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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:
-
ⓐ \(\ -1+3=\underline{\hspace{1.5em}}\)
-
ⓑ \(\ 4\cdot 9=\underline{\hspace{1.5em}}\)
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 \(x=\frac{7}{8}.\)
- ⓐ \(\ x+0.37+(-x)\)
- ⓑ \(\ x+(-x)+0.37\)
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 \(n=17.\)
-
ⓐ \(\ \frac{4}{3}(\frac{3}{4}n)\)
-
ⓑ \(\ (\frac{4}{3}\cdot \frac{3}{4})n\)
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 \(a⋅b=b⋅a\)
- Commutative Property of Addition:
- 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 \((a⋅b)⋅c=a⋅(b⋅c)\)
- Associative Property of Addition:
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:
- ⓐ \(\ y+0.49+(-y)\)
- ⓑ \(\ y+(-y)+0.49\)
Try it.
If \(z=\frac{7}{8},\) evaluate:
- ⓐ \(\ z+0.97+(-z)\)
- ⓑ \(\ z+(-z)+0.97\)
Solution
- ⓐ 0.97
- ⓑ 0.97
Try it.
If \(c=-\frac{11}{4},\) evaluate:
- ⓐ \(\ c+3.125+(-c)\)
- ⓑ \(\ c+(-c)+3.125\)
Try it.
If \(d=-\frac{9}{4},\) evaluate:
- ⓐ \(\ d+2.375+(-d)\)
- ⓑ \(\ d+(-d)+2.375\)
Solution
- ⓐ 2.375
- ⓑ 2.375
Try it.
If \(j=11,\) evaluate:
- ⓐ \(\ \frac{5}{6}(\frac{6}{5}j)\)
- ⓑ \(\ (\frac{5}{6}\cdot \frac{6}{5})j\)
Try it.
If \(k=21,\) evaluate:
- ⓐ \(\ \frac{4}{13}(\frac{13}{4}k)\)
- ⓑ \(\ (\frac{4}{13}\cdot \frac{13}{4})k\)
Solution
- ⓐ 21
- ⓑ 21
Try it.
If \(m=-25,\) evaluate:
- ⓐ \(\ -\frac{3}{7}(\frac{7}{3}m)\)
- ⓑ \(\ (-\frac{3}{7}\cdot \frac{7}{3})m\)
Try it.
If \(n=-8,\) evaluate:
- ⓐ \(\ -\frac{5}{21}(\frac{21}{5}n)\)
- ⓑ \(\ (-\frac{5}{21}\cdot \frac{21}{5})n\)
Solution
- ⓐ 8
- ⓑ 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.
Haddaad Ma jiro xisaabis oo ku nool midkan, laakiin qaybaha ka mid ah waa la xisaabi karaa. Ku raaxayso mid ka hooseeya, ama ku qor adiga kuu gaar ah.
Tababar (40)
Ku day mid kasta oo ka mid ah oo ku saabsan buugaag hore. Muuji jawaabta si aad u hubiso; kuwa la xaqiijiyay waxaa la furi karaa in xalka ah tallaabo kasta.
