Handling opposites when adding and subtracting rational expressions
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I'm following example 8.47 from the OpenStax book Elementary Algebra.
When trying to create a common denominator it is sometimes necessary to handle opposites on either side of an equation.
In the example, $(2-n)$ needs to be converted to $(n-2)$ as follows:
$$-frac{(n+3)}{(2-n)}$$
To do this we multiply the numerator and denominator by $-1$.
$$-frac{(-1)(n+3)}{(-1)(2-n)}$$
Which gives:
$$+frac{(n+3)}{(n-2)}$$
I understand why the denominator changes but not why the numerator stays the same and the sign of the whole expression changes. Could someone explain this, please?
algebra-precalculus rational-numbers
asked Dec 18 '18 at 12:05
Brendan CostiganBrendan Costigan
82
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add a comment |
$begingroup$
I'm following example 8.47 from the OpenStax book Elementary Algebra.
When trying to create a common denominator it is sometimes necessary to handle opposites on either side of an equation.
In the example, $(2-n)$ needs to be converted to $(n-2)$ as follows:
$$-frac{(n+3)}{(2-n)}$$
To do this we multiply the numerator and denominator by $-1$.
$$-frac{(-1)(n+3)}{(-1)(2-n)}$$
Which gives:
$$+frac{(n+3)}{(n-2)}$$
I understand why the denominator changes but not why the numerator stays the same and the sign of the whole expression changes. Could someone explain this, please?
algebra-precalculus rational-numbers
asked Dec 18 '18 at 12:05
Brendan CostiganBrendan Costigan
82
$endgroup$
add a comment |
$begingroup$
I'm following example 8.47 from the OpenStax book Elementary Algebra.
When trying to create a common denominator it is sometimes necessary to handle opposites on either side of an equation.
In the example, $(2-n)$ needs to be converted to $(n-2)$ as follows:
$$-frac{(n+3)}{(2-n)}$$
To do this we multiply the numerator and denominator by $-1$.
$$-frac{(-1)(n+3)}{(-1)(2-n)}$$
Which gives:
$$+frac{(n+3)}{(n-2)}$$
I understand why the denominator changes but not why the numerator stays the same and the sign of the whole expression changes. Could someone explain this, please?
algebra-precalculus rational-numbers
asked Dec 18 '18 at 12:05
Brendan CostiganBrendan Costigan
82
$endgroup$
I'm following example 8.47 from the OpenStax book Elementary Algebra.
When trying to create a common denominator it is sometimes necessary to handle opposites on either side of an equation.
In the example, $(2-n)$ needs to be converted to $(n-2)$ as follows:
$$-frac{(n+3)}{(2-n)}$$
To do this we multiply the numerator and denominator by $-1$.
$$-frac{(-1)(n+3)}{(-1)(2-n)}$$
Which gives:
$$+frac{(n+3)}{(n-2)}$$
I understand why the denominator changes but not why the numerator stays the same and the sign of the whole expression changes. Could someone explain this, please?
algebra-precalculus rational-numbers
algebra-precalculus rational-numbers
asked Dec 18 '18 at 12:05
Brendan CostiganBrendan Costigan
82
asked Dec 18 '18 at 12:05
Brendan CostiganBrendan Costigan
82
asked Dec 18 '18 at 12:05
Brendan CostiganBrendan Costigan
82
asked Dec 18 '18 at 12:05
Brendan CostiganBrendan Costigan
82
asked Dec 18 '18 at 12:05
Brendan CostiganBrendan Costigan
82
82
add a comment |
add a comment |
1 Answer
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oldest
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Don't forget the minus sign 'outside' the fraction disappeared as well:
$$-frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$-1frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$frac{-1cdot-1cdot(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(n-2)}$$
answered Dec 18 '18 at 12:10
GlorfindelGlorfindel
3,41581830
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Sorry to be stupid but why has the -1 moved from the numerator to the 'whole' expression?
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– Brendan Costigan
Dec 18 '18 at 12:16
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That's how fractions work; everything 'in front of' the fraction is actually part of the numerator.
$endgroup$
– Glorfindel
Dec 18 '18 at 12:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Don't forget the minus sign 'outside' the fraction disappeared as well:
$$-frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$-1frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$frac{-1cdot-1cdot(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(n-2)}$$
answered Dec 18 '18 at 12:10
GlorfindelGlorfindel
3,41581830
$endgroup$
$begingroup$
Sorry to be stupid but why has the -1 moved from the numerator to the 'whole' expression?
$endgroup$
– Brendan Costigan
Dec 18 '18 at 12:16
$begingroup$
That's how fractions work; everything 'in front of' the fraction is actually part of the numerator.
$endgroup$
– Glorfindel
Dec 18 '18 at 12:18
add a comment |
$begingroup$
Don't forget the minus sign 'outside' the fraction disappeared as well:
$$-frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$-1frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$frac{-1cdot-1cdot(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(n-2)}$$
answered Dec 18 '18 at 12:10
GlorfindelGlorfindel
3,41581830
$endgroup$
$begingroup$
Sorry to be stupid but why has the -1 moved from the numerator to the 'whole' expression?
$endgroup$
– Brendan Costigan
Dec 18 '18 at 12:16
$begingroup$
That's how fractions work; everything 'in front of' the fraction is actually part of the numerator.
$endgroup$
– Glorfindel
Dec 18 '18 at 12:18
add a comment |
$begingroup$
Don't forget the minus sign 'outside' the fraction disappeared as well:
$$-frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$-1frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$frac{-1cdot-1cdot(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(n-2)}$$
answered Dec 18 '18 at 12:10
GlorfindelGlorfindel
3,41581830
$endgroup$
Don't forget the minus sign 'outside' the fraction disappeared as well:
$$-frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$-1frac{(-1)(n+3)}{(-1)(2-n)}=$$
$$frac{-1cdot-1cdot(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(-1)(2-n)}=$$
$$frac{(n+3)}{(n-2)}$$
answered Dec 18 '18 at 12:10
GlorfindelGlorfindel
3,41581830
edited Dec 18 '18 at 12:17
answered Dec 18 '18 at 12:10
GlorfindelGlorfindel
3,41581830
answered Dec 18 '18 at 12:10
GlorfindelGlorfindel
3,41581830
answered Dec 18 '18 at 12:10
GlorfindelGlorfindel
3,41581830
3,41581830
$begingroup$
Sorry to be stupid but why has the -1 moved from the numerator to the 'whole' expression?
$endgroup$
– Brendan Costigan
Dec 18 '18 at 12:16
$begingroup$
That's how fractions work; everything 'in front of' the fraction is actually part of the numerator.
$endgroup$
– Glorfindel
Dec 18 '18 at 12:18
add a comment |
$begingroup$
Sorry to be stupid but why has the -1 moved from the numerator to the 'whole' expression?
$endgroup$
– Brendan Costigan
Dec 18 '18 at 12:16
$begingroup$
That's how fractions work; everything 'in front of' the fraction is actually part of the numerator.
$endgroup$
– Glorfindel
Dec 18 '18 at 12:18
$begingroup$
Sorry to be stupid but why has the -1 moved from the numerator to the 'whole' expression?
$endgroup$
– Brendan Costigan
Dec 18 '18 at 12:16
$begingroup$
Sorry to be stupid but why has the -1 moved from the numerator to the 'whole' expression?
$endgroup$
– Brendan Costigan
Dec 18 '18 at 12:16
$begingroup$
That's how fractions work; everything 'in front of' the fraction is actually part of the numerator.
$endgroup$
– Glorfindel
Dec 18 '18 at 12:18
$begingroup$
That's how fractions work; everything 'in front of' the fraction is actually part of the numerator.
$endgroup$
– Glorfindel
Dec 18 '18 at 12:18
add a comment |
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