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Commutative and Associative Properties

Use the commutative and associative properties

Use the Commutative and Associative Properties

Think about adding two numbers, such as \(5\) and \(3.\)

\[\begin{array}{llll}5+3 & & & 3+5 \\ 8 & & & 8\end{array}\]

The results are the same. \(5+3=3+5\)

Notice, the order in which we add does not matter. The same is true when multiplying \(5\) and \(3.\)

\[\begin{array}{llll}5\cdot 3 & & & 3\cdot 5 \\ 15 & & & 15\end{array}\]

Again, the results are the same! \(5\cdot 3=3\cdot 5.\) The order in which we multiply does not matter.

These examples illustrate the commutative properties of addition and multiplication.

The commutative properties have to do with order. If you change the order of the numbers when adding or multiplying, the result is the same.

Example

Try it.

Use the commutative properties to rewrite the following expressions:

  1. ⓐ \(\ -1+3=\underline{\hspace{1.5em}}\)

  2. ⓑ \(\ 4\cdot 9=\underline{\hspace{1.5em}}\)

Solution
ⓐ
\(-1+3=\underline{\hspace{1.5em}}\)
Use the commutative property of addition to change the order.\(-1+3=3+(-1)\)
ⓑ
\(4\cdot 9=\underline{\hspace{1.5em}}\)
Use the commutative property of multiplication to change the order.\(4\cdot 9=9\cdot 4\)

What about subtraction? Does order matter when we subtract numbers? Does \(7-3\) give the same result as \(3-7?\)

\[\begin{array}{lll}7-3 & & 3-7 \\ 4 & & -4 \\ & 4\ne -4 & \end{array}\]\[\text{The results are not the same.}\ 7-3\ne 3-7\]\[\begin{array}{lll}12\div 4 & & 4\div 12 \\ \frac{12}{4} & & \frac{4}{12} \\ 3 & & \frac{1}{3} \\ & 3\ne \frac{1}{3} & \end{array}\]\[\text{The results are not the same. So}\ 12\div 4\ne 4\div 12\]\[7+8+2\]\[5\cdot \frac{1}{3}\cdot 3\]

Condensed: the full section is in OpenStax Prealgebra 2e.

Evaluate Expressions using the Commutative and Associative Properties

The commutative and associative properties can make it easier to evaluate some algebraic expressions. Since order does not matter when adding or multiplying three or more terms, we can rearrange and re-group terms to make our work easier, as the next several examples illustrate.

Example

Try it.

Evaluate each expression when \(x=\frac{7}{8}.\)

  1. ⓐ \(\ x+0.37+(-x)\)
  2. ⓑ \(\ x+(-x)+0.37\)
Solution
ⓐ
Substitute \(\frac{7}{8}\) for \(x\).
Convert fractions to decimals.
Add left to right.
Subtract.
ⓑ
Substitute \(\frac{7}{8}\) for x.
Add opposites first.

What was the difference between part ⓐ and part ⓑ ? Only the order changed. By the Commutative Property of Addition, \(x+0.37+(-x)=x+(-x)+0.37.\) But wasn’t part ⓑ much easier?

Let’s do one more, this time with multiplication.

Example

Try it.

Evaluate each expression when \(n=17.\)

  1. ⓐ \(\ \frac{4}{3}(\frac{3}{4}n)\)

  2. ⓑ \(\ (\frac{4}{3}\cdot \frac{3}{4})n\)

Solution
ⓐ
Substitute 17 for n.
Multiply in the parentheses first.
Multiply again.
ⓑ
Substitute 17 for n.
Multiply. The product of reciprocals is 1.
Multiply again.

What was the difference between part ⓐ and part ⓑ here? Only the grouping changed. By the Associative Property of Multiplication, \(\frac{4}{3}(\frac{3}{4}n)=(\frac{4}{3}\cdot \frac{3}{4})n.\) By carefully choosing how to group the factors, we can make the work easier.

