How does rearrangement work for pulling out factors
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Given 10(6*3+2) we can rearrange the equation by pulling out the 3 and getting 10((6*3)/3+(2/3))3
Okay I understand that it works. But I don't understand how. I would have assumed to do 10((6*3)/3+2)3 since we are only removing the 3 as a factor to 6. Why does 3 need to be divided by the 2 as well? 3 has nothing to do with the 2.
algebra-precalculus arithmetic
asked Jan 7 at 21:10
SephSeph
1685
$endgroup$
add a comment |
$begingroup$
Given 10(6*3+2) we can rearrange the equation by pulling out the 3 and getting 10((6*3)/3+(2/3))3
Okay I understand that it works. But I don't understand how. I would have assumed to do 10((6*3)/3+2)3 since we are only removing the 3 as a factor to 6. Why does 3 need to be divided by the 2 as well? 3 has nothing to do with the 2.
algebra-precalculus arithmetic
asked Jan 7 at 21:10
SephSeph
1685
$endgroup$
$begingroup$
The only way the value of the expression will be unchanged is if you both multiply and divide by $3$.
$endgroup$
– saulspatz
Jan 7 at 21:14
1
$begingroup$
Notice that everything in the parentheses is being multiplied by $10$. If you extract a $3$ from the parentheses, each term left in the parentheses is being multiplied by $30$.
$endgroup$
– N. F. Taussig
Jan 7 at 21:21
$begingroup$
Please use MathJax.
$endgroup$
– N. F. Taussig
Jan 7 at 21:24
add a comment |
$begingroup$
Given 10(6*3+2) we can rearrange the equation by pulling out the 3 and getting 10((6*3)/3+(2/3))3
Okay I understand that it works. But I don't understand how. I would have assumed to do 10((6*3)/3+2)3 since we are only removing the 3 as a factor to 6. Why does 3 need to be divided by the 2 as well? 3 has nothing to do with the 2.
algebra-precalculus arithmetic
asked Jan 7 at 21:10
SephSeph
1685
$endgroup$
Given 10(6*3+2) we can rearrange the equation by pulling out the 3 and getting 10((6*3)/3+(2/3))3
Okay I understand that it works. But I don't understand how. I would have assumed to do 10((6*3)/3+2)3 since we are only removing the 3 as a factor to 6. Why does 3 need to be divided by the 2 as well? 3 has nothing to do with the 2.
algebra-precalculus arithmetic
algebra-precalculus arithmetic
asked Jan 7 at 21:10
SephSeph
1685
asked Jan 7 at 21:10
SephSeph
1685
asked Jan 7 at 21:10
SephSeph
1685
asked Jan 7 at 21:10
SephSeph
1685
asked Jan 7 at 21:10
SephSeph
1685
1685
$begingroup$
The only way the value of the expression will be unchanged is if you both multiply and divide by $3$.
$endgroup$
– saulspatz
Jan 7 at 21:14
1
$begingroup$
Notice that everything in the parentheses is being multiplied by $10$. If you extract a $3$ from the parentheses, each term left in the parentheses is being multiplied by $30$.
$endgroup$
– N. F. Taussig
Jan 7 at 21:21
$begingroup$
Please use MathJax.
$endgroup$
– N. F. Taussig
Jan 7 at 21:24
add a comment |
$begingroup$
The only way the value of the expression will be unchanged is if you both multiply and divide by $3$.
$endgroup$
– saulspatz
Jan 7 at 21:14
1
$begingroup$
Notice that everything in the parentheses is being multiplied by $10$. If you extract a $3$ from the parentheses, each term left in the parentheses is being multiplied by $30$.
$endgroup$
– N. F. Taussig
Jan 7 at 21:21
$begingroup$
Please use MathJax.
$endgroup$
– N. F. Taussig
Jan 7 at 21:24
$begingroup$
The only way the value of the expression will be unchanged is if you both multiply and divide by $3$.
$endgroup$
– saulspatz
Jan 7 at 21:14
$begingroup$
The only way the value of the expression will be unchanged is if you both multiply and divide by $3$.
$endgroup$
– saulspatz
Jan 7 at 21:14
1
1
$begingroup$
Notice that everything in the parentheses is being multiplied by $10$. If you extract a $3$ from the parentheses, each term left in the parentheses is being multiplied by $30$.
$endgroup$
– N. F. Taussig
Jan 7 at 21:21
$begingroup$
Notice that everything in the parentheses is being multiplied by $10$. If you extract a $3$ from the parentheses, each term left in the parentheses is being multiplied by $30$.
