Equation solving?
I've got this system of equations.
(a,b,c) ∈ ℤ
$a*b+1 = c$
$a^2 + b^2 +1 = 2c$
$2a + b = c$
I tried to substitute a little bit:
$a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$
Ultimately:
$a(a-2-b) + b(b-1) = 0$
Now I'm not sure how to substitute b into a in order to go on.
Maybe I didn't even start well.
Can someone help?
Thx
systems-of-equations quadratics
edited Nov 23 '18 at 12:00
Harry Peter
5,46111439
asked Nov 15 '18 at 11:30
calculatormathematicalcalculatormathematical
3811
add a comment |
I've got this system of equations.
(a,b,c) ∈ ℤ
$a*b+1 = c$
$a^2 + b^2 +1 = 2c$
$2a + b = c$
I tried to substitute a little bit:
$a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$
Ultimately:
$a(a-2-b) + b(b-1) = 0$
Now I'm not sure how to substitute b into a in order to go on.
Maybe I didn't even start well.
Can someone help?
Thx
systems-of-equations quadratics
edited Nov 23 '18 at 12:00
Harry Peter
5,46111439
asked Nov 15 '18 at 11:30
calculatormathematicalcalculatormathematical
3811
add a comment |
I've got this system of equations.
(a,b,c) ∈ ℤ
$a*b+1 = c$
$a^2 + b^2 +1 = 2c$
$2a + b = c$
I tried to substitute a little bit:
$a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$
Ultimately:
$a(a-2-b) + b(b-1) = 0$
Now I'm not sure how to substitute b into a in order to go on.
Maybe I didn't even start well.
Can someone help?
Thx
systems-of-equations quadratics
edited Nov 23 '18 at 12:00
Harry Peter
5,46111439
asked Nov 15 '18 at 11:30
calculatormathematicalcalculatormathematical
3811
I've got this system of equations.
(a,b,c) ∈ ℤ
$a*b+1 = c$
$a^2 + b^2 +1 = 2c$
$2a + b = c$
I tried to substitute a little bit:
$a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$
Ultimately:
$a(a-2-b) + b(b-1) = 0$
Now I'm not sure how to substitute b into a in order to go on.
Maybe I didn't even start well.
Can someone help?
Thx
systems-of-equations quadratics
systems-of-equations quadratics
edited Nov 23 '18 at 12:00
Harry Peter
5,46111439
asked Nov 15 '18 at 11:30
calculatormathematicalcalculatormathematical
3811
edited Nov 23 '18 at 12:00
Harry Peter
5,46111439
asked Nov 15 '18 at 11:30
calculatormathematicalcalculatormathematical
3811
edited Nov 23 '18 at 12:00
Harry Peter
5,46111439
edited Nov 23 '18 at 12:00
Harry Peter
5,46111439
edited Nov 23 '18 at 12:00
Harry Peter
5,46111439
5,46111439
asked Nov 15 '18 at 11:30
calculatormathematicalcalculatormathematical
3811
asked Nov 15 '18 at 11:30
calculatormathematicalcalculatormathematical
3811
asked Nov 15 '18 at 11:30
calculatormathematicalcalculatormathematical
3811
3811
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.
answered Nov 15 '18 at 11:41
LacramioaraLacramioara
32126
add a comment |
Eliminating $c$ from the first two equations,
$$(a-b)^2=1$$ or $$a=bpm1.$$
Then eliminating $a$ and $c$ from the first and third, we get two quadratic,
$$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$
$$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$
answered Nov 15 '18 at 11:38
Yves DaoustYves Daoust
124k671222
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 '18 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 '18 at 14:44
add a comment |
Here is a possible way to solve your problem.
From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.
Hope it works.
answered Nov 15 '18 at 11:40
Sujit BhattacharyyaSujit Bhattacharyya
945318
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.
answered Nov 15 '18 at 11:41
LacramioaraLacramioara
32126
add a comment |
You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.
answered Nov 15 '18 at 11:41
LacramioaraLacramioara
32126
add a comment |
You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.
answered Nov 15 '18 at 11:41
LacramioaraLacramioara
32126
You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.
answered Nov 15 '18 at 11:41
LacramioaraLacramioara
32126
answered Nov 15 '18 at 11:41
LacramioaraLacramioara
32126
answered Nov 15 '18 at 11:41
LacramioaraLacramioara
32126
answered Nov 15 '18 at 11:41
LacramioaraLacramioara
32126
32126
add a comment |
add a comment |
Eliminating $c$ from the first two equations,
$$(a-b)^2=1$$ or $$a=bpm1.$$
Then eliminating $a$ and $c$ from the first and third, we get two quadratic,
$$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$
$$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$
answered Nov 15 '18 at 11:38
Yves DaoustYves Daoust
124k671222
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 '18 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 '18 at 14:44
add a comment |
Eliminating $c$ from the first two equations,
$$(a-b)^2=1$$ or $$a=bpm1.$$
Then eliminating $a$ and $c$ from the first and third, we get two quadratic,
$$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$
$$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$
answered Nov 15 '18 at 11:38
Yves DaoustYves Daoust
124k671222
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 '18 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 '18 at 14:44
add a comment |
Eliminating $c$ from the first two equations,
$$(a-b)^2=1$$ or $$a=bpm1.$$
Then eliminating $a$ and $c$ from the first and third, we get two quadratic,
$$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$
$$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$
answered Nov 15 '18 at 11:38
Yves DaoustYves Daoust
124k671222
Eliminating $c$ from the first two equations,
$$(a-b)^2=1$$ or $$a=bpm1.$$
Then eliminating $a$ and $c$ from the first and third, we get two quadratic,
$$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$
$$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$
answered Nov 15 '18 at 11:38
Yves DaoustYves Daoust
124k671222
answered Nov 15 '18 at 11:38
Yves DaoustYves Daoust
124k671222
answered Nov 15 '18 at 11:38
Yves DaoustYves Daoust
124k671222
answered Nov 15 '18 at 11:38
Yves DaoustYves Daoust
124k671222
124k671222
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 '18 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 '18 at 14:44
add a comment |
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 '18 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 '18 at 14:44
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 '18 at 11:54
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 '18 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 '18 at 14:44
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 '18 at 14:44
add a comment |
Here is a possible way to solve your problem.
From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.
Hope it works.
answered Nov 15 '18 at 11:40
Sujit BhattacharyyaSujit Bhattacharyya
945318
add a comment |
Here is a possible way to solve your problem.
From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.
Hope it works.
answered Nov 15 '18 at 11:40
Sujit BhattacharyyaSujit Bhattacharyya
945318
add a comment |
Here is a possible way to solve your problem.
From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.
Hope it works.
answered Nov 15 '18 at 11:40
Sujit BhattacharyyaSujit Bhattacharyya
945318
Here is a possible way to solve your problem.
From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.
Hope it works.
answered Nov 15 '18 at 11:40
Sujit BhattacharyyaSujit Bhattacharyya
945318
answered Nov 15 '18 at 11:40
Sujit BhattacharyyaSujit Bhattacharyya
945318
answered Nov 15 '18 at 11:40
Sujit BhattacharyyaSujit Bhattacharyya
945318
answered Nov 15 '18 at 11:40
Sujit BhattacharyyaSujit Bhattacharyya
945318
945318
add a comment |
add a comment |
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