-
Simplify: \(7y+2+y+13.\)
Jawaabta muuji
\(8y+15\)
-
Multiply: \(\frac{2}{3}\cdot 18.\)
Jawaabta muuji
\(12\)
-
Find the opposite of \(15.\)
Jawaabta muuji
\(-15\)
-
Use the commutative properties to rewrite the following expressions:
-
ⓐ \(\ -1+3=\underline{\hspace{1.5em}}\)
-
ⓑ \(\ 4\cdot 9=\underline{\hspace{1.5em}}\)
Jawaabta muuji
ⓐ \(-1+3=\underline{\hspace{1.5em}}\) 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. \(4\cdot 9=9\cdot 4\) -
-
Use the commutative properties to rewrite the following:
- ⓐ \(\ -4+7=\underline{\hspace{1.5em}}\)
- ⓑ \(\ 6\cdot 12=\underline{\hspace{1.5em}}\)
Jawaabta muuji
- ⓐ −4 + 7 = 7 + (−4)
- ⓑ 6 · 12 = 12 · 6
-
Use the commutative properties to rewrite the following:
- ⓐ \(\ 14+(-2)=\underline{\hspace{1.5em}}\)
- ⓑ \(\ 3(-5)=\underline{\hspace{1.5em}}\)
Jawaabta muuji
- ⓐ 14 + (−2) = −2 + 14
- ⓑ 3(−5) = (−5)3
-
Use the associative properties to rewrite the following:
-
ⓐ \(\ (3+0.6)+0.4=\underline{\hspace{1.5em}}\)
-
ⓑ \(\ (-4\cdot \frac{2}{5})\cdot 15=\underline{\hspace{1.5em}}\)
Jawaabta muuji
ⓐ \((3+0.6)+0.4=\underline{\hspace{1.5em}}\) 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.
-
-
Use the associative properties to rewrite the following:
ⓐ \((1+0.7)+0.3=\underline{\hspace{1.5em}}\) ⓑ \(\ (-9\cdot 8)\cdot \frac{3}{4}=\underline{\hspace{1.5em}}\)Jawaabta muuji
- ⓐ \(\ (1+0.7)+0.3=1+(0.7+0.3)\)
- ⓑ \(\ (-9\cdot 8)\cdot \frac{3}{4}=-9(8\cdot \frac{3}{4})\)
-
Use the associative properties to rewrite the following:
ⓐ \((4+0.6)+0.4=\underline{\hspace{1.5em}}\) ⓑ \(\ (-2\cdot 12)\cdot \frac{5}{6}=\underline{\hspace{1.5em}}\)Jawaabta muuji
- ⓐ \(\ (4+0.6)+0.4=4+(0.6+0.4)\)
- ⓑ \(\ (-2\cdot 12)\cdot \frac{5}{6}=-2(12\cdot \frac{5}{6})\)
-
Use the Associative Property of Multiplication to simplify: \(6(3x).\)
Jawaabta muuji
\(6(3x)\) 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 \(x.\)
-
Use the Associative Property of Multiplication to simplify the given expression: \(8(4x).\)
Jawaabta muuji
8(4x) = (8 · 4)x = 32x
-
Use the Associative Property of Multiplication to simplify the given expression: \(-9(7y).\)
Jawaabta muuji
−9(7y) = (−9 · 7)y = −63y
-
Evaluate each expression when \(x=\frac{7}{8}.\)
- ⓐ \(\ x+0.37+(-x)\)
- ⓑ \(\ x+(-x)+0.37\)
Jawaabta muuji
ⓐ 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?
-
Evaluate each expression when \(y=\frac{3}{8}:\)ⓐ \(\ y+0.84+(-y)\) ⓑ \(\ y+(-y)+0.84.\)
Jawaabta muuji
- ⓐ 0.84
- ⓑ 0.84
-
Evaluate each expression when \(f=\frac{17}{20}:\)ⓐ \(\ f+0.975+(-f)\) ⓑ \(\ f+(-f)+0.975.\)
Jawaabta muuji
- ⓐ 0.975
- ⓑ 0.975
-
Evaluate each expression when \(n=17.\)
-
ⓐ \(\ \frac{4}{3}(\frac{3}{4}n)\)
-
ⓑ \(\ (\frac{4}{3}\cdot \frac{3}{4})n\)
Jawaabta muuji
ⓐ 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.
