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6th Grade · 7th Grade · 6.EE.A.4 · 7.EE.A.2

Equivalent expressions in 6th and 7th grade

Equivalent expressions are expressions that always have the same value, whatever number the letter stands for. For example, n+n+n+nn + n + n + n and 4n4n are equivalent. In 6th grade, your child learns to test whether two expressions are equivalent. In 7th grade, writing equivalent expressions becomes a tool for understanding a real problem.

What the standards say

The official Common Core wording, quoted exactly. The plain version follows.

6.EE.A.4 Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them).

7.EE.A.2 Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related.

What it means, in plain words

6th grade: telling whether two expressions are equivalent 6.EE.A.4

An expression is a math phrase made of numbers, letters and operation signs, with no equals sign. The letter is a variable: it stands for a number. Two expressions are equivalent when they give the same value for every number you put in for the variable. n+n+n+nn + n + n + n and 4n4n are equivalent, because four of a number is always four times that number.

Identifying equivalent expressions in 6th grade happens in two ways. One way is to rewrite an expression with rules your child already knows. The distributive property turns 5(y+1)5(y + 1) into 5y+55y + 5. Combining like terms (terms with the same letter) turns 2y+3y2y + 3y into 5y5y. If one expression can be rewritten into the other, the two are equivalent.

The other way is to test numbers: put the same value into both expressions. If even one value gives different answers, the expressions are not equivalent. Matching answers for a few values suggest the expressions are equivalent, but they do not prove it. Only rewriting proves it, because the rules work for every number.

Your child should be able to:

  • Explain that equivalent expressions give the same value for every number
  • Test two expressions by putting the same number into both
  • Use one value that gives different answers to show two expressions are not equivalent
  • Rewrite an expression with the distributive property or by combining like terms to show two expressions are equivalent

Example

Are 5y−2y+65y - 2y + 6 and 3(y+2)3(y + 2) equivalent expressions?

  1. Combine like terms in the first one: 5y−2y=3y5y - 2y = 3y, so it becomes 3y+63y + 6.
  2. Expand the second one: 3(y+2)=3y+63(y + 2) = 3y + 6.
  3. Both become 3y+63y + 6. A test with y=4y = 4 agrees: 20−8+6=1820 - 8 + 6 = 18 and 3×6=183 \times 6 = 18.
  4. Yes, the two expressions are equivalent.

Example

Are y2y^2 and 2y2y equivalent expressions?

  1. Test y=2y = 2: 22=42^2 = 4 and 2×2=42 \times 2 = 4. They match.
  2. Test y=3y = 3: 32=93^2 = 9 and 2×3=62 \times 3 = 6. They do not match.
  3. One value with different answers is enough.
  4. No, the expressions are not equivalent, even though they agree when y=2y = 2.

7th grade: rewriting expressions to understand a problem 7.EE.A.2

Equivalent expressions in 7th grade come with a purpose. Rewriting expressions in a new form can show something about a situation that the first form hides. The value stays the same. What changes is what you can read from the expression.

Take a price with a 20% discount. If the price is pp dollars, the discount is 0.2p0.2p dollars, so the sale price is p−0.2pp - 0.2p. Rewritten, that is 0.8p0.8p. The first form shows the money taken off. The second form shows that you pay 80% of the price.

The perimeter of a rectangle (the distance around it) works the same way. It can be written 2l+2w2l + 2w or 2(l+w)2(l + w), where ll is the length and ww is the width. The first form adds two lengths and two widths. The second form adds one length and one width, then doubles the result.

Your child should be able to:

  • Rewrite an expression in a different form, such as p−0.2pp - 0.2p as 0.8p0.8p
  • Explain what each form of an expression shows about a real situation
  • Explain what each number in an expression means in the problem
  • Choose the form of an expression that best answers the question asked

Example

A class trip costs 12 dollars per student for the bus and 8 dollars per student for lunch. Write the total cost for ss students in two ways. What does each way show?

  1. Add the two costs: 12s+8s12s + 8s. This form shows the bus total and the lunch total separately.
  2. Combine like terms: 12s+8s=20s12s + 8s = 20s. This form shows that each student costs 20 dollars in all.
  3. For 25 students, both give the same total: 12×25+8×25=300+200=50012 \times 25 + 8 \times 25 = 300 + 200 = 500, and 20×25=50020 \times 25 = 500.
  4. The two expressions are 12s+8s12s + 8s and 20s20s, and for 25 students the trip costs 500 dollars.

Where children usually get stuck

  • Your child tests x+4x + 4 and 3x3x with x=2x = 2, gets 6 for both and says they are equivalent.

    One match feels like proof. Two different expressions can still agree at one number.

    What helps. Ask your child to try two more numbers, including 0. When x=0x = 0, x+4=4x + 4 = 4 but 3x=03x = 0, so the expressions are not equivalent.

  • Your child says 10−n10 - n and n−10n - 10 are equivalent because they use the same numbers.

    Changing the order works for adding and multiplying, so it seems to work for subtracting too.

    What helps. Test n=4n = 4 together. 10−4=610 - 4 = 6, but 4−10=−64 - 10 = -6. Changing the order changed the value.

  • Your child rewrites p−0.2pp - 0.2p as 0.2p0.2p and says the sale price is 20% of the price.

