Mixing
Solve for $r$ and a from systems-of-equations
$begingroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
algebra-precalculus systems-of-equations
edited Jan 27 at 10:39
kelalaka
3351314
asked Jan 27 at 9:51
nazmulnazmul
1
$endgroup$
add a comment |
$begingroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
algebra-precalculus systems-of-equations
edited Jan 27 at 10:39
kelalaka
3351314
asked Jan 27 at 9:51
nazmulnazmul
1
$endgroup$
$begingroup$
divides the second equation by the first one, you'll deduce $r$.
$endgroup$
– Atmos
Jan 27 at 9:55
add a comment |
$begingroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
algebra-precalculus systems-of-equations
edited Jan 27 at 10:39
kelalaka
3351314
asked Jan 27 at 9:51
nazmulnazmul
1
$endgroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
algebra-precalculus systems-of-equations
algebra-precalculus systems-of-equations
edited Jan 27 at 10:39
kelalaka
3351314
asked Jan 27 at 9:51
nazmulnazmul
1
edited Jan 27 at 10:39
kelalaka
3351314
asked Jan 27 at 9:51
nazmulnazmul
1
edited Jan 27 at 10:39
kelalaka
3351314
edited Jan 27 at 10:39
kelalaka
3351314
edited Jan 27 at 10:39
kelalaka
3351314
3351314
asked Jan 27 at 9:51
nazmulnazmul
1
asked Jan 27 at 9:51
nazmulnazmul
1
asked Jan 27 at 9:51
nazmulnazmul
1
1
$begingroup$
divides the second equation by the first one, you'll deduce $r$.
$endgroup$
– Atmos
Jan 27 at 9:55
add a comment |
$begingroup$
divides the second equation by the first one, you'll deduce $r$.
$endgroup$
– Atmos
Jan 27 at 9:55
$begingroup$
divides the second equation by the first one, you'll deduce $r$.
$endgroup$
– Atmos
Jan 27 at 9:55
$begingroup$
divides the second equation by the first one, you'll deduce $r$.
$endgroup$
– Atmos
Jan 27 at 9:55
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: We have $$a(1-r^7)=86(1-r)$$ and $$a(1-r^{10})=-682(1-r)$$. Dividing both equations, simplifying and factorizing we get
$$-2 (r-1) (r+2) left(43 r^8-43 r^7+129 r^6+126 r^5+132 r^4+120 r^3+144 r^2+96 r+192right)=0$$
answered Jan 27 at 9:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
add a comment |
$begingroup$
Make the ratio of the two expressions to get
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}$$ In absolute value, the rhs is large and then $r$ can be "large"; so may be, for the time being we could write
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}simeq frac{r^{10}}{r^7}=r^3 $$ Use you calculator to get, as an approximation
$$r_0=-sqrt[3]{frac{341}{43}}approx -1.99417$$
If you dont want to jumpt to the obvious conclusion, now consider that you look for the zero of function
$$f(r)=43(1-r^{10})+341(1-r^7)$$ and build the Taylor series around $r=r_0$; this would give
$$f(r)=left(-43 r_0^{10}-341 r_0^7+384right)-left(430 r_0^9+2387 r_0^6right)
(r-r_0)+Oleft((r-r_0)^2right)$$ Ignoring the higher order tems, solve for $r$ to get
$$r=r_0-frac{43 r_0^{10}+341 r_0^7-384}{r_0^6 left(430 r_0^3+2387right)}$$ You know the exact value of $r_0^3$, then the exact value of $r_0^6$; now approximate $r_0^7simeq -2 r_0^6$ and $r_0^{10}simeq-2 r_0^3, r_0^6$ . This then gives
$$r_1=r_0-frac{236672}{39651821}=-2.00014$$
You do not need to solve a polynomial of degree $10$. Be practical !
answered Jan 27 at 17:21
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
Cancel out $ 1 - r$ from the denominator by substitution.
$$a = 86frac{1-r}{1-r^7}$$
Now substitute:
$$ frac{86( 1-r^{10} )}{1-r^7} = -682 Rightarrow {86 - 86r^{10}} = -682 + 682r^7$$
Let $t = r^7$
$$768 = 862t + 86r^3$$
Forms a Cubic Equation, which needs to be solved.
answered Jan 28 at 14:48
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
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',
autoActivateHeartbeat: false,
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%2f3089350%2fsolve-for-r-and-a-from-systems-of-equations%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
$begingroup$
Hint: We have $$a(1-r^7)=86(1-r)$$ and $$a(1-r^{10})=-682(1-r)$$. Dividing both equations, simplifying and factorizing we get
$$-2 (r-1) (r+2) left(43 r^8-43 r^7+129 r^6+126 r^5+132 r^4+120 r^3+144 r^2+96 r+192right)=0$$
answered Jan 27 at 9:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
add a comment |
$begingroup$
Hint: We have $$a(1-r^7)=86(1-r)$$ and $$a(1-r^{10})=-682(1-r)$$. Dividing both equations, simplifying and factorizing we get
$$-2 (r-1) (r+2) left(43 r^8-43 r^7+129 r^6+126 r^5+132 r^4+120 r^3+144 r^2+96 r+192right)=0$$
answered Jan 27 at 9:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
add a comment |
$begingroup$
Hint: We have $$a(1-r^7)=86(1-r)$$ and $$a(1-r^{10})=-682(1-r)$$. Dividing both equations, simplifying and factorizing we get
$$-2 (r-1) (r+2) left(43 r^8-43 r^7+129 r^6+126 r^5+132 r^4+120 r^3+144 r^2+96 r+192right)=0$$
answered Jan 27 at 9:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
Hint: We have $$a(1-r^7)=86(1-r)$$ and $$a(1-r^{10})=-682(1-r)$$. Dividing both equations, simplifying and factorizing we get
$$-2 (r-1) (r+2) left(43 r^8-43 r^7+129 r^6+126 r^5+132 r^4+120 r^3+144 r^2+96 r+192right)=0$$
answered Jan 27 at 9:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 27 at 9:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 27 at 9:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 27 at 9:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
78k42866
add a comment |
add a comment |
$begingroup$
Make the ratio of the two expressions to get
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}$$ In absolute value, the rhs is large and then $r$ can be "large"; so may be, for the time being we could write
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}simeq frac{r^{10}}{r^7}=r^3 $$ Use you calculator to get, as an approximation
$$r_0=-sqrt[3]{frac{341}{43}}approx -1.99417$$
If you dont want to jumpt to the obvious conclusion, now consider that you look for the zero of function
$$f(r)=43(1-r^{10})+341(1-r^7)$$ and build the Taylor series around $r=r_0$; this would give
$$f(r)=left(-43 r_0^{10}-341 r_0^7+384right)-left(430 r_0^9+2387 r_0^6right)
(r-r_0)+Oleft((r-r_0)^2right)$$ Ignoring the higher order tems, solve for $r$ to get
$$r=r_0-frac{43 r_0^{10}+341 r_0^7-384}{r_0^6 left(430 r_0^3+2387right)}$$ You know the exact value of $r_0^3$, then the exact value of $r_0^6$; now approximate $r_0^7simeq -2 r_0^6$ and $r_0^{10}simeq-2 r_0^3, r_0^6$ . This then gives
$$r_1=r_0-frac{236672}{39651821}=-2.00014$$
You do not need to solve a polynomial of degree $10$. Be practical !
answered Jan 27 at 17:21
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Make the ratio of the two expressions to get
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}$$ In absolute value, the rhs is large and then $r$ can be "large"; so may be, for the time being we could write
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}simeq frac{r^{10}}{r^7}=r^3 $$ Use you calculator to get, as an approximation
$$r_0=-sqrt[3]{frac{341}{43}}approx -1.99417$$
If you dont want to jumpt to the obvious conclusion, now consider that you look for the zero of function
$$f(r)=43(1-r^{10})+341(1-r^7)$$ and build the Taylor series around $r=r_0$; this would give
$$f(r)=left(-43 r_0^{10}-341 r_0^7+384right)-left(430 r_0^9+2387 r_0^6right)
(r-r_0)+Oleft((r-r_0)^2right)$$ Ignoring the higher order tems, solve for $r$ to get
$$r=r_0-frac{43 r_0^{10}+341 r_0^7-384}{r_0^6 left(430 r_0^3+2387right)}$$ You know the exact value of $r_0^3$, then the exact value of $r_0^6$; now approximate $r_0^7simeq -2 r_0^6$ and $r_0^{10}simeq-2 r_0^3, r_0^6$ . This then gives
$$r_1=r_0-frac{236672}{39651821}=-2.00014$$
You do not need to solve a polynomial of degree $10$. Be practical !
answered Jan 27 at 17:21
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Make the ratio of the two expressions to get
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}$$ In absolute value, the rhs is large and then $r$ can be "large"; so may be, for the time being we could write
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}simeq frac{r^{10}}{r^7}=r^3 $$ Use you calculator to get, as an approximation
$$r_0=-sqrt[3]{frac{341}{43}}approx -1.99417$$
If you dont want to jumpt to the obvious conclusion, now consider that you look for the zero of function
$$f(r)=43(1-r^{10})+341(1-r^7)$$ and build the Taylor series around $r=r_0$; this would give
$$f(r)=left(-43 r_0^{10}-341 r_0^7+384right)-left(430 r_0^9+2387 r_0^6right)
(r-r_0)+Oleft((r-r_0)^2right)$$ Ignoring the higher order tems, solve for $r$ to get
$$r=r_0-frac{43 r_0^{10}+341 r_0^7-384}{r_0^6 left(430 r_0^3+2387right)}$$ You know the exact value of $r_0^3$, then the exact value of $r_0^6$; now approximate $r_0^7simeq -2 r_0^6$ and $r_0^{10}simeq-2 r_0^3, r_0^6$ . This then gives
$$r_1=r_0-frac{236672}{39651821}=-2.00014$$
You do not need to solve a polynomial of degree $10$. Be practical !
answered Jan 27 at 17:21
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Make the ratio of the two expressions to get
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}$$ In absolute value, the rhs is large and then $r$ can be "large"; so may be, for the time being we could write
$$frac{1-r^{10}}{1-r^7}=-frac{341}{43}simeq frac{r^{10}}{r^7}=r^3 $$ Use you calculator to get, as an approximation
$$r_0=-sqrt[3]{frac{341}{43}}approx -1.99417$$
If you dont want to jumpt to the obvious conclusion, now consider that you look for the zero of function
$$f(r)=43(1-r^{10})+341(1-r^7)$$ and build the Taylor series around $r=r_0$; this would give
$$f(r)=left(-43 r_0^{10}-341 r_0^7+384right)-left(430 r_0^9+2387 r_0^6right)
(r-r_0)+Oleft((r-r_0)^2right)$$ Ignoring the higher order tems, solve for $r$ to get
$$r=r_0-frac{43 r_0^{10}+341 r_0^7-384}{r_0^6 left(430 r_0^3+2387right)}$$ You know the exact value of $r_0^3$, then the exact value of $r_0^6$; now approximate $r_0^7simeq -2 r_0^6$ and $r_0^{10}simeq-2 r_0^3, r_0^6$ . This then gives
$$r_1=r_0-frac{236672}{39651821}=-2.00014$$
You do not need to solve a polynomial of degree $10$. Be practical !
answered Jan 27 at 17:21
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 27 at 17:21
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 27 at 17:21
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 27 at 17:21
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
$begingroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
Cancel out $ 1 - r$ from the denominator by substitution.
$$a = 86frac{1-r}{1-r^7}$$
Now substitute:
$$ frac{86( 1-r^{10} )}{1-r^7} = -682 Rightarrow {86 - 86r^{10}} = -682 + 682r^7$$
Let $t = r^7$
$$768 = 862t + 86r^3$$
Forms a Cubic Equation, which needs to be solved.
answered Jan 28 at 14:48
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
add a comment |
$begingroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
Cancel out $ 1 - r$ from the denominator by substitution.
$$a = 86frac{1-r}{1-r^7}$$
Now substitute:
$$ frac{86( 1-r^{10} )}{1-r^7} = -682 Rightarrow {86 - 86r^{10}} = -682 + 682r^7$$
Let $t = r^7$
$$768 = 862t + 86r^3$$
Forms a Cubic Equation, which needs to be solved.
answered Jan 28 at 14:48
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
add a comment |
$begingroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
Cancel out $ 1 - r$ from the denominator by substitution.
$$a = 86frac{1-r}{1-r^7}$$
Now substitute:
$$ frac{86( 1-r^{10} )}{1-r^7} = -682 Rightarrow {86 - 86r^{10}} = -682 + 682r^7$$
Let $t = r^7$
$$768 = 862t + 86r^3$$
Forms a Cubic Equation, which needs to be solved.
answered Jan 28 at 14:48
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
Can anybody calculate the value of $r$ and a from systems-of-equations
given below?
$$frac{a(1-r^7)}{1-r}=86$$
$$frac{a(1-r^{10})}{1-r}=-682$$
Cancel out $ 1 - r$ from the denominator by substitution.
$$a = 86frac{1-r}{1-r^7}$$
Now substitute:
$$ frac{86( 1-r^{10} )}{1-r^7} = -682 Rightarrow {86 - 86r^{10}} = -682 + 682r^7$$
Let $t = r^7$
$$768 = 862t + 86r^3$$
Forms a Cubic Equation, which needs to be solved.
answered Jan 28 at 14:48
Abhas Kumar SinhaAbhas Kumar Sinha
304115
answered Jan 28 at 14:48
Abhas Kumar SinhaAbhas Kumar Sinha
304115
answered Jan 28 at 14:48
Abhas Kumar SinhaAbhas Kumar Sinha
304115
answered Jan 28 at 14:48
Abhas Kumar SinhaAbhas Kumar Sinha
304115
304115
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.
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%2f3089350%2fsolve-for-r-and-a-from-systems-of-equations%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
$begingroup$
divides the second equation by the first one, you'll deduce $r$.
$endgroup$
– Atmos
Jan 27 at 9:55