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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.
Tani ti. Asnjë kalkulator nuk e zgjidh këtë, por pjesët e saj janë të llogaritura. Provo një më poshtë, ose shkruaj tënde.
Praktiko (40)
Provo secilin në letër së pari. Zbulo përgjigjen për të kontrolluar; ato të verifikuara mund të hapen në zgjidhës për çdo hap.
-
Simplify: \(7y+2+y+13.\)
Zbulo përgjigjen
\(8y+15\)
-
Multiply: \(\frac{2}{3}\cdot 18.\)
Zbulo përgjigjen
\(12\)
-
Find the opposite of \(15.\)
Zbulo përgjigjen
\(-15\)
-
Use the commutative properties to rewrite the following expressions:
-
ⓐ \(\ -1+3=\underline{\hspace{1.5em}}\)
-
ⓑ \(\ 4\cdot 9=\underline{\hspace{1.5em}}\)
Zbulo përgjigjen
ⓐ \(-1+3=\underline{\hspace{1.5em}}\) 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:
- ⓐ \(\ -4+7=\underline{\hspace{1.5em}}\)
- ⓑ \(\ 6\cdot 12=\underline{\hspace{1.5em}}\)
Zbulo përgjigjen
- ⓐ −4 + 7 = 7 + (−4)
- ⓑ 6 · 12 = 12 · 6
-
Use the commutative properties to rewrite the following:
- ⓐ \(\ 14+(-2)=\underline{\hspace{1.5em}}\)
- ⓑ \(\ 3(-5)=\underline{\hspace{1.5em}}\)
Zbulo përgjigjen
- ⓐ 14 + (−2) = −2 + 14
- ⓑ 3(−5) = (−5)3
-
Use the associative properties to rewrite the following:
-
ⓐ \(\ (3+0.6)+0.4=\underline{\hspace{1.5em}}\)
-
ⓑ \(\ (-4\cdot \frac{2}{5})\cdot 15=\underline{\hspace{1.5em}}\)
Zbulo përgjigjen
ⓐ \((3+0.6)+0.4=\underline{\hspace{1.5em}}\) 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:
ⓐ \((1+0.7)+0.3=\underline{\hspace{1.5em}}\) ⓑ \(\ (-9\cdot 8)\cdot \frac{3}{4}=\underline{\hspace{1.5em}}\)Zbulo përgjigjen
- ⓐ \(\ (1+0.7)+0.3=1+(0.7+0.3)\)
- ⓑ \(\ (-9\cdot 8)\cdot \frac{3}{4}=-9(8\cdot \frac{3}{4})\)
-
Use the associative properties to rewrite the following:
ⓐ \((4+0.6)+0.4=\underline{\hspace{1.5em}}\) ⓑ \(\ (-2\cdot 12)\cdot \frac{5}{6}=\underline{\hspace{1.5em}}\)Zbulo përgjigjen
- ⓐ \(\ (4+0.6)+0.4=4+(0.6+0.4)\)
- ⓑ \(\ (-2\cdot 12)\cdot \frac{5}{6}=-2(12\cdot \frac{5}{6})\)
-
Use the Associative Property of Multiplication to simplify: \(6(3x).\)
Zbulo përgjigjen
\(6(3x)\) 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: \(8(4x).\)
Zbulo përgjigjen
8(4x) = (8 · 4)x = 32x
-
Use the Associative Property of Multiplication to simplify the given expression: \(-9(7y).\)
Zbulo përgjigjen
−9(7y) = (−9 · 7)y = −63y
-
Evaluate each expression when \(x=\frac{7}{8}.\)
- ⓐ \(\ x+0.37+(-x)\)
- ⓑ \(\ x+(-x)+0.37\)
Zbulo përgjigjen
ⓐ 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 \(y=\frac{3}{8}:\)ⓐ \(\ y+0.84+(-y)\) ⓑ \(\ y+(-y)+0.84.\)
Zbulo përgjigjen
- ⓐ 0.84
- ⓑ 0.84
-
Evaluate each expression when \(f=\frac{17}{20}:\)ⓐ \(\ f+0.975+(-f)\) ⓑ \(\ f+(-f)+0.975.\)
Zbulo përgjigjen
- ⓐ 0.975
- ⓑ 0.975
-
Evaluate each expression when \(n=17.\)
-
ⓐ \(\ \frac{4}{3}(\frac{3}{4}n)\)
-
ⓑ \(\ (\frac{4}{3}\cdot \frac{3}{4})n\)
Zbulo përgjigjen
ⓐ 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 \(p=24\text{:}\)ⓐ \(\ \frac{5}{9}(\frac{9}{5}p)\) ⓑ \(\ (\frac{5}{9}\cdot \frac{9}{5})p.\)
Zbulo përgjigjen
- ⓐ 24
- ⓑ 24
-
Evaluate each expression when \(q=15\text{:}\)ⓐ \(\ \frac{7}{11}(\frac{11}{7}q)\) ⓑ \(\ (\frac{7}{11}\cdot \frac{11}{7})q\)
Zbulo përgjigjen
- ⓐ 15
- ⓑ 15
-
Simplify: \(-84n+(-73n)+84n.\)
Zbulo përgjigjen
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: \(-27a+(-48a)+27a.\)
Zbulo përgjigjen
−48a
-
Simplify: \(39x+(-92x)+(-39x).\)
Zbulo përgjigjen
−92x
-
Simplify: \(\frac{7}{15}\cdot \frac{8}{23}\cdot \frac{15}{7}.\)
Zbulo përgjigjen
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: \(\frac{9}{16}\cdot \frac{5}{49}\cdot \frac{16}{9}.\)
Zbulo përgjigjen
\(\frac{5}{49}\)
-
Simplify: \(\frac{6}{17}\cdot \frac{11}{25}\cdot \frac{17}{6}.\)
Zbulo përgjigjen
\(\frac{11}{25}\)
-
Simplify: \((\frac{5}{13}+\frac{3}{4})+\frac{1}{4}.\)
Zbulo përgjigjen
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: \((\frac{7}{15}+\frac{5}{8})+\frac{3}{8}.\)
Zbulo përgjigjen
\(\frac{22}{15}\)
-
Simplify: \((\frac{2}{9}+\frac{7}{12})+\frac{5}{12}.\)
Zbulo përgjigjen
\(\frac{11}{9}\)
-
Simplify: \((6.47q+9.99q)+1.01q.\)
Zbulo përgjigjen
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: \((5.58c+8.75c)+1.25c.\)
Zbulo përgjigjen
15.58c
-
Simplify: \((8.79d+3.55d)+5.45d.\)
Zbulo përgjigjen
17.79d
-
Simplify the expression: \([1.67(8)](0.25).\)
Zbulo përgjigjen
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: \([1.17(4)](2.25).\)
Zbulo përgjigjen
10.53
-
Simplify: \([3.52(8)](2.5).\)
Zbulo përgjigjen
70.4
-
Simplify: \(6(9x).\)
Zbulo përgjigjen
\(6(9x)\) Use the associative property of multiplication to re-group. \((6\cdot 9)x\) Multiply in the parentheses. \(54x\) -
Simplify: \(8(3y).\)
Zbulo përgjigjen
24y
-
Simplify: \(12(5z).\)
Zbulo përgjigjen
60z
-
Simplify: \(18p+6q+(-15p)+5q.\)
Zbulo përgjigjen
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: \(23r+14s+9r+(-15s).\)
Zbulo përgjigjen
32r − s
-
Simplify: \(37m+21n+4m+(-15n).\)
Zbulo përgjigjen
41m + 6n
-
\(8(-12)=\underline{\hspace{1.5em}}\)
Një llogari e lirë shtohet shënime në çdo mësim, një regjistrim të asaj që ju keni përfunduar, problemet tuaja të zgjidhura në një vend, dhe një mësues që ju mund të pyesni rreth kësaj faqeje. Matematika vetë është e hapur për të gjithë, të regjistruar apo jo.
Regjistrohu HyrSimbolet e përdorura këtu
Prek çdo simbol për përkufizimin e plotë, një fotografi dhe se çfarë do të thotë çdo shkronjë në të.
Si: 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
Pyetja që bëjnë njerëzit
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.
Pjesa e kësaj faqeje është adaptuar nga OpenStax Prealgebra 2e (CC BY-NC-SA 4.0). E përmbledhur dhe ri-shkruar këtu; gabimet janë tona. Për shkak se burimi i tij është licencuar për përdorim jo komercial, ky mësim është i lirë, nuk shet asgjë dhe ndahet nën të njëjtën licencë.