    The pp on its own has no number in front, so it seems to disappear. In fact pp means 1p1p, and 1p−0.2p=0.8p1p - 0.2p = 0.8p.

    What helps. Use a real price. For 50 dollars, the discount is 10 dollars, so the sale price is 40 dollars. That is 0.8×500.8 \times 50, not 0.2×500.2 \times 50.

Try four practice questions

Real 6th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

  1. 1Warm-up
    Look at Expression A and Expression B below. Are they equivalent?
    Expression A: 3(x+2)+2xExpression B: 5x+6\displaystyle \begin{array}{l} \text{Expression A: } 3(x + 2) + 2x \\ \text{Expression B: } 5x + 6 \end{array}
    Select the correct answer.
    A
    Equivalent
    B
    Not equivalent
    Show the answer and how to solve it

    The answer is Equivalent because Expression A simplifies to 5x + 6, which matches Expression B.

    1. 1Expression A has parentheses, so it needs to be simplified. Expression B is already simplified.
    2. 2Use the distributive property on Expression A: multiply 3 by both x and 2.
    3. 3Combine the like terms 3x and 2x to get 5x + 6. This matches Expression B exactly, so the expressions are equivalent.

    Common mistake. A common mistake is distributing 3 only to x and not to 2. That gives 3x+2+2x3x + 2 + 2x instead of 3x+6+2x3x + 6 + 2x, so the expressions seem different even though they are equivalent.

    The idea underneath. Equivalent expressions represent the same value for every x, so if their simplified forms match term-for-term, they are equivalent.

  2. 2On level
    Each bag has 2m stickers and n markers. A student wrote an expression for 3 bags, as shown below. Which statement best describes the student's rewrite?
    3(2m+n)\displaystyle 3(2m + n)
    Student's work
    =3m+3n\displaystyle = 3m + 3n
    A
    The student should keep 2m + n together as one quantity instead of splitting it apart.
    B
    The student should combine 2m and n first before rewriting the expression.
    C
    The rewrite is correct because the 3 was applied to both terms inside the parentheses.
    D
    The 3 should multiply the whole term 2m, so the first term should be 6m, not 3m.
    Show the answer and how to solve it

    The correct statement is that the 3 must multiply the whole term 2m. Since 3×2m=6m3 \times 2m = 6m and 3×n=3n3 \times n = 3n, the correct expansion is 6m+3n6m + 3n.

    1. 1Look at the displayed work. The outside factor is 3, so it must be applied to every entire term inside the parentheses.
    2. 2Distribute 3 to both terms, 2m and n.
    3. 3Multiply each product. The term 2m has a coefficient of 2, so 3×2m=6m3 \times 2m = 6m, and 3×n=3n3 \times n = 3n.
    4. 4Compare that to the student's line, 3m + 3n. The n-term is correct, but the m-term is not. The coefficient 2 inside 2m also had to be multiplied by 3, giving 6m, not 3m.
    5. 5The mistake was ignoring the coefficient inside the term 2m. The best statement says the first term should be 6m, not 3m.

    Common mistake. A common mistake is multiplying the outside factor by only the variable letter and forgetting the coefficient already attached to it. In 2m, the 2 is part of the term, so 3×2m=6m3 \times 2m = 6m, not 3m.

    The idea underneath. When you distribute, you multiply the outside factor by each entire term inside the parentheses, including any coefficient already attached to a variable.

  3. 3On level
    Look at the two expressions below. Are they equivalent for all values of the variables?
    4x+b+5b+5+4x\displaystyle \begin{array}{l} 4x + b + 5 \\ b + 5 + 4x \end{array}
    Select Equivalent or Not equivalent.
    A
    Equivalent
    B
    Not equivalent
    Show the answer and how to solve it

    The answer is Equivalent because both expressions contain 4x, b, and 5 in a different order.

    1. 1Look at the first line. Its whole terms are 4x, b, and 5.
    2. 2The second line also has 4x, b, and 5. Nothing was changed, added, or removed — only the order is different.
    3. 3Because the same whole terms appear in both lines, the expressions are equivalent.

    Common mistake. A common mistake is thinking that writing b before 4x changes the expression. Moving terms around in a sum does not change its value.

    The idea underneath. Equivalent expressions have the same value for every choice of the variables. Reordering or regrouping the same added terms keeps that value.

  4. 4Stretch
    What multiplier represents a 0.5% increase?
    A
    0.005
    B
    0.995
    C
    1.05
    D
    1.005
    Show the answer and how to solve it

    The answer is 1.005 because 0.5% = 0.005, and an increase multiplier is 1+0.0051 + 0.005.

    1. 1An increase multiplier starts with 1 because the original amount is still included.
    2. 2Convert 0.5% to a decimal by dividing by 100. Because 0.5 already has a decimal point, this shifts the decimal two places to the left, giving 0.005—not 0.05.
    3. 3Add that decimal to 1.
    4. 4So the multiplier is 1.005.

    Common mistake. A common mistake is choosing 1.05, which treats 0.5% as if it were 5%. Dividing 0.5 by 100 moves the decimal two places left to get 0.005, not 0.05.

    The idea underneath. A percent increase multiplier always includes the original whole 1 plus the increase written as a decimal.