Simplify Expressions Using the Commutative and Associative Properties

When we have to simplify algebraic expressions, we can often make the work easier by applying the Commutative or Associative Property first instead of automatically following the order of operations. Notice that in part ⓑ was easier to simplify than part ⓐ because the opposites were next to each other and their sum is \(0.\) Likewise, part ⓑ in was easier, with the reciprocals grouped together, because their product is \(1.\) In the next few examples, we’ll use our number sense to look for ways to apply these properties to make our work easier.

Example

Try it.

Simplify: \(-84n+(-73n)+84n.\)

Solution

Notice the first and third terms are opposites, so we can use the commutative property of addition to reorder the terms.

\(-84n+(-73n)+84n\)
Re-order the terms.\(-84n+84n+(-73n)\)
Add left to right.\(0+(-73n)\)
Add.\(-73n\)

Now we will see how recognizing reciprocals is helpful. Before multiplying left to right, look for reciprocals, their product is \(1.\)

Example

Try it.

Simplify: \(\frac{7}{15}\cdot \frac{8}{23}\cdot \frac{15}{7}.\)

Solution

Notice the first and third terms are reciprocals, so we can use the Commutative Property of Multiplication to reorder the factors.

\(\frac{7}{15}\cdot \frac{8}{23}\cdot \frac{15}{7}\)
Re-order the terms.\(\frac{7}{15}\cdot \frac{15}{7}\cdot \frac{8}{23}\)
Multiply left to right.\(1\cdot \frac{8}{23}\)
Multiply.\(\frac{8}{23}\)

In expressions where we need to add or subtract three or more fractions, combine those with a common denominator first.

Example

Try it.

Simplify: \((\frac{5}{13}+\frac{3}{4})+\frac{1}{4}.\)

Solution

Notice that the second and third terms have a common denominator, so this work will be easier if we change the grouping.

\((\frac{5}{13}+\frac{3}{4})+\frac{1}{4}\)
Group the terms with a common denominator.\(\frac{5}{13}+(\frac{3}{4}+\frac{1}{4})\)
Add in the parentheses first.\(\frac{5}{13}+(\frac{4}{4})\)
Simplify the fraction.\(\frac{5}{13}+1\)
Add.\(1\frac{5}{13}\)
Convert to an improper fraction.\(\frac{18}{13}\)

When adding and subtracting three or more terms involving decimals, look for terms that combine to give whole numbers.

Condensed: the full section is in OpenStax Prealgebra 2e.

Key Concepts

  • Commutative Properties
    • Commutative Property of Addition:
      • If \(a,b\) are real numbers, then \(a+b=b+a\)
    • Commutative Property of Multiplication:
      • If \(a,b\) are real numbers, then \(a⋅b=b⋅a\)
  • Associative Properties
    • Associative Property of Addition:
      • If \(a,b,c\) are real numbers then \((a+b)+c=a+(b+c)\)
    • Associative Property of Multiplication:
      • If \(a,b,c\) are real numbers then \((a⋅b)⋅c=a⋅(b⋅c)\)

Commutative and Associative Properties

Use the Commutative and Associative Properties

In the following exercises, use the commutative properties to rewrite the given expression.

Try it.

\(8(-12)=\underline{\hspace{1.5em}}\)

Try it.

\(7(-13)=\underline{\hspace{1.5em}}\)

Solution

7(−13) = (−13)7

Try it.

\((-19)(-14)=\underline{\hspace{1.5em}}\)

Try it.

\((-12)(-18)=\underline{\hspace{1.5em}}\)

Solution

(−12)(−18) = (−18)(−12)

Try it.

\(-11+8=\underline{\hspace{1.5em}}\)

Try it.

\(-15+7=\underline{\hspace{1.5em}}\)

Solution

−15 + 7 = 7 + (−15)

In the following exercises, use the associative properties to rewrite the given expression.

Try it.

\((11+9)+14=\underline{\hspace{1.5em}}\)

Try it.

\((21+14)+9=\underline{\hspace{1.5em}}\)

Solution

(21 + 14) + 9 = 21 + (14 + 9)

Try it.

\((12\cdot 5)\cdot 7=\underline{\hspace{1.5em}}\)

Try it.

