplete: collect w terms to get 30w + 15x. |
| 3j + 6k + 9h + 12 | 3(j + 2k + 3h + 4) | Equivalent factored form; the original is already an expanded combined form. |
| 4(2r + 3s + 5) | −20 − 8r − 12s | 8r + 12s + 20; a positive 4 does not reverse signs. |
In the last two rows, read the original expression and the proposed result as separate columns. They are not one longer expression. For 3j + 6k + 9h + 12, factoring out 3 is a valid equality. It simply changes the form; it does not make the value wrong. “Simplest” needs a stated aim—here, remove brackets and collect like terms.
Within the linear expressions in this chapter, a fully collected form can have one term for each surviving letter plus a constant term. Some letter terms may cancel completely. For example, 3b − 2b − b contains b initially but simplifies to zero. Do not predict the final term count just by counting the letters in the original writing.
Testing several numerical cases supports a conjecture, but does not prove equivalence for every value. Use valid transformations for a general explanation. To disprove equivalence, one correctly calculated counterexample is sufficient.
Check Your Understanding
Use the ideas from this lesson to choose an answer. Explain your choice to yourself before opening the explanations below.
Quiz
Which statement best defines equivalent expressions?
At x = 2, x + 2 and 2x both equal 4. What follows?
Expand 10(y − 3).
Simplify 2y + (3y − 6).
Which judgement about 3j + 6k + 9h + 12 = 3(j + 2k + 3h + 4) is correct?
Equivalence concerns all allowed assignments, not appearance or a single case.
Practice Problems
- Simplify the total of two 5y entries, two x entries, −6, and 2.
- A picture has four 2p entries, four 3q entries, and constants 3, −2, −2, 3. Find its total.
- A picture has twelve 5k entries and four −5g entries. Find its total.
- Correct 6(p + 2) = 6p + 8 and 5 − (2 − 6z) = 3 − 6z.
- Simplify 5(2w + 3x + 4w) and explain whether 10w + 15x + 20w is wrong.
- Simplify 4(2r + 3s + 5). Is −20 − 8r − 12s equivalent to it?
- Show that 2(x + 3) and 2x + 6 are equivalent without checking individual values.
- Find a counterexample to the claim that 3a + 2b always equals 5.
10y + 2x − 4.
Key Takeaways
• Equivalent expressions agree for every allowed assignment. • One matching numerical case does not establish equivalence. • One counterexample can disprove equivalence. • Brackets determine which terms a multiplier or minus sign acts on. • Different counting routes give equivalent expressions when every entry is counted once. • An equivalent but unfinished simplification differs from an invalid equality.