Determine all possible integers $x$ and $y$ s.t. $3x + 7y equiv 14 pmod28$ and $x + 3y equiv 8 pmod28$.

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Determine all possible ints $x$ and $y$ that satisfy $3x + 7y equiv 14 pmod 28$ and $x + 3y equiv 8 pmod 28$. The answer should be in terms of $xequiv r$ and $y equiv s$ where $r$ and $s$ are remainders.

Here is what I have tried. Am I headed in the right direction?

$8cdot (3x + 7y) equiv 14 pmod 28$ and $14 cdot (x + 3y) equiv 8 pmod 28$

$24x + 56y equiv 112 pmod 28$ and $14x + 42y equiv 112 pmod 28$

$24x + 56y equiv 14x + 42y pmod 28$

This is where I start to get unsure,

$24x + 28y + 28y equiv 14x + 14y + 28y pmod 28$

$24x equiv 14x + 14y pmod 28$

$48x equiv 28x + 28y pmod 28$

So then...?

$20x equiv 1 pmod 28$?

Stuck here. Is this even the right idea?

elementary-number-theory modular-arithmetic

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edited Jul 19 at 20:35

David G. Stork

7,6632929

asked Jul 19 at 19:54

SolidSnackDrive

1107

add a comment | 

up vote
3
down vote

favorite

Determine all possible ints $x$ and $y$ that satisfy $3x + 7y equiv 14 pmod 28$ and $x + 3y equiv 8 pmod 28$. The answer should be in terms of $xequiv r$ and $y equiv s$ where $r$ and $s$ are remainders.

Here is what I have tried. Am I headed in the right direction?

$8cdot (3x + 7y) equiv 14 pmod 28$ and $14 cdot (x + 3y) equiv 8 pmod 28$

$24x + 56y equiv 112 pmod 28$ and $14x + 42y equiv 112 pmod 28$

$24x + 56y equiv 14x + 42y pmod 28$

This is where I start to get unsure,

$24x + 28y + 28y equiv 14x + 14y + 28y pmod 28$

$24x equiv 14x + 14y pmod 28$

$48x equiv 28x + 28y pmod 28$

So then...?

$20x equiv 1 pmod 28$?

Stuck here. Is this even the right idea?

elementary-number-theory modular-arithmetic

share|cite|improve this question

edited Jul 19 at 20:35

David G. Stork

7,6632929

asked Jul 19 at 19:54

SolidSnackDrive

1107

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Determine all possible ints $x$ and $y$ that satisfy $3x + 7y equiv 14 pmod 28$ and $x + 3y equiv 8 pmod 28$. The answer should be in terms of $xequiv r$ and $y equiv s$ where $r$ and $s$ are remainders.

Here is what I have tried. Am I headed in the right direction?

$8cdot (3x + 7y) equiv 14 pmod 28$ and $14 cdot (x + 3y) equiv 8 pmod 28$

$24x + 56y equiv 112 pmod 28$ and $14x + 42y equiv 112 pmod 28$

$24x + 56y equiv 14x + 42y pmod 28$

This is where I start to get unsure,

$24x + 28y + 28y equiv 14x + 14y + 28y pmod 28$

$24x equiv 14x + 14y pmod 28$

$48x equiv 28x + 28y pmod 28$

So then...?

$20x equiv 1 pmod 28$?

Stuck here. Is this even the right idea?

elementary-number-theory modular-arithmetic

share|cite|improve this question

edited Jul 19 at 20:35

David G. Stork

7,6632929

asked Jul 19 at 19:54

SolidSnackDrive

1107

Determine all possible ints $x$ and $y$ that satisfy $3x + 7y equiv 14 pmod 28$ and $x + 3y equiv 8 pmod 28$. The answer should be in terms of $xequiv r$ and $y equiv s$ where $r$ and $s$ are remainders.

Here is what I have tried. Am I headed in the right direction?

$8cdot (3x + 7y) equiv 14 pmod 28$ and $14 cdot (x + 3y) equiv 8 pmod 28$

$24x + 56y equiv 112 pmod 28$ and $14x + 42y equiv 112 pmod 28$

$24x + 56y equiv 14x + 42y pmod 28$

This is where I start to get unsure,

$24x + 28y + 28y equiv 14x + 14y + 28y pmod 28$

$24x equiv 14x + 14y pmod 28$

$48x equiv 28x + 28y pmod 28$

So then...?

$20x equiv 1 pmod 28$?

Stuck here. Is this even the right idea?

elementary-number-theory modular-arithmetic

share|cite|improve this question

edited Jul 19 at 20:35

David G. Stork

7,6632929

asked Jul 19 at 19:54

SolidSnackDrive

1107

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 20:35

David G. Stork

7,6632929

edited Jul 19 at 20:35

David G. Stork

7,6632929

edited Jul 19 at 20:35

David G. Stork

7,6632929

7,6632929

asked Jul 19 at 19:54

SolidSnackDrive

1107

asked Jul 19 at 19:54

SolidSnackDrive

1107

asked Jul 19 at 19:54

SolidSnackDrive

1107

1107

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
6
down vote



accepted

We have
begineqnarray*
3x+7y equiv 14 pmod28 \
x+3y equiv 8 pmod28.
endeqnarray*
Multiply the second equation by $3$ and then subtract the first
begineqnarray*
3x+9y equiv 24 pmod28 \
2y equiv 10 pmod28.
endeqnarray*
So
begineqnarray*
y &equiv 5 pmod28 \
x & equiv 21 pmod28
endeqnarray*
orbegineqnarray*
y &equiv 19 pmod28 \
x & equiv 7 pmod28.
endeqnarray*

share|cite|improve this answer

edited Jul 19 at 20:06

answered Jul 19 at 20:02

Donald Splutterwit

21.3k21243

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  and what about the case $yequiv 19$, since $38equiv 10bmod 28$?
  – Joffan
  Jul 19 at 20:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Crap, I did way too much work!
  – SolidSnackDrive
  Jul 19 at 20:05
  

  
  
  
  

add a comment | 

Your Answer

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote



accepted

We have
begineqnarray*
3x+7y equiv 14 pmod28 \
x+3y equiv 8 pmod28.
endeqnarray*
Multiply the second equation by $3$ and then subtract the first
begineqnarray*
3x+9y equiv 24 pmod28 \
2y equiv 10 pmod28.
endeqnarray*
So
begineqnarray*
y &equiv 5 pmod28 \
x & equiv 21 pmod28
endeqnarray*
orbegineqnarray*
y &equiv 19 pmod28 \
x & equiv 7 pmod28.
endeqnarray*

share|cite|improve this answer

edited Jul 19 at 20:06

answered Jul 19 at 20:02

Donald Splutterwit

21.3k21243

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  and what about the case $yequiv 19$, since $38equiv 10bmod 28$?
  – Joffan
  Jul 19 at 20:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Crap, I did way too much work!
  – SolidSnackDrive
  Jul 19 at 20:05
  

  
  
  
  

add a comment | 

up vote
6
down vote



accepted

We have
begineqnarray*
3x+7y equiv 14 pmod28 \
x+3y equiv 8 pmod28.
endeqnarray*
Multiply the second equation by $3$ and then subtract the first
begineqnarray*
3x+9y equiv 24 pmod28 \
2y equiv 10 pmod28.
endeqnarray*
So
begineqnarray*
y &equiv 5 pmod28 \
x & equiv 21 pmod28
endeqnarray*
orbegineqnarray*
y &equiv 19 pmod28 \
x & equiv 7 pmod28.
endeqnarray*

share|cite|improve this answer

edited Jul 19 at 20:06

answered Jul 19 at 20:02

Donald Splutterwit

21.3k21243

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  and what about the case $yequiv 19$, since $38equiv 10bmod 28$?
  – Joffan
  Jul 19 at 20:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Crap, I did way too much work!
  – SolidSnackDrive
  Jul 19 at 20:05
  

  
  
  
  

add a comment | 

up vote
6
down vote



accepted

up vote
6
down vote



accepted

We have
begineqnarray*
3x+7y equiv 14 pmod28 \
x+3y equiv 8 pmod28.
endeqnarray*
Multiply the second equation by $3$ and then subtract the first
begineqnarray*
3x+9y equiv 24 pmod28 \
2y equiv 10 pmod28.
endeqnarray*
So
begineqnarray*
y &equiv 5 pmod28 \
x & equiv 21 pmod28
endeqnarray*
orbegineqnarray*
y &equiv 19 pmod28 \
x & equiv 7 pmod28.
endeqnarray*

share|cite|improve this answer

edited Jul 19 at 20:06

answered Jul 19 at 20:02

Donald Splutterwit

21.3k21243

We have
begineqnarray*
3x+7y equiv 14 pmod28 \
x+3y equiv 8 pmod28.
endeqnarray*
Multiply the second equation by $3$ and then subtract the first
begineqnarray*
3x+9y equiv 24 pmod28 \
2y equiv 10 pmod28.
endeqnarray*
So
begineqnarray*
y &equiv 5 pmod28 \
x & equiv 21 pmod28
endeqnarray*
orbegineqnarray*
y &equiv 19 pmod28 \
x & equiv 7 pmod28.
endeqnarray*

share|cite|improve this answer

edited Jul 19 at 20:06

answered Jul 19 at 20:02

Donald Splutterwit

21.3k21243

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 20:06

edited Jul 19 at 20:06

edited Jul 19 at 20:06

answered Jul 19 at 20:02

Donald Splutterwit

21.3k21243

answered Jul 19 at 20:02

Donald Splutterwit

21.3k21243

answered Jul 19 at 20:02

Donald Splutterwit

21.3k21243

21.3k21243

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  and what about the case $yequiv 19$, since $38equiv 10bmod 28$?
  – Joffan
  Jul 19 at 20:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Crap, I did way too much work!
  – SolidSnackDrive
  Jul 19 at 20:05
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  and what about the case $yequiv 19$, since $38equiv 10bmod 28$?
  – Joffan
  Jul 19 at 20:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Crap, I did way too much work!
  – SolidSnackDrive
  Jul 19 at 20:05
  

  
  
  
  

2

2

and what about the case $yequiv 19$, since $38equiv 10bmod 28$?
– Joffan
Jul 19 at 20:04

and what about the case $yequiv 19$, since $38equiv 10bmod 28$?
– Joffan
Jul 19 at 20:04

Crap, I did way too much work!
– SolidSnackDrive
Jul 19 at 20:05

Crap, I did way too much work!
– SolidSnackDrive
Jul 19 at 20:05

add a comment | 

 

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