\((14\cdot 6)\cdot 9=\underline{\hspace{1.5em}}\)

Solution

(14 · 6) · 9 = 14(6 · 9)

Try it.

\((-7+9)+8=\underline{\hspace{1.5em}}\)

Try it.

\((-2+6)+7=\underline{\hspace{1.5em}}\)

Solution

(−2 + 6) + 7 = −2 + (6 + 7)

Try it.

\((16\cdot \frac{4}{5})\cdot 15=\underline{\hspace{1.5em}}\)

Try it.

\((13\cdot \frac{2}{3})\cdot 18=\underline{\hspace{1.5em}}\)

Solution

\((13\cdot \frac{2}{3})\cdot 18=13(\frac{2}{3}\cdot 18)\)

Try it.

\(3(4x)=\underline{\hspace{1.5em}}\)

Try it.

\(4(7x)=\underline{\hspace{1.5em}}\)

Solution

4(7x) = (4 · 7)x

Try it.

\((12+x)+28=\underline{\hspace{1.5em}}\)

Try it.

\((17+y)+33=\underline{\hspace{1.5em}}\)

Solution

(17 + y) + 33 = 17 + (y + 33)

Evaluate Expressions using the Commutative and Associative Properties

In the following exercises, evaluate each expression for the given value.

Try it.

If \(y=\frac{5}{8},\) evaluate:

  1. ⓐ \(\ y+0.49+(-y)\)
  2. ⓑ \(\ y+(-y)+0.49\)

Try it.

If \(z=\frac{7}{8},\) evaluate:

  • ⓐ \(\ z+0.97+(-z)\)
  • ⓑ \(\ z+(-z)+0.97\)

Solution
  1. ⓐ 0.97
  2. ⓑ 0.97

Try it.

If \(c=-\frac{11}{4},\) evaluate:

  1. ⓐ \(\ c+3.125+(-c)\)
  2. ⓑ \(\ c+(-c)+3.125\)

Try it.

If \(d=-\frac{9}{4},\) evaluate:

  1. ⓐ \(\ d+2.375+(-d)\)
  2. ⓑ \(\ d+(-d)+2.375\)

Solution

  1. ⓐ 2.375
  2. ⓑ 2.375

Try it.

If \(j=11,\) evaluate:

  1. ⓐ \(\ \frac{5}{6}(\frac{6}{5}j)\)
  2. ⓑ \(\ (\frac{5}{6}\cdot \frac{6}{5})j\)

Try it.

If \(k=21,\) evaluate:

  1. ⓐ \(\ \frac{4}{13}(\frac{13}{4}k)\)
  2. ⓑ \(\ (\frac{4}{13}\cdot \frac{13}{4})k\)

Solution

  1. ⓐ 21
  2. ⓑ 21

Try it.

If \(m=-25,\) evaluate:

  1. ⓐ \(\ -\frac{3}{7}(\frac{7}{3}m)\)
  2. ⓑ \(\ (-\frac{3}{7}\cdot \frac{7}{3})m\)

Try it.

If \(n=-8,\) evaluate:

  1. ⓐ \(\ -\frac{5}{21}(\frac{21}{5}n)\)
  2. ⓑ \(\ (-\frac{5}{21}\cdot \frac{21}{5})n\)

Solution

  1. ⓐ 8
  2. ⓑ 8

Simplify Expressions Using the Commutative and Associative Properties

In the following exercises, simplify.

Try it.

\(-45a+15+45a\)

Try it.

\(9y+23+(-9y)\)

Solution

23

Try it.

\(\frac{1}{2}+\frac{7}{8}+(-\frac{1}{2})\)

Try it.

\(\frac{2}{5}+\frac{5}{12}+(-\frac{2}{5})\)

Solution

\(\frac{5}{12}\)

Try it.

\(\frac{3}{20}\cdot \frac{49}{11}\cdot \frac{20}{3}\)

Try it.

\(\frac{13}{18}\cdot \frac{25}{7}\cdot \frac{18}{13}\)

Solution

\(\frac{25}{7}\)

Try it.