-
-
Evaluate each expression when \(p=24\text{:}\)ⓐ \(\ \frac{5}{9}(\frac{9}{5}p)\) ⓑ \(\ (\frac{5}{9}\cdot \frac{9}{5})p.\)
Jawaabta muuji
- ⓐ 24
- ⓑ 24
-
Evaluate each expression when \(q=15\text{:}\)ⓐ \(\ \frac{7}{11}(\frac{11}{7}q)\) ⓑ \(\ (\frac{7}{11}\cdot \frac{11}{7})q\)
Jawaabta muuji
- ⓐ 15
- ⓑ 15
-
Simplify: \(-84n+(-73n)+84n.\)
Jawaabta muuji
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. \(-73n\) -
Simplify: \(-27a+(-48a)+27a.\)
Jawaabta muuji
−48a
-
Simplify: \(39x+(-92x)+(-39x).\)
Jawaabta muuji
−92x
-
Simplify: \(\frac{7}{15}\cdot \frac{8}{23}\cdot \frac{15}{7}.\)
Jawaabta muuji
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. \(\frac{8}{23}\) -
Simplify: \(\frac{9}{16}\cdot \frac{5}{49}\cdot \frac{16}{9}.\)
Jawaabta muuji
\(\frac{5}{49}\)
-
Simplify: \(\frac{6}{17}\cdot \frac{11}{25}\cdot \frac{17}{6}.\)
Jawaabta muuji
\(\frac{11}{25}\)
-
Simplify: \((\frac{5}{13}+\frac{3}{4})+\frac{1}{4}.\)
Jawaabta muuji
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. \(\frac{18}{13}\) -
Simplify: \((\frac{7}{15}+\frac{5}{8})+\frac{3}{8}.\)
Jawaabta muuji
\(\frac{22}{15}\)
-
Simplify: \((\frac{2}{9}+\frac{7}{12})+\frac{5}{12}.\)
Jawaabta muuji
\(\frac{11}{9}\)
-
Simplify: \((6.47q+9.99q)+1.01q.\)
Jawaabta muuji
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 \(9.99+1.01?\)
-
Simplify: \((5.58c+8.75c)+1.25c.\)
Jawaabta muuji
15.58c
-
Simplify: \((8.79d+3.55d)+5.45d.\)
Jawaabta muuji
17.79d
-
Simplify the expression: \([1.67(8)](0.25).\)
Jawaabta muuji
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. \(3.34\) -
Simplify: \([1.17(4)](2.25).\)
Jawaabta muuji
10.53
-
Simplify: \([3.52(8)](2.5).\)
Jawaabta muuji
70.4
-
Simplify: \(6(9x).\)
Jawaabta muuji
\(6(9x)\) Use the associative property of multiplication to re-group. \((6\cdot 9)x\) Multiply in the parentheses. \(54x\) -
Simplify: \(8(3y).\)
Jawaabta muuji
24y
-
Simplify: \(12(5z).\)
Jawaabta muuji
60z
-
Simplify: \(18p+6q+(-15p)+5q.\)
Jawaabta muuji
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. \(3p+11q\) -
Simplify: \(23r+14s+9r+(-15s).\)
Jawaabta muuji
32r − s
-
Simplify: \(37m+21n+4m+(-15n).\)
Jawaabta muuji
41m + 6n
-
\(8(-12)=\underline{\hspace{1.5em}}\)
A xisaab bilaash ah oo ku daray qoraalada waxbarasho kasta, diiwaanka waxa aad dhamaatay, dhibaatooyinka aad xal u tahay meel kaliya, iyo macallinka aad weydiisan kartaa oo ku saabsan bogga. Xisaabta oo keliya waa u furan yahay qof kasta, la diiwaan geliyey ama aan.
Ku soo biir Ku soo galCalaamadaha la isticmaalay halkan
Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.
Sida loo: Commutative and Associative Properties
- Use the commutative and associative properties
- Evaluate expressions using the commutative and associative properties
- Simplify expressions using the commutative and associative properties
Su'aalaha dadku waydiiyaan
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.
Qaybaha ee boggaan waxaa la beddelay ka OpenStax Prealgebra 2e (CC BY-NC-SA 4.0). Condensed iyo dib-oo halkan lagu sharaxay; khaladaadka waa inoo. Sababtoo ah asalka waxaa loo oggolaaday in aan loo isticmaalin ganacsi, fasalkan waa bilaash, waxba ma iibiyo, iyo waxaa lagu wadaagayaa liisan la mid ah.