Solving a linear system of reciprocals.Solving system of multivariable 2nd-degree polynomialsProblem with system of equationsSolving a system of equations with variables in denominator.Solving system of non-linear equations.Solving a combined system of linear and bilinear equationsSolving system of linear equations ( 4 variables, 3 equations)Solving a system of 4 non-linear equationsSolving a linear system of congruencesEasier solution for system of equations.Solve the following system of equations - (4).
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Solving a linear system of reciprocals.
Solving system of multivariable 2nd-degree polynomialsProblem with system of equationsSolving a system of equations with variables in denominator.Solving system of non-linear equations.Solving a combined system of linear and bilinear equationsSolving system of linear equations ( 4 variables, 3 equations)Solving a system of 4 non-linear equationsSolving a linear system of congruencesEasier solution for system of equations.Solve the following system of equations - (4).
$begingroup$
Solve for $begincasesfrac1x +frac1y+frac1z=0\frac4x +frac3y+frac2z=5\frac3x +frac2y+frac4z=-4endcases$
I turn the equations into $begincasesyz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzendcases$
Not sure if I am doing fine
systems-of-equations
edited 18 mins ago
Cameron Buie
87.6k773162
asked 21 mins ago
DavidDavid
764
$endgroup$
add a comment |
$begingroup$
Solve for $begincasesfrac1x +frac1y+frac1z=0\frac4x +frac3y+frac2z=5\frac3x +frac2y+frac4z=-4endcases$
I turn the equations into $begincasesyz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzendcases$
Not sure if I am doing fine
systems-of-equations
edited 18 mins ago
Cameron Buie
87.6k773162
asked 21 mins ago
DavidDavid
764
$endgroup$
add a comment |
$begingroup$
Solve for $begincasesfrac1x +frac1y+frac1z=0\frac4x +frac3y+frac2z=5\frac3x +frac2y+frac4z=-4endcases$
I turn the equations into $begincasesyz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzendcases$
Not sure if I am doing fine
systems-of-equations
edited 18 mins ago
Cameron Buie
87.6k773162
asked 21 mins ago
DavidDavid
764
$endgroup$
Solve for $begincasesfrac1x +frac1y+frac1z=0\frac4x +frac3y+frac2z=5\frac3x +frac2y+frac4z=-4endcases$
I turn the equations into $begincasesyz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzendcases$
Not sure if I am doing fine
systems-of-equations
systems-of-equations
edited 18 mins ago
Cameron Buie
87.6k773162
asked 21 mins ago
DavidDavid
764
edited 18 mins ago
Cameron Buie
87.6k773162
asked 21 mins ago
DavidDavid
764
edited 18 mins ago
Cameron Buie
87.6k773162
edited 18 mins ago
Cameron Buie
87.6k773162
edited 18 mins ago
Cameron Buie
87.6k773162
87.6k773162
asked 21 mins ago
DavidDavid
764
asked 21 mins ago
DavidDavid
764
asked 21 mins ago
DavidDavid
764
764
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.
answered 15 mins ago
Cameron BuieCameron Buie
87.6k773162
$endgroup$
add a comment |
$begingroup$
It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.
answered 19 mins ago
Matt SamuelMatt Samuel
39.6k63870
$endgroup$
add a comment |
$begingroup$
how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$
answered 14 mins ago
Henry LeeHenry Lee
2,211319
$endgroup$
add a comment |
$begingroup$
Observe that the system's matrix of the reciprocal of the unknowns is:
$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&-1&-4endpmatrixstackrelR_3-R_2longrightarrow$$
$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&1&-9endpmatrix$$
Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$
answered 13 mins ago
DonAntonioDonAntonio
180k1495234
$endgroup$
$begingroup$
There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
$endgroup$
– Bernard
8 mins ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.
answered 15 mins ago
Cameron BuieCameron Buie
87.6k773162
$endgroup$
add a comment |
$begingroup$
If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.
answered 15 mins ago
Cameron BuieCameron Buie
87.6k773162
$endgroup$
add a comment |
$begingroup$
If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.