\(\frac{7}{12}\cdot \frac{9}{17}\cdot \frac{24}{7}\)

Try it.

\(\frac{3}{10}\cdot \frac{13}{23}\cdot \frac{50}{3}\)

Solution

\(\frac{65}{23}\)

Try it.

\(-24\cdot 7\cdot \frac{3}{8}\)

Try it.

\(-36\cdot 11\cdot \frac{4}{9}\)

Solution

−176

Try it.

\((\frac{5}{6}+\frac{8}{15})+\frac{7}{15}\)

Try it.

\((\frac{1}{12}+\frac{4}{9})+\frac{5}{9}\)

Solution

\(\frac{13}{12}\)

Try it.

\(\frac{5}{13}+\frac{3}{4}+\frac{1}{4}\)

Try it.

\(\frac{8}{15}+\frac{5}{7}+\frac{2}{7}\)

Solution

\(\frac{23}{15}\)

Try it.

\((4.33p+1.09p)+3.91p\)

Try it.

\((5.89d+2.75d)+1.25d\)

Solution

9.89d

Try it.

\(17(0.25)(4)\)

Try it.

\(36(0.2)(5)\)

Solution

36

Try it.

\([2.48(12)](0.5)\)

Try it.

\([9.731(4)](0.75)\)

Solution

29.193

Try it.

\(12(\frac{5}{6}p)\)

Try it.

\(20(\frac{3}{5}q)\)

Solution

12q

Try it.

\(14x+19y+25x+3y\)

Try it.

\(15u+11v+27u+19v\)

Solution

42u + 30v

Try it.

\(43m+(-12n)+\text{}(-16m)+(-9n)\)

Try it.

\(-22p+17q+(-35p)+(-27q)\)

Solution

−57p + (−10q)

Try it.

\(\frac{3}{8}g+\frac{1}{12}h+\frac{7}{8}g+\frac{5}{12}h\)

Try it.

\(\frac{5}{6}a+\frac{3}{10}b+\frac{1}{6}a+\frac{9}{10}b\)

Solution

\(a+\frac{6}{5}b\)

Try it.

\(6.8p+9.14q+(-4.37p)+(-0.88q)\)

Try it.

\(9.6m+7.22n+(-2.19m)+(-0.65n)\)

Solution

7.41m + 6.57n

Condensed: the full section is in OpenStax Prealgebra 2e.

Sada ti Nema kalkulatora koji bi riješio ovaj problem, ali njegovi dijelovi se mogu izračunati. Pokušajte jedan od gore navedenih ili upišite svoj.

Vježba (40)

Prvo isprobajte svaki na papiru. Otkrijte odgovor da provjerite; provjereni se mogu otvoriti u rješavaču za svaki korak.

  1. Simplify: \(7y+2+y+13.\)

    Otkrij odgovor

    \(8y+15\)

  2. Multiply: \(\frac{2}{3}\cdot 18.\)

    Otkrij odgovor

    \(12\)

  3. Find the opposite of \(15.\)

    Otkrij odgovor

    \(-15\)

  4. Use the commutative properties to rewrite the following expressions:

    1. ⓐ \(\ -1+3=\underline{\hspace{1.5em}}\)

    2. ⓑ \(\ 4\cdot 9=\underline{\hspace{1.5em}}\)

    Otkrij odgovor
    ⓐ
    \(-1+3=\underline{\hspace{1.5em}}\)
    Use the commutative property of addition to change the order.\(-1+3=3+(-1)\)
    ⓑ
    \(4\cdot 9=\underline{\hspace{1.5em}}\)
    Use the commutative property of multiplication to change the order.\(4\cdot 9=9\cdot 4\)
  5. Use the commutative properties to rewrite the following:

    1. ⓐ \(\ -4+7=\underline{\hspace{1.5em}}\)
    2. ⓑ \(\ 6\cdot 12=\underline{\hspace{1.5em}}\)

    Otkrij odgovor
    1. ⓐ −4 + 7 = 7 + (−4)
    2. ⓑ 6 · 12 = 12 · 6
  6. Use the commutative properties to rewrite the following:

    1. ⓐ \(\ 14+(-2)=\underline{\hspace{1.5em}}\)
    2. ⓑ \(\ 3(-5)=\underline{\hspace{1.5em}}\)

    Otkrij odgovor
    1. ⓐ 14 + (−2) = −2 + 14
    2. ⓑ 3(−5) = (−5)3
  7. Use the associative properties to rewrite the following:

    1. ⓐ \(\ (3+0.6)+0.4=\underline{\hspace{1.5em}}\)

    2. ⓑ \(\ (-4\cdot \frac{2}{5})\cdot 15=\underline{\hspace{1.5em}}\)

    Otkrij odgovor
    ⓐ
    \((3+0.6)+0.4=\underline{\hspace{1.5em}}\)
    Change the grouping.\((3+0.6)+0.4=3+(0.6+0.4)\)

    Notice that \(0.6+0.4\) is \(1,\) so the addition will be easier if we group as shown on the right.

    ⓑ
    \((-4\cdot \frac{2}{5})\cdot 15=\underline{\hspace{1.5em}}\)
    Change the grouping.\((-4\cdot \frac{2}{5})\cdot 15=-4\cdot (\frac{2}{5}\cdot 15)\)

    Notice that \(\frac{2}{5}\cdot 15\) is \(6.\) The multiplication will be easier if we group as shown on the right.

  8. Use the associative properties to rewrite the following:
    ⓐ \((1+0.7)+0.3=\underline{\hspace{1.5em}}\) ⓑ \(\ (-9\cdot 8)\cdot \frac{3}{4}=\underline{\hspace{1.5em}}\)

    Otkrij odgovor
    1. ⓐ \(\ (1+0.7)+0.3=1+(0.7+0.3)\)
    2. ⓑ \(\ (-9\cdot 8)\cdot \frac{3}{4}=-9(8\cdot \frac{3}{4})\)
  9. Use the associative properties to rewrite the following:
    ⓐ \((4+0.6)+0.4=\underline{\hspace{1.5em}}\) ⓑ \(\ (-2\cdot 12)\cdot \frac{5}{6}=\underline{\hspace{1.5em}}\)

    Otkrij odgovor
    1. ⓐ \(\ (4+0.6)+0.4=4+(0.6+0.4)\)
    2. ⓑ \(\ (-2\cdot 12)\cdot \frac{5}{6}=-2(12\cdot \frac{5}{6})\)
  10. Use the Associative Property of Multiplication to simplify: \(6(3x).\)

    Otkrij odgovor
    \(6(3x)\)
    Change the grouping.\((6\cdot 3)x\)
    Multiply in the parentheses.\(18x\)

    Notice that we can multiply \(6\cdot 3,\) but we could not multiply \(3\cdot x\) without having a value for \(x.\)

  11. Use the Associative Property of Multiplication to simplify the given expression: \(8(4x).\)

    Otkrij odgovor

    8(4x) = (8 · 4)x = 32x

  12. Use the Associative Property of Multiplication to simplify the given expression: \(-9(7y).\)

    Otkrij odgovor

    −9(7y) = (−9 · 7)y = −63y

  13. Evaluate each expression when \(x=\frac{7}{8}.\)

    1. ⓐ \(\ x+0.37+(-x)\)
    2. ⓑ \(\ x+(-x)+0.37\)
    Otkrij odgovor
    ⓐ
    Substitute \(\frac{7}{8}\) for \(x\).
    Convert fractions to decimals.
    Add left to right.
    Subtract.
    ⓑ
    Substitute \(\frac{7}{8}\) for x.
    Add opposites first.

    What was the difference between part ⓐ and part ⓑ ? Only the order changed. By the Commutative Property of Addition, \(x+0.37+(-x)=x+(-x)+0.37.\) But wasn’t part ⓑ much easier?

  14. Evaluate each expression when \(y=\frac{3}{8}:\)ⓐ \(\ y+0.84+(-y)\) ⓑ \(\ y+(-y)+0.84.\)

    Otkrij odgovor

    1. ⓐ 0.84
    2. ⓑ 0.84

  15. Evaluate each expression when \(f=\frac{17}{20}:\)ⓐ \(\ f+0.975+(-f)\) ⓑ \(\ f+(-f)+0.975.\)

    Otkrij odgovor

    1. ⓐ 0.975
    2. ⓑ 0.975

  16. Evaluate each expression when \(n=17.\)

    1. ⓐ \(\ \frac{4}{3}(\frac{3}{4}n)\)

    2. ⓑ \(\ (\frac{4}{3}\cdot \frac{3}{4})n\)

    Otkrij odgovor
    ⓐ
    Substitute 17 for n.
    Multiply in the parentheses first.
    Multiply again.
    ⓑ
    Substitute 17 for n.
    Multiply. The product of reciprocals is 1.
    Multiply again.

    What was the difference between part ⓐ and part ⓑ here? Only the grouping changed. By the Associative Property of Multiplication, \(\frac{4}{3}(\frac{3}{4}n)=(\frac{4}{3}\cdot \frac{3}{4})n.\) By carefully choosing how to group the factors, we can make the work easier.

  17. Evaluate each expression when \(p=24\text{:}\)ⓐ \(\ \frac{5}{9}(\frac{9}{5}p)\) ⓑ \(\ (\frac{5}{9}\cdot \frac{9}{5})p.\)

    Otkrij odgovor

    1. ⓐ 24
    2. ⓑ 24

  18. Evaluate each expression when \(q=15\text{:}\)ⓐ \(\ \frac{7}{11}(\frac{11}{7}q)\) ⓑ \(\ (\frac{7}{11}\cdot \frac{11}{7})q\)

    Otkrij odgovor

    1. ⓐ 15
    2. ⓑ 15

  19. Simplify: \(-84n+(-73n)+84n.\)

    Otkrij odgovor

    Notice the first and third terms are opposites, so we can use the commutative property of addition to reorder the terms.

    \(-84n+(-73n)+84n\)
    Re-order the terms.\(-84n+84n+(-73n)\)
    Add left to right.\(0+(-73n)\)
    Add.\(-73n\)
  20. Simplify: \(-27a+(-48a)+27a.\)

    Otkrij odgovor

    −48a

  21. Simplify: \(39x+(-92x)+(-39x).\)

    Otkrij odgovor

    −92x

  22. Simplify: \(\frac{7}{15}\cdot \frac{8}{23}\cdot \frac{15}{7}.\)

    Otkrij odgovor

    Notice the first and third terms are reciprocals, so we can use the Commutative Property of Multiplication to reorder the factors.

    \(\frac{7}{15}\cdot \frac{8}{23}\cdot \frac{15}{7}\)
    Re-order the terms.\(\frac{7}{15}\cdot \frac{15}{7}\cdot \frac{8}{23}\)
    Multiply left to right.\(1\cdot \frac{8}{23}\)
    Multiply.\(\frac{8}{23}\)
  23. Simplify: \(\frac{9}{16}\cdot \frac{5}{49}\cdot \frac{16}{9}.\)

    Otkrij odgovor

    \(\frac{5}{49}\)

  24. Simplify: \(\frac{6}{17}\cdot \frac{11}{25}\cdot \frac{17}{6}.\)

    Otkrij odgovor

    \(\frac{11}{25}\)

  25. Simplify: \((\frac{5}{13}+\frac{3}{4})+\frac{1}{4}.\)

    Otkrij odgovor

    Notice that the second and third terms have a common denominator, so this work will be easier if we change the grouping.

    \((\frac{5}{13}+\frac{3}{4})+\frac{1}{4}\)
    Group the terms with a common denominator.\(\frac{5}{13}+(\frac{3}{4}+\frac{1}{4})\)
    Add in the parentheses first.\(\frac{5}{13}+(\frac{4}{4})\)
    Simplify the fraction.\(\frac{5}{13}+1\)
    Add.\(1\frac{5}{13}\)
    Convert to an improper fraction.\(\frac{18}{13}\)
  26. Simplify: \((\frac{7}{15}+\frac{5}{8})+\frac{3}{8}.\)

    Otkrij odgovor

    \(\frac{22}{15}\)

  27. Simplify: \((\frac{2}{9}+\frac{7}{12})+\frac{5}{12}.\)

    Otkrij odgovor

    \(\frac{11}{9}\)

  28. Simplify: \((6.47q+9.99q)+1.01q.\)

    Otkrij odgovor

    Notice that the sum of the second and third coefficients is a whole number.

    \((6.47q+9.99q)+1.01q\)
    Change the grouping.\(6.47q+(9.99q+1.01q)\)
    Add in the parentheses first.\(6.47q+(11.00q)\)
    Add.\(17.47q\)

    Many people have good number sense when they deal with money. Think about adding \(99\) cents and \(1\) cent. Do you see how this applies to adding \(9.99+1.01?\)

  29. Simplify: \((5.58c+8.75c)+1.25c.\)

    Otkrij odgovor

    15.58c

  30. Simplify: \((8.79d+3.55d)+5.45d.\)

    Otkrij odgovor

    17.79d

  31. Simplify the expression: \([1.67(8)](0.25).\)

    Otkrij odgovor

    Notice that multiplying \((8)(0.25)\) is easier than multiplying \(1.67(8)\) because it gives a whole number. (Think about having \(8\) quarters, that makes \(\text{\$2.)}\)

    \([1.67(8)](0.25)\)
    Regroup.\(1.67[(8)(0.25)]\)
    Multiply in the brackets first.\(1.67[2]\)
    Multiply.\(3.34\)
  32. Simplify: \([1.17(4)](2.25).\)

    Otkrij odgovor

    10.53

  33. Simplify: \([3.52(8)](2.5).\)

    Otkrij odgovor

    70.4

  34. Simplify: \(6(9x).\)

    Otkrij odgovor
    \(6(9x)\)
    Use the associative property of multiplication to re-group.\((6\cdot 9)x\)
    Multiply in the parentheses.\(54x\)
  35. Simplify: \(8(3y).\)

    Otkrij odgovor

    24y

  36. Simplify: \(12(5z).\)

    Otkrij odgovor

    60z

  37. Simplify: \(18p+6q+(-15p)+5q.\)

    Otkrij odgovor

    Use the Commutative Property of Addition to re-order so that like terms are together.

    \(18p+6q+(-15p)+5q\)
    Re-order terms.\(18p+(-15p)+6q+5q\)
    Combine like terms.\(3p+11q\)
  38. Simplify: \(23r+14s+9r+(-15s).\)

    Otkrij odgovor

    32r − s

  39. Simplify: \(37m+21n+4m+(-15n).\)

    Otkrij odgovor

    41m + 6n

  40. \(8(-12)=\underline{\hspace{1.5em}}\)

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Simboli korišteni ovdje

Dodirnite bilo koji simbol za potpunu definiciju, sliku i što svako slovo u njemu znači.

Kako: Commutative and Associative Properties

  1. Use the commutative and associative properties
  2. Evaluate expressions using the commutative and associative properties
  3. Simplify expressions using the commutative and associative properties

Pitanja koja ljudi postavljaju

Why does the order of operations matter?

Because 2 + 3 × 4 would otherwise be two different numbers. The convention (brackets, exponents, multiplication and division, addition and subtraction) exists so every reader gets the same value from the same expression.

How do I check an arithmetic answer?

Estimate first (round every number and compute roughly), then compare. If the estimate and the exact answer disagree by more than a little, one of them is wrong. The solver shows every operation, so you can find which line went astray.

Why are fractions harder than decimals?

They are not harder, they are more exact: 1/3 is a precise number, 0.333 is an approximation. Fractions need a common denominator to add, which is the one extra step people trip on.

Dio ove stranice je prilagođen sa OpenStax Prealgebra 2e (CC BY-NC-SA 4.0). U ovom slučaju, greške su naše. Budući da je izvor licenciran za nekomercijalnu upotrebu, ova lekcija je besplatna, ne prodaje ništa i dijeli se pod istom licencom.

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