Solving system of equations with $3$ variables: [closed]
I am looking for some detail about how the following system of equations is solved, I checked using wolfram and got the answer but I am not sure the details of how to get there.
$$x+y=1$$ $$xw+y=frac{1}{2}$$ $$xw^2+y = frac{1}{3}$$
and the solution is $w = frac{1}{3}, x = frac{3}{4}, y = frac{1}{4}$
algebra-precalculus systems-of-equations
closed as off-topic by Nosrati, Alexander Gruber♦ Dec 4 '18 at 4:25
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- "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." – Nosrati, Alexander Gruber
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add a comment |
I am looking for some detail about how the following system of equations is solved, I checked using wolfram and got the answer but I am not sure the details of how to get there.
$$x+y=1$$ $$xw+y=frac{1}{2}$$ $$xw^2+y = frac{1}{3}$$
and the solution is $w = frac{1}{3}, x = frac{3}{4}, y = frac{1}{4}$
algebra-precalculus systems-of-equations
closed as off-topic by Nosrati, Alexander Gruber♦ Dec 4 '18 at 4:25
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." – Nosrati, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
Plug first equation into second and third to get 2 by 2 I'm $x,w$.
– Zachary Selk
Nov 22 '18 at 23:13
add a comment |
I am looking for some detail about how the following system of equations is solved, I checked using wolfram and got the answer but I am not sure the details of how to get there.
$$x+y=1$$ $$xw+y=frac{1}{2}$$ $$xw^2+y = frac{1}{3}$$
and the solution is $w = frac{1}{3}, x = frac{3}{4}, y = frac{1}{4}$
algebra-precalculus systems-of-equations
I am looking for some detail about how the following system of equations is solved, I checked using wolfram and got the answer but I am not sure the details of how to get there.
$$x+y=1$$ $$xw+y=frac{1}{2}$$ $$xw^2+y = frac{1}{3}$$
and the solution is $w = frac{1}{3}, x = frac{3}{4}, y = frac{1}{4}$
algebra-precalculus systems-of-equations
algebra-precalculus systems-of-equations
edited Nov 22 '18 at 23:17
José Carlos Santos
152k22123226
152k22123226
asked Nov 22 '18 at 23:05
mt12345mt12345
958
958
closed as off-topic by Nosrati, Alexander Gruber♦ Dec 4 '18 at 4:25
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." – Nosrati, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, Alexander Gruber♦ Dec 4 '18 at 4:25
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." – Nosrati, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
Plug first equation into second and third to get 2 by 2 I'm $x,w$.
– Zachary Selk
Nov 22 '18 at 23:13
add a comment |
Plug first equation into second and third to get 2 by 2 I'm $x,w$.
– Zachary Selk
Nov 22 '18 at 23:13
Plug first equation into second and third to get 2 by 2 I'm $x,w$.
– Zachary Selk
Nov 22 '18 at 23:13
Plug first equation into second and third to get 2 by 2 I'm $x,w$.
– Zachary Selk
Nov 22 '18 at 23:13
add a comment |
2 Answers
2
active
oldest
votes
From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…
so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
– mt12345
Nov 22 '18 at 23:11
Yes, that's it.
– José Carlos Santos
Nov 22 '18 at 23:13
when you say but now dividng the second equality by the first could you show details of that?
– mt12345
Nov 22 '18 at 23:13
$$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
– José Carlos Santos
Nov 22 '18 at 23:15
and then after this I can plug x into y = 1 -x to find y?
– mt12345
Nov 22 '18 at 23:16
|
show 2 more comments
Subtract the second equation from the first equation
begin{eqnarray*}
x(1-w)=frac{1}{2}
end{eqnarray*}
Subtract the third equation from the second equation
begin{eqnarray*}
xw(1-w)=frac{1}{6}
end{eqnarray*}
So $w=?$ ....
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…
so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
– mt12345
Nov 22 '18 at 23:11
Yes, that's it.
– José Carlos Santos
Nov 22 '18 at 23:13
when you say but now dividng the second equality by the first could you show details of that?
– mt12345
Nov 22 '18 at 23:13
$$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
– José Carlos Santos
Nov 22 '18 at 23:15
and then after this I can plug x into y = 1 -x to find y?
– mt12345
Nov 22 '18 at 23:16
|
show 2 more comments
From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…
so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
– mt12345
Nov 22 '18 at 23:11
Yes, that's it.
– José Carlos Santos
Nov 22 '18 at 23:13
when you say but now dividng the second equality by the first could you show details of that?
– mt12345
Nov 22 '18 at 23:13
$$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
– José Carlos Santos
Nov 22 '18 at 23:15
and then after this I can plug x into y = 1 -x to find y?
– mt12345
Nov 22 '18 at 23:16
|
show 2 more comments
From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…
From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…
answered Nov 22 '18 at 23:10
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
– mt12345
Nov 22 '18 at 23:11
Yes, that's it.
– José Carlos Santos
Nov 22 '18 at 23:13
when you say but now dividng the second equality by the first could you show details of that?
– mt12345
Nov 22 '18 at 23:13
$$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
– José Carlos Santos
Nov 22 '18 at 23:15
and then after this I can plug x into y = 1 -x to find y?
– mt12345
Nov 22 '18 at 23:16
|
show 2 more comments
so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
– mt12345
Nov 22 '18 at 23:11
Yes, that's it.
– José Carlos Santos
Nov 22 '18 at 23:13
when you say but now dividng the second equality by the first could you show details of that?
– mt12345
Nov 22 '18 at 23:13
$$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
– José Carlos Santos
Nov 22 '18 at 23:15
and then after this I can plug x into y = 1 -x to find y?
– mt12345
Nov 22 '18 at 23:16
so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
– mt12345
Nov 22 '18 at 23:11
so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
– mt12345
Nov 22 '18 at 23:11
Yes, that's it.
– José Carlos Santos
Nov 22 '18 at 23:13
Yes, that's it.
– José Carlos Santos
Nov 22 '18 at 23:13
when you say but now dividng the second equality by the first could you show details of that?
– mt12345
Nov 22 '18 at 23:13
when you say but now dividng the second equality by the first could you show details of that?
– mt12345
Nov 22 '18 at 23:13
$$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
– José Carlos Santos
Nov 22 '18 at 23:15
$$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
– José Carlos Santos
Nov 22 '18 at 23:15
and then after this I can plug x into y = 1 -x to find y?
– mt12345
Nov 22 '18 at 23:16
and then after this I can plug x into y = 1 -x to find y?
– mt12345
Nov 22 '18 at 23:16
|
show 2 more comments
Subtract the second equation from the first equation
begin{eqnarray*}
x(1-w)=frac{1}{2}
end{eqnarray*}
Subtract the third equation from the second equation
begin{eqnarray*}
xw(1-w)=frac{1}{6}
end{eqnarray*}
So $w=?$ ....
add a comment |
Subtract the second equation from the first equation
begin{eqnarray*}
x(1-w)=frac{1}{2}
end{eqnarray*}
Subtract the third equation from the second equation
begin{eqnarray*}
xw(1-w)=frac{1}{6}
end{eqnarray*}
So $w=?$ ....
add a comment |
Subtract the second equation from the first equation
begin{eqnarray*}
x(1-w)=frac{1}{2}
end{eqnarray*}
Subtract the third equation from the second equation
begin{eqnarray*}
xw(1-w)=frac{1}{6}
end{eqnarray*}
So $w=?$ ....
Subtract the second equation from the first equation
begin{eqnarray*}
x(1-w)=frac{1}{2}
end{eqnarray*}
Subtract the third equation from the second equation
begin{eqnarray*}
xw(1-w)=frac{1}{6}
end{eqnarray*}
So $w=?$ ....
answered Nov 22 '18 at 23:12
Donald SplutterwitDonald Splutterwit
22.3k21344
22.3k21344
add a comment |
add a comment |
Plug first equation into second and third to get 2 by 2 I'm $x,w$.
– Zachary Selk
Nov 22 '18 at 23:13