Find the intersection between the following two curves in 3 unknown variables. [closed]

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There are two curves:
$$S1: xy^2z - yz + 2xz =39$$
and
$$S2: 2x - 3y + 5z = 26$$



I tried substituting x, y, z into both equations alternatively to get an equation in two variables and then find the dependence of one variable on another. But, I am not able to get the solution.
Any help will be much appreciated.



Thank You.



I am able to substitute for x, y, and z, and as in this answer: Find the intersection between the following two curves in 3 unknown variables., I even get that equation, but from here I am having a problem.




algebra-precalculus geometry




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closed as off-topic by amWhy, Strants, Isaac Browne, Szeto, Parcly Taxel Jul 20 at 4:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Strants, Isaac Browne, Szeto, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.












  • The method you described should work... Share your solution so we can take a look at it
    – Green.H
    Jul 19 at 13:05














up vote
1
down vote

favorite












There are two curves:
$$S1: xy^2z - yz + 2xz =39$$
and
$$S2: 2x - 3y + 5z = 26$$



I tried substituting x, y, z into both equations alternatively to get an equation in two variables and then find the dependence of one variable on another. But, I am not able to get the solution.
Any help will be much appreciated.



Thank You.



I am able to substitute for x, y, and z, and as in this answer: Find the intersection between the following two curves in 3 unknown variables., I even get that equation, but from here I am having a problem.




algebra-precalculus geometry




share|cite|improve this question













closed as off-topic by amWhy, Strants, Isaac Browne, Szeto, Parcly Taxel Jul 20 at 4:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Strants, Isaac Browne, Szeto, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.












  • The method you described should work... Share your solution so we can take a look at it
    – Green.H
    Jul 19 at 13:05












up vote
1
down vote

favorite









up vote
1
down vote

favorite











There are two curves:
$$S1: xy^2z - yz + 2xz =39$$
and
$$S2: 2x - 3y + 5z = 26$$



I tried substituting x, y, z into both equations alternatively to get an equation in two variables and then find the dependence of one variable on another. But, I am not able to get the solution.
Any help will be much appreciated.



Thank You.



I am able to substitute for x, y, and z, and as in this answer: Find the intersection between the following two curves in 3 unknown variables., I even get that equation, but from here I am having a problem.




algebra-precalculus geometry




share|cite|improve this question













There are two curves:
$$S1: xy^2z - yz + 2xz =39$$
and
$$S2: 2x - 3y + 5z = 26$$



I tried substituting x, y, z into both equations alternatively to get an equation in two variables and then find the dependence of one variable on another. But, I am not able to get the solution.
Any help will be much appreciated.



Thank You.



I am able to substitute for x, y, and z, and as in this answer: Find the intersection between the following two curves in 3 unknown variables., I even get that equation, but from here I am having a problem.





algebra-precalculus geometry





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 19 at 15:39
























asked Jul 19 at 13:02









thelogicalkoan

154




154




closed as off-topic by amWhy, Strants, Isaac Browne, Szeto, Parcly Taxel Jul 20 at 4:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Strants, Isaac Browne, Szeto, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Strants, Isaac Browne, Szeto, Parcly Taxel Jul 20 at 4:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Strants, Isaac Browne, Szeto, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.











  • The method you described should work... Share your solution so we can take a look at it
    – Green.H
    Jul 19 at 13:05
















  • The method you described should work... Share your solution so we can take a look at it
    – Green.H
    Jul 19 at 13:05















The method you described should work... Share your solution so we can take a look at it
– Green.H
Jul 19 at 13:05




The method you described should work... Share your solution so we can take a look at it
– Green.H
Jul 19 at 13:05










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Hint: From the first erquation we get
$$z=frac39xy^2+2x-y$$
plugging this in the second equation we obtain



$$2x^2y^2-3xy^3-26xy^2+4x^2-8xy+3y^2-52x+26y+195=0$$
this can be solved for $$x$$ or $$y$$



$$x^2(2y^2+4)+x(-3y^3+26y^2-8y-52)+3y^2+26y+195=0$$
Can you solve this?






share|cite|improve this answer























  • You should of course consider the case that your denominator vanishes separately ;-)
    – James
    Jul 19 at 13:18










  • Oh yes this should belong to my hint, thank you!
    – Dr. Sonnhard Graubner
    Jul 19 at 13:20










  • I am able to get this, but from here, I am having problems reducing.
    – thelogicalkoan
    Jul 19 at 15:34










  • You can solve the equation above for $x$ or $y$, $x$ is the better choice, since it is a quadratic equation.
    – Dr. Sonnhard Graubner
    Jul 19 at 16:19

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Hint: From the first erquation we get
$$z=frac39xy^2+2x-y$$
plugging this in the second equation we obtain



$$2x^2y^2-3xy^3-26xy^2+4x^2-8xy+3y^2-52x+26y+195=0$$
this can be solved for $$x$$ or $$y$$



$$x^2(2y^2+4)+x(-3y^3+26y^2-8y-52)+3y^2+26y+195=0$$
Can you solve this?






