Mixing
Linear systems of equations with 4 unknowns
$begingroup$
I tried to solve these systems of equations in my book:
begin{align}
7&x+4y+3z+2w=46\
5&x-y+4w=23\
&x+z=6\
3&x+7w=15
end{align}
I tried to solve it in many different ways, but I still haven't gotten $x=5$, $y=2$, $z=1$ and $w=0$ which are the solutions of these systems of equations.
Did I make a mistake? How did the book get $x=5$, $y=2$, $z=1$ and $w=0$?
linear-algebra systems-of-equations
asked Oct 6 '18 at 13:30
lincong wanglincong wang
175
$endgroup$
add a comment |
$begingroup$
I tried to solve these systems of equations in my book:
begin{align}
7&x+4y+3z+2w=46\
5&x-y+4w=23\
&x+z=6\
3&x+7w=15
end{align}
I tried to solve it in many different ways, but I still haven't gotten $x=5$, $y=2$, $z=1$ and $w=0$ which are the solutions of these systems of equations.
Did I make a mistake? How did the book get $x=5$, $y=2$, $z=1$ and $w=0$?
linear-algebra systems-of-equations
asked Oct 6 '18 at 13:30
lincong wanglincong wang
175
$endgroup$
1
$begingroup$
Linear systems of equations are very tedious when you have to do them by hand. Just check you work. Again. Again. And you need a very steady hand too.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:31
$begingroup$
Or maybe my book is wrong.
$endgroup$
– lincong wang
Oct 6 '18 at 13:34
1
$begingroup$
No it's not. The given values satisfy the equations.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:35
add a comment |
$begingroup$
I tried to solve these systems of equations in my book:
begin{align}
7&x+4y+3z+2w=46\
5&x-y+4w=23\
&x+z=6\
3&x+7w=15
end{align}
I tried to solve it in many different ways, but I still haven't gotten $x=5$, $y=2$, $z=1$ and $w=0$ which are the solutions of these systems of equations.
Did I make a mistake? How did the book get $x=5$, $y=2$, $z=1$ and $w=0$?
linear-algebra systems-of-equations
asked Oct 6 '18 at 13:30
lincong wanglincong wang
175
$endgroup$
I tried to solve these systems of equations in my book:
begin{align}
7&x+4y+3z+2w=46\
5&x-y+4w=23\
&x+z=6\
3&x+7w=15
end{align}
I tried to solve it in many different ways, but I still haven't gotten $x=5$, $y=2$, $z=1$ and $w=0$ which are the solutions of these systems of equations.
Did I make a mistake? How did the book get $x=5$, $y=2$, $z=1$ and $w=0$?
linear-algebra systems-of-equations
linear-algebra systems-of-equations
asked Oct 6 '18 at 13:30
lincong wanglincong wang
175
asked Oct 6 '18 at 13:30
lincong wanglincong wang
175
edited Jan 27 at 22:30
lincong wang
asked Oct 6 '18 at 13:30
lincong wanglincong wang
175
asked Oct 6 '18 at 13:30
lincong wanglincong wang
175
asked Oct 6 '18 at 13:30
lincong wanglincong wang
175
175
1
$begingroup$
Linear systems of equations are very tedious when you have to do them by hand. Just check you work. Again. Again. And you need a very steady hand too.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:31
$begingroup$
Or maybe my book is wrong.
$endgroup$
– lincong wang
Oct 6 '18 at 13:34
1
$begingroup$
No it's not. The given values satisfy the equations.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:35
add a comment |
1
$begingroup$
Linear systems of equations are very tedious when you have to do them by hand. Just check you work. Again. Again. And you need a very steady hand too.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:31
$begingroup$
Or maybe my book is wrong.
$endgroup$
– lincong wang
Oct 6 '18 at 13:34
1
$begingroup$
No it's not. The given values satisfy the equations.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:35
1
1
$begingroup$
Linear systems of equations are very tedious when you have to do them by hand. Just check you work. Again. Again. And you need a very steady hand too.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:31
$begingroup$
Linear systems of equations are very tedious when you have to do them by hand. Just check you work. Again. Again. And you need a very steady hand too.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:31
$begingroup$
Or maybe my book is wrong.
$endgroup$
– lincong wang
Oct 6 '18 at 13:34
$begingroup$
Or maybe my book is wrong.
$endgroup$
– lincong wang
Oct 6 '18 at 13:34
1
1
$begingroup$
No it's not. The given values satisfy the equations.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:35
$begingroup$
No it's not. The given values satisfy the equations.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
With $z = 6 - x$ we can eliminate $z$ so we get:
$$7x+4y+3(6-x)+2w=46$$
Then distributing the $3$:
$$7x+4y+18-3x+2w=46$$
Moving variable terms to left of the left side and constant term to right of left side then combining like terms we get:
$$4x+4y+2w+18=46$$
Subtracting $18$ from both sides so we get
$$4x+4y+2w=28$$
Dividing by $2$, then it becomes:
$$2x+2y+w=28$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Then deriving from the first equation we get
$$w=14-2x-2y$$
Substituting $w$ we get:
$$5x-y+4(14-2x-2y)=23$$
$$3x+7(14-2x-2y)=15$$
By the process the steps for the first one are below:
$$5x-y+56-8x-8y=23$$
$$5x-y-8x-8y+56=23$$
$$-3x-9y+56=23$$
$$-3x-9y=-33$$
Let's get rid of the negatives:
$$3x+9y=33$$
And the steps for the second one are below:
$$3x+98-14x-14y=15$$
$$-11x-14y+98=15$$
$$-11x-14y=-83$$
Get rid of the negatives, there are the 2 equations:
$$3x+9y=33$$
$$11x+14y=83$$
We could do elimination here:
$$11(3x+9y)=11times 33$$
$$3(11x+14y)=3times 83$$
We get then:
$$33x+99y=363$$
$$33x+42y=249$$
Subtracting one from the other we get:
$$(33x+99y)-(33x+42y)=363-249$$
We can distribute the minus sign on the left side:
$$33x+99y-33x-42y=363-249$$
Simplifying we get:
$$57y=114$$
Dividing both sides by $57$ we get:
$$y=2$$
Now let's plug this in into any one of the equations before multiplying (in this case, I used $3x+9y=33$ just for simplicity, simplifying, through the steps:
$$3x+9times 2=33$$
$$3x+18=33$$
$$3x=15$$
$$x=5$$
We've got $x=5$ and $y=2$. Can you finish the rest to get $z=1$ and $w=0$?
edited Jan 27 at 22:45
lincong wang
175
answered Jan 27 at 22:12
zixuanzixuan
13410
$endgroup$
$begingroup$
That's the longest answer I've ever saw!
$endgroup$
– lincong wang
Jan 27 at 22:13
1
$begingroup$
Only because the equation you gave is a equation with 4 unknowns.
$endgroup$
– zixuan
Jan 27 at 22:19
$begingroup$
Yep got it and got exactly what the book told
$endgroup$
– lincong wang
Jan 27 at 22:25
add a comment |
$begingroup$
With $z=6-x$ we can eliminate the variable $z$ so
we get
$$2x+2y+w=14$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Using
$$w=14-2x-2y$$
we get
$$x+3y=11$$
$$11x+14y=83$$
with
$$x=11-3y$$ we get
$$-11(11-3y)-14y=-83$$
so
$$y=2$$
Can you finish?
edited Oct 6 '18 at 13:53
lincong wang
175
answered Oct 6 '18 at 13:49
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
$begingroup$
Yes, and then I get exactly x=5, y=2, z=1 and w=0.
$endgroup$
– lincong wang
Oct 6 '18 at 13:55
$begingroup$
That is nice, so your problem is solved now?
$endgroup$
– Dr. Sonnhard Graubner
Oct 6 '18 at 13:55
$begingroup$
Yes. I thought so that I made a mistake
$endgroup$
– lincong wang
Oct 6 '18 at 13:57
$begingroup$
Sorry, I had the more longer answer
$endgroup$
– lincong wang
Jan 27 at 22:16
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With $z = 6 - x$ we can eliminate $z$ so we get:
$$7x+4y+3(6-x)+2w=46$$
Then distributing the $3$:
$$7x+4y+18-3x+2w=46$$
Moving variable terms to left of the left side and constant term to right of left side then combining like terms we get:
$$4x+4y+2w+18=46$$
Subtracting $18$ from both sides so we get
$$4x+4y+2w=28$$
Dividing by $2$, then it becomes:
$$2x+2y+w=28$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Then deriving from the first equation we get
$$w=14-2x-2y$$
Substituting $w$ we get:
$$5x-y+4(14-2x-2y)=23$$
$$3x+7(14-2x-2y)=15$$
By the process the steps for the first one are below:
$$5x-y+56-8x-8y=23$$
$$5x-y-8x-8y+56=23$$
$$-3x-9y+56=23$$
$$-3x-9y=-33$$
Let's get rid of the negatives:
$$3x+9y=33$$
And the steps for the second one are below:
$$3x+98-14x-14y=15$$
$$-11x-14y+98=15$$
$$-11x-14y=-83$$
Get rid of the negatives, there are the 2 equations:
$$3x+9y=33$$
$$11x+14y=83$$
We could do elimination here:
$$11(3x+9y)=11times 33$$
$$3(11x+14y)=3times 83$$
We get then:
$$33x+99y=363$$
$$33x+42y=249$$
Subtracting one from the other we get:
$$(33x+99y)-(33x+42y)=363-249$$
We can distribute the minus sign on the left side:
$$33x+99y-33x-42y=363-249$$
Simplifying we get:
$$57y=114$$
Dividing both sides by $57$ we get:
$$y=2$$
Now let's plug this in into any one of the equations before multiplying (in this case, I used $3x+9y=33$ just for simplicity, simplifying, through the steps:
$$3x+9times 2=33$$
$$3x+18=33$$
$$3x=15$$
$$x=5$$
We've got $x=5$ and $y=2$. Can you finish the rest to get $z=1$ and $w=0$?
edited Jan 27 at 22:45
lincong wang
175
answered Jan 27 at 22:12
zixuanzixuan
13410
$endgroup$
$begingroup$
That's the longest answer I've ever saw!
$endgroup$
– lincong wang
Jan 27 at 22:13
1
$begingroup$
Only because the equation you gave is a equation with 4 unknowns.
$endgroup$
– zixuan
Jan 27 at 22:19
$begingroup$
Yep got it and got exactly what the book told
$endgroup$
– lincong wang
Jan 27 at 22:25
add a comment |
$begingroup$
With $z = 6 - x$ we can eliminate $z$ so we get:
$$7x+4y+3(6-x)+2w=46$$
Then distributing the $3$:
$$7x+4y+18-3x+2w=46$$
Moving variable terms to left of the left side and constant term to right of left side then combining like terms we get:
$$4x+4y+2w+18=46$$
Subtracting $18$ from both sides so we get
$$4x+4y+2w=28$$
Dividing by $2$, then it becomes:
$$2x+2y+w=28$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Then deriving from the first equation we get
$$w=14-2x-2y$$
Substituting $w$ we get:
$$5x-y+4(14-2x-2y)=23$$
$$3x+7(14-2x-2y)=15$$
By the process the steps for the first one are below:
$$5x-y+56-8x-8y=23$$
$$5x-y-8x-8y+56=23$$
$$-3x-9y+56=23$$
$$-3x-9y=-33$$
Let's get rid of the negatives:
$$3x+9y=33$$
And the steps for the second one are below:
$$3x+98-14x-14y=15$$
$$-11x-14y+98=15$$
$$-11x-14y=-83$$
Get rid of the negatives, there are the 2 equations:
$$3x+9y=33$$
$$11x+14y=83$$
We could do elimination here:
$$11(3x+9y)=11times 33$$
$$3(11x+14y)=3times 83$$
We get then:
$$33x+99y=363$$
$$33x+42y=249$$
Subtracting one from the other we get:
$$(33x+99y)-(33x+42y)=363-249$$
We can distribute the minus sign on the left side:
$$33x+99y-33x-42y=363-249$$
Simplifying we get:
$$57y=114$$
Dividing both sides by $57$ we get:
$$y=2$$
Now let's plug this in into any one of the equations before multiplying (in this case, I used $3x+9y=33$ just for simplicity, simplifying, through the steps:
$$3x+9times 2=33$$
$$3x+18=33$$
$$3x=15$$
$$x=5$$
We've got $x=5$ and $y=2$. Can you finish the rest to get $z=1$ and $w=0$?
edited Jan 27 at 22:45
lincong wang
175
answered Jan 27 at 22:12
zixuanzixuan
13410
$endgroup$
$begingroup$
That's the longest answer I've ever saw!
$endgroup$
– lincong wang
Jan 27 at 22:13
1
$begingroup$
Only because the equation you gave is a equation with 4 unknowns.
$endgroup$
– zixuan
Jan 27 at 22:19
$begingroup$
Yep got it and got exactly what the book told
$endgroup$
– lincong wang
Jan 27 at 22:25
add a comment |
$begingroup$
With $z = 6 - x$ we can eliminate $z$ so we get:
$$7x+4y+3(6-x)+2w=46$$
Then distributing the $3$:
$$7x+4y+18-3x+2w=46$$
Moving variable terms to left of the left side and constant term to right of left side then combining like terms we get:
$$4x+4y+2w+18=46$$
Subtracting $18$ from both sides so we get
$$4x+4y+2w=28$$
Dividing by $2$, then it becomes:
$$2x+2y+w=28$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Then deriving from the first equation we get
$$w=14-2x-2y$$
Substituting $w$ we get:
$$5x-y+4(14-2x-2y)=23$$
$$3x+7(14-2x-2y)=15$$
By the process the steps for the first one are below:
$$5x-y+56-8x-8y=23$$
$$5x-y-8x-8y+56=23$$
$$-3x-9y+56=23$$
$$-3x-9y=-33$$
Let's get rid of the negatives:
$$3x+9y=33$$
And the steps for the second one are below:
$$3x+98-14x-14y=15$$
$$-11x-14y+98=15$$
$$-11x-14y=-83$$
Get rid of the negatives, there are the 2 equations:
$$3x+9y=33$$
$$11x+14y=83$$
We could do elimination here:
$$11(3x+9y)=11times 33$$
$$3(11x+14y)=3times 83$$
We get then:
$$33x+99y=363$$
$$33x+42y=249$$
Subtracting one from the other we get:
$$(33x+99y)-(33x+42y)=363-249$$
We can distribute the minus sign on the left side:
$$33x+99y-33x-42y=363-249$$
Simplifying we get:
$$57y=114$$
Dividing both sides by $57$ we get:
$$y=2$$
Now let's plug this in into any one of the equations before multiplying (in this case, I used $3x+9y=33$ just for simplicity, simplifying, through the steps:
$$3x+9times 2=33$$
$$3x+18=33$$
$$3x=15$$
$$x=5$$
We've got $x=5$ and $y=2$. Can you finish the rest to get $z=1$ and $w=0$?
edited Jan 27 at 22:45
lincong wang
175
answered Jan 27 at 22:12
zixuanzixuan
13410
$endgroup$
With $z = 6 - x$ we can eliminate $z$ so we get:
$$7x+4y+3(6-x)+2w=46$$
Then distributing the $3$:
$$7x+4y+18-3x+2w=46$$
Moving variable terms to left of the left side and constant term to right of left side then combining like terms we get:
$$4x+4y+2w+18=46$$
Subtracting $18$ from both sides so we get
$$4x+4y+2w=28$$
Dividing by $2$, then it becomes:
$$2x+2y+w=28$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Then deriving from the first equation we get
$$w=14-2x-2y$$
Substituting $w$ we get:
$$5x-y+4(14-2x-2y)=23$$
$$3x+7(14-2x-2y)=15$$
By the process the steps for the first one are below:
$$5x-y+56-8x-8y=23$$
$$5x-y-8x-8y+56=23$$
$$-3x-9y+56=23$$
$$-3x-9y=-33$$
Let's get rid of the negatives:
$$3x+9y=33$$
And the steps for the second one are below:
$$3x+98-14x-14y=15$$
$$-11x-14y+98=15$$
$$-11x-14y=-83$$
Get rid of the negatives, there are the 2 equations:
$$3x+9y=33$$
$$11x+14y=83$$
We could do elimination here:
$$11(3x+9y)=11times 33$$
$$3(11x+14y)=3times 83$$
We get then:
$$33x+99y=363$$
$$33x+42y=249$$
Subtracting one from the other we get:
$$(33x+99y)-(33x+42y)=363-249$$
We can distribute the minus sign on the left side:
$$33x+99y-33x-42y=363-249$$
Simplifying we get:
$$57y=114$$
Dividing both sides by $57$ we get:
$$y=2$$
Now let's plug this in into any one of the equations before multiplying (in this case, I used $3x+9y=33$ just for simplicity, simplifying, through the steps:
$$3x+9times 2=33$$
$$3x+18=33$$
$$3x=15$$
$$x=5$$
We've got $x=5$ and $y=2$. Can you finish the rest to get $z=1$ and $w=0$?
edited Jan 27 at 22:45
lincong wang
175
answered Jan 27 at 22:12
zixuanzixuan
13410
edited Jan 27 at 22:45
lincong wang
175
edited Jan 27 at 22:45
lincong wang
175
edited Jan 27 at 22:45
lincong wang
175
175
answered Jan 27 at 22:12
zixuanzixuan
13410
answered Jan 27 at 22:12
zixuanzixuan
13410
answered Jan 27 at 22:12
zixuanzixuan
13410
13410
$begingroup$
That's the longest answer I've ever saw!
$endgroup$
– lincong wang
Jan 27 at 22:13
1
$begingroup$
Only because the equation you gave is a equation with 4 unknowns.
$endgroup$
– zixuan
Jan 27 at 22:19
$begingroup$
Yep got it and got exactly what the book told
$endgroup$
– lincong wang
Jan 27 at 22:25
add a comment |
$begingroup$
That's the longest answer I've ever saw!
$endgroup$
– lincong wang
Jan 27 at 22:13
1
$begingroup$
Only because the equation you gave is a equation with 4 unknowns.
$endgroup$
– zixuan
Jan 27 at 22:19
$begingroup$
Yep got it and got exactly what the book told
$endgroup$
– lincong wang
Jan 27 at 22:25
$begingroup$
That's the longest answer I've ever saw!
$endgroup$
– lincong wang
Jan 27 at 22:13
$begingroup$
That's the longest answer I've ever saw!
$endgroup$
– lincong wang
Jan 27 at 22:13
1
1
$begingroup$
Only because the equation you gave is a equation with 4 unknowns.
$endgroup$
– zixuan
Jan 27 at 22:19
$begingroup$
Only because the equation you gave is a equation with 4 unknowns.
$endgroup$
– zixuan
Jan 27 at 22:19
$begingroup$
Yep got it and got exactly what the book told
$endgroup$
– lincong wang
Jan 27 at 22:25
$begingroup$
Yep got it and got exactly what the book told
$endgroup$
– lincong wang
Jan 27 at 22:25
add a comment |
$begingroup$
With $z=6-x$ we can eliminate the variable $z$ so
we get
$$2x+2y+w=14$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Using
$$w=14-2x-2y$$
we get
$$x+3y=11$$
$$11x+14y=83$$
with
$$x=11-3y$$ we get
$$-11(11-3y)-14y=-83$$
so
$$y=2$$
Can you finish?
edited Oct 6 '18 at 13:53
lincong wang
175
answered Oct 6 '18 at 13:49
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
$begingroup$
Yes, and then I get exactly x=5, y=2, z=1 and w=0.
$endgroup$
– lincong wang
Oct 6 '18 at 13:55
$begingroup$
That is nice, so your problem is solved now?
$endgroup$
– Dr. Sonnhard Graubner
Oct 6 '18 at 13:55
$begingroup$
Yes. I thought so that I made a mistake
$endgroup$
– lincong wang
Oct 6 '18 at 13:57
$begingroup$
Sorry, I had the more longer answer
$endgroup$
– lincong wang
Jan 27 at 22:16
add a comment |
$begingroup$
With $z=6-x$ we can eliminate the variable $z$ so
we get
$$2x+2y+w=14$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Using
$$w=14-2x-2y$$
we get
$$x+3y=11$$
$$11x+14y=83$$
with
$$x=11-3y$$ we get
$$-11(11-3y)-14y=-83$$
so
$$y=2$$
Can you finish?
edited Oct 6 '18 at 13:53
lincong wang
175
answered Oct 6 '18 at 13:49
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
$begingroup$
Yes, and then I get exactly x=5, y=2, z=1 and w=0.
$endgroup$
– lincong wang
Oct 6 '18 at 13:55
$begingroup$
That is nice, so your problem is solved now?
$endgroup$
– Dr. Sonnhard Graubner
Oct 6 '18 at 13:55
$begingroup$
Yes. I thought so that I made a mistake
$endgroup$
– lincong wang
Oct 6 '18 at 13:57
$begingroup$
Sorry, I had the more longer answer
$endgroup$
– lincong wang
Jan 27 at 22:16
add a comment |
$begingroup$
With $z=6-x$ we can eliminate the variable $z$ so
we get
$$2x+2y+w=14$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Using
$$w=14-2x-2y$$
we get
$$x+3y=11$$
$$11x+14y=83$$
with
$$x=11-3y$$ we get
$$-11(11-3y)-14y=-83$$
so
$$y=2$$
Can you finish?
edited Oct 6 '18 at 13:53
lincong wang
175
answered Oct 6 '18 at 13:49
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
With $z=6-x$ we can eliminate the variable $z$ so
we get
$$2x+2y+w=14$$
$$5x-y+4w=23$$
$$3x+7w=15$$
Using
$$w=14-2x-2y$$
we get
$$x+3y=11$$
$$11x+14y=83$$
with
$$x=11-3y$$ we get
$$-11(11-3y)-14y=-83$$
so
$$y=2$$
Can you finish?
edited Oct 6 '18 at 13:53
lincong wang
175
answered Oct 6 '18 at 13:49
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
edited Oct 6 '18 at 13:53
lincong wang
175
edited Oct 6 '18 at 13:53
lincong wang
175
edited Oct 6 '18 at 13:53
lincong wang
175
175
answered Oct 6 '18 at 13:49
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered Oct 6 '18 at 13:49
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered Oct 6 '18 at 13:49
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
78.1k42867
$begingroup$
Yes, and then I get exactly x=5, y=2, z=1 and w=0.
$endgroup$
– lincong wang
Oct 6 '18 at 13:55
$begingroup$
That is nice, so your problem is solved now?
$endgroup$
– Dr. Sonnhard Graubner
Oct 6 '18 at 13:55
$begingroup$
Yes. I thought so that I made a mistake
$endgroup$
– lincong wang
Oct 6 '18 at 13:57
$begingroup$
Sorry, I had the more longer answer
$endgroup$
– lincong wang
Jan 27 at 22:16
add a comment |
$begingroup$
Yes, and then I get exactly x=5, y=2, z=1 and w=0.
$endgroup$
– lincong wang
Oct 6 '18 at 13:55
$begingroup$
That is nice, so your problem is solved now?
$endgroup$
– Dr. Sonnhard Graubner
Oct 6 '18 at 13:55
$begingroup$
Yes. I thought so that I made a mistake
$endgroup$
– lincong wang
Oct 6 '18 at 13:57
$begingroup$
Sorry, I had the more longer answer
$endgroup$
– lincong wang
Jan 27 at 22:16
$begingroup$
Yes, and then I get exactly x=5, y=2, z=1 and w=0.
$endgroup$
– lincong wang
Oct 6 '18 at 13:55
$begingroup$
Yes, and then I get exactly x=5, y=2, z=1 and w=0.
$endgroup$
– lincong wang
Oct 6 '18 at 13:55
$begingroup$
That is nice, so your problem is solved now?
$endgroup$
– Dr. Sonnhard Graubner
Oct 6 '18 at 13:55
$begingroup$
That is nice, so your problