Equation solving?
up vote
0
down vote
favorite
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 at 12:00
Harry Peter
5,43111439
asked Nov 15 at 11:30
calculatormathematical
3811
add a comment |
up vote
0
down vote
favorite
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 at 12:00
Harry Peter
5,43111439
asked Nov 15 at 11:30
calculatormathematical
3811
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 12:00
Harry Peter
5,43111439
asked Nov 15 at 11:30
calculatormathematical
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 at 12:00
Harry Peter
5,43111439
asked Nov 15 at 11:30
calculatormathematical
3811
edited Nov 23 at 12:00
Harry Peter
5,43111439
asked Nov 15 at 11:30
calculatormathematical
3811
edited Nov 23 at 12:00
Harry Peter
5,43111439
edited Nov 23 at 12:00
Harry Peter
5,43111439
edited Nov 23 at 12:00
Harry Peter
5,43111439
5,43111439
asked Nov 15 at 11:30
calculatormathematical
3811
asked Nov 15 at 11:30
calculatormathematical
3811
asked Nov 15 at 11:30
calculatormathematical
3811
3811
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
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 at 11:41
Lacramioara
32126
add a comment |
up vote
0
down vote
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 at 11:38
Yves Daoust
123k668219
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 at 14:44
add a comment |
up vote
0
down vote
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 at 11:40
Sujit Bhattacharyya
945317
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999578%2fequation-solving%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
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 at 11:41
Lacramioara
32126
add a comment |
up vote
2
down vote
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 at 11:41
Lacramioara
32126
add a comment |
up vote
2
down vote
up vote
2
down vote
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 at 11:41
Lacramioara
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 at 11:41
Lacramioara
32126
answered Nov 15 at 11:41
Lacramioara
32126
answered Nov 15 at 11:41
Lacramioara
32126
answered Nov 15 at 11:41
Lacramioara
32126
32126
add a comment |
add a comment |
up vote
0
down vote
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 at 11:38
Yves Daoust
123k668219
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 at 14:44
add a comment |
up vote
0
down vote
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 at 11:38
Yves Daoust
123k668219
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 at 14:44
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 11:38
Yves Daoust
123k668219
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 at 11:38
Yves Daoust
123k668219
answered Nov 15 at 11:38
Yves Daoust
123k668219
answered Nov 15 at 11:38
Yves Daoust
123k668219
answered Nov 15 at 11:38
Yves Daoust
123k668219
123k668219
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 at 14:44
add a comment |
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 at 14:44
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 at 11:54
Ok but what if (a,b,c) ∈ ℤ
– calculatormathematical
Nov 15 at 11:54
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 at 14:44
@calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
– Yves Daoust
Nov 15 at 14:44
add a comment |
up vote
0
down vote
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 at 11:40
Sujit Bhattacharyya
945317
add a comment |
up vote
0
down vote
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 at 11:40
Sujit Bhattacharyya
945317
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 11:40
Sujit Bhattacharyya
945317
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 at 11:40
Sujit Bhattacharyya
945317
answered Nov 15 at 11:40
Sujit Bhattacharyya
945317
answered Nov 15 at 11:40
Sujit Bhattacharyya
945317
answered Nov 15 at 11:40
Sujit Bhattacharyya
945317
945317
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999578%2fequation-solving%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