share|cite|improve this answer























  • You should of course consider the case that your denominator vanishes separately ;-)
    – James
    Jul 19 at 13:18










  • Oh yes this should belong to my hint, thank you!
    – Dr. Sonnhard Graubner
    Jul 19 at 13:20










  • I am able to get this, but from here, I am having problems reducing.
    – thelogicalkoan
    Jul 19 at 15:34










  • You can solve the equation above for $x$ or $y$, $x$ is the better choice, since it is a quadratic equation.
    – Dr. Sonnhard Graubner
    Jul 19 at 16:19














up vote
1
down vote



accepted










Hint: From the first erquation we get
$$z=frac39xy^2+2x-y$$
plugging this in the second equation we obtain



$$2x^2y^2-3xy^3-26xy^2+4x^2-8xy+3y^2-52x+26y+195=0$$
this can be solved for $$x$$ or $$y$$



$$x^2(2y^2+4)+x(-3y^3+26y^2-8y-52)+3y^2+26y+195=0$$
Can you solve this?






share|cite|improve this answer























  • You should of course consider the case that your denominator vanishes separately ;-)
    – James
    Jul 19 at 13:18










  • Oh yes this should belong to my hint, thank you!
    – Dr. Sonnhard Graubner
    Jul 19 at 13:20










  • I am able to get this, but from here, I am having problems reducing.
    – thelogicalkoan
    Jul 19 at 15:34










  • You can solve the equation above for $x$ or $y$, $x$ is the better choice, since it is a quadratic equation.
    – Dr. Sonnhard Graubner
    Jul 19 at 16:19












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Hint: From the first erquation we get
$$z=frac39xy^2+2x-y$$
plugging this in the second equation we obtain



$$2x^2y^2-3xy^3-26xy^2+4x^2-8xy+3y^2-52x+26y+195=0$$
this can be solved for $$x$$ or $$y$$



$$x^2(2y^2+4)+x(-3y^3+26y^2-8y-52)+3y^2+26y+195=0$$
Can you solve this?






share|cite|improve this answer















Hint: From the first erquation we get
$$z=frac39xy^2+2x-y$$
plugging this in the second equation we obtain



$$2x^2y^2-3xy^3-26xy^2+4x^2-8xy+3y^2-52x+26y+195=0$$
this can be solved for $$x$$ or $$y$$



$$x^2(2y^2+4)+x(-3y^3+26y^2-8y-52)+3y^2+26y+195=0$$
Can you solve this?







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 19 at 16:21


























answered Jul 19 at 13:17









Dr. Sonnhard Graubner

66.8k32659




66.8k32659











  • You should of course consider the case that your denominator vanishes separately ;-)
    – James
    Jul 19 at 13:18










  • Oh yes this should belong to my hint, thank you!
    – Dr. Sonnhard Graubner
    Jul 19 at 13:20










  • I am able to get this, but from here, I am having problems reducing.
    – thelogicalkoan
    Jul 19 at 15:34










  • You can solve the equation above for $x$ or $y$, $x$ is the better choice, since it is a quadratic equation.
    – Dr. Sonnhard Graubner
    Jul 19 at 16:19
















  • You should of course consider the case that your denominator vanishes separately ;-)
    – James
    Jul 19 at 13:18










  • Oh yes this should belong to my hint, thank you!
    – Dr. Sonnhard Graubner
    Jul 19 at 13:20










  • I am able to get this, but from here, I am having problems reducing.
    – thelogicalkoan
    Jul 19 at 15:34










  • You can solve the equation above for $x$ or $y$, $x$ is the better choice, since it is a quadratic equation.
    – Dr. Sonnhard Graubner
    Jul 19 at 16:19















You should of course consider the case that your denominator vanishes separately ;-)
– James
Jul 19 at 13:18




You should of course consider the case that your denominator vanishes separately ;-)
– James
Jul 19 at 13:18












Oh yes this should belong to my hint, thank you!
– Dr. Sonnhard Graubner
Jul 19 at 13:20




Oh yes this should belong to my hint, thank you!
– Dr. Sonnhard Graubner
Jul 19 at 13:20












I am able to get this, but from here, I am having problems reducing.
– thelogicalkoan
Jul 19 at 15:34




I am able to get this, but from here, I am having problems reducing.
– thelogicalkoan
Jul 19 at 15:34












You can solve the equation above for $x$ or $y$, $x$ is the better choice, since it is a quadratic equation.
– Dr. Sonnhard Graubner
Jul 19 at 16:19




You can solve the equation above for $x$ or $y$, $x$ is the better choice, since it is a quadratic equation.
– Dr. Sonnhard Graubner
Jul 19 at 16:19


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