answered 15 mins ago
Cameron BuieCameron Buie
87.6k773162
$endgroup$
If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.
answered 15 mins ago
Cameron BuieCameron Buie
87.6k773162
answered 15 mins ago
Cameron BuieCameron Buie
87.6k773162
answered 15 mins ago
Cameron BuieCameron Buie
87.6k773162
answered 15 mins ago
Cameron BuieCameron Buie
87.6k773162
87.6k773162
add a comment |
add a comment |
$begingroup$
It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.
answered 19 mins ago
Matt SamuelMatt Samuel
39.6k63870
$endgroup$
add a comment |
$begingroup$
It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.
answered 19 mins ago
Matt SamuelMatt Samuel
39.6k63870
$endgroup$
add a comment |
$begingroup$
It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.
answered 19 mins ago
Matt SamuelMatt Samuel
39.6k63870
$endgroup$
It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.
answered 19 mins ago
Matt SamuelMatt Samuel
39.6k63870
answered 19 mins ago
Matt SamuelMatt Samuel
39.6k63870
answered 19 mins ago
Matt SamuelMatt Samuel
39.6k63870
answered 19 mins ago
Matt SamuelMatt Samuel
39.6k63870
39.6k63870
add a comment |
add a comment |
$begingroup$
how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$
answered 14 mins ago
Henry LeeHenry Lee
2,211319
$endgroup$
add a comment |
$begingroup$
how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$
answered 14 mins ago
Henry LeeHenry Lee
2,211319
$endgroup$
add a comment |
$begingroup$
how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$
answered 14 mins ago
Henry LeeHenry Lee
2,211319
$endgroup$
how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$
answered 14 mins ago
Henry LeeHenry Lee
2,211319
answered 14 mins ago
Henry LeeHenry Lee
2,211319
answered 14 mins ago
Henry LeeHenry Lee
2,211319
answered 14 mins ago
Henry LeeHenry Lee
2,211319
2,211319
add a comment |
add a comment |
$begingroup$
Observe that the system's matrix of the reciprocal of the unknowns is:
$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&-1&-4endpmatrixstackrelR_3-R_2longrightarrow$$
$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&1&-9endpmatrix$$
Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$
answered 13 mins ago
DonAntonioDonAntonio
180k1495234
$endgroup$
$begingroup$
There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
$endgroup$
– Bernard
8 mins ago
add a comment |
$begingroup$
Observe that the system's matrix of the reciprocal of the unknowns is:
$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&-1&-4endpmatrixstackrelR_3-R_2longrightarrow$$
$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&1&-9endpmatrix$$
Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$
answered 13 mins ago
DonAntonioDonAntonio
180k1495234
$endgroup$
$begingroup$
There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
$endgroup$
– Bernard
8 mins ago
add a comment |
$begingroup$
Observe that the system's matrix of the reciprocal of the unknowns is:
$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&-1&-4endpmatrixstackrelR_3-R_2longrightarrow$$
$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&1&-9endpmatrix$$
Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$
answered 13 mins ago
DonAntonioDonAntonio
180k1495234
$endgroup$
Observe that the system's matrix of the reciprocal of the unknowns is:
$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&-1&-4endpmatrixstackrelR_3-R_2longrightarrow$$
$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&1&-9endpmatrix$$
Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$
answered 13 mins ago
DonAntonioDonAntonio
180k1495234
answered 13 mins ago
DonAntonioDonAntonio
180k1495234
answered 13 mins ago
DonAntonioDonAntonio
180k1495234
answered 13 mins ago
DonAntonioDonAntonio
180k1495234
180k1495234
$begingroup$
There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
$endgroup$
– Bernard
8 mins ago
add a comment |
$begingroup$
There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
$endgroup$
– Bernard
8 mins ago
$begingroup$
There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
$endgroup$
– Bernard
8 mins ago
$begingroup$
There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
$endgroup$
– Bernard
8 mins ago
add a comment |
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