$endgroup$
– N. F. Taussig
Jan 7 at 21:21
$begingroup$
Please use MathJax.
$endgroup$
– N. F. Taussig
Jan 7 at 21:24
$begingroup$
Please use MathJax.
$endgroup$
– N. F. Taussig
Jan 7 at 21:24
add a comment |
1 Answer
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$begingroup$
According to the distributive law
$$a(b+c)=ab+ac$$
In your problem
$$a=3,~b=((6cdot3)/3),~c=(2/3)$$
$$((6cdot3)/3+(2/3))cdot3=((6cdot3)/3) cdot3 +(2/3)cdot3=6cdot3+2$$
On the other hand, you are wrong to say that
$$((6cdot3)/3+2)cdot3=6cdot3+2$$
Since according to the distributive law
$$((6cdot3)/3+2)cdot3=(6cdot3/3)cdot3+2cdot3=6cdot3+6$$
answered Jan 7 at 21:37
LarryLarry
2,54531131
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
According to the distributive law
$$a(b+c)=ab+ac$$
In your problem
$$a=3,~b=((6cdot3)/3),~c=(2/3)$$
$$((6cdot3)/3+(2/3))cdot3=((6cdot3)/3) cdot3 +(2/3)cdot3=6cdot3+2$$
On the other hand, you are wrong to say that
$$((6cdot3)/3+2)cdot3=6cdot3+2$$
Since according to the distributive law
$$((6cdot3)/3+2)cdot3=(6cdot3/3)cdot3+2cdot3=6cdot3+6$$
answered Jan 7 at 21:37
LarryLarry
2,54531131
$endgroup$
add a comment |
$begingroup$
According to the distributive law
$$a(b+c)=ab+ac$$
In your problem
$$a=3,~b=((6cdot3)/3),~c=(2/3)$$
$$((6cdot3)/3+(2/3))cdot3=((6cdot3)/3) cdot3 +(2/3)cdot3=6cdot3+2$$
On the other hand, you are wrong to say that
$$((6cdot3)/3+2)cdot3=6cdot3+2$$
Since according to the distributive law
$$((6cdot3)/3+2)cdot3=(6cdot3/3)cdot3+2cdot3=6cdot3+6$$
answered Jan 7 at 21:37
LarryLarry
2,54531131
$endgroup$
add a comment |
$begingroup$
According to the distributive law
$$a(b+c)=ab+ac$$
In your problem
$$a=3,~b=((6cdot3)/3),~c=(2/3)$$
$$((6cdot3)/3+(2/3))cdot3=((6cdot3)/3) cdot3 +(2/3)cdot3=6cdot3+2$$
On the other hand, you are wrong to say that
$$((6cdot3)/3+2)cdot3=6cdot3+2$$
Since according to the distributive law
$$((6cdot3)/3+2)cdot3=(6cdot3/3)cdot3+2cdot3=6cdot3+6$$
answered Jan 7 at 21:37
LarryLarry
2,54531131
$endgroup$
According to the distributive law
$$a(b+c)=ab+ac$$
In your problem
$$a=3,~b=((6cdot3)/3),~c=(2/3)$$
$$((6cdot3)/3+(2/3))cdot3=((6cdot3)/3) cdot3 +(2/3)cdot3=6cdot3+2$$
On the other hand, you are wrong to say that
$$((6cdot3)/3+2)cdot3=6cdot3+2$$
Since according to the distributive law
$$((6cdot3)/3+2)cdot3=(6cdot3/3)cdot3+2cdot3=6cdot3+6$$
answered Jan 7 at 21:37
LarryLarry
2,54531131
answered Jan 7 at 21:37
LarryLarry
2,54531131
answered Jan 7 at 21:37
LarryLarry
2,54531131
answered Jan 7 at 21:37
LarryLarry
2,54531131
2,54531131
add a comment |
add a comment |
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$begingroup$
The only way the value of the expression will be unchanged is if you both multiply and divide by $3$.
$endgroup$
– saulspatz
Jan 7 at 21:14
1
$begingroup$
Notice that everything in the parentheses is being multiplied by $10$. If you extract a $3$ from the parentheses, each term left in the parentheses is being multiplied by $30$.
$endgroup$
– N. F. Taussig
Jan 7 at 21:21
$begingroup$
Please use MathJax.
$endgroup$
– N. F. Taussig
Jan 7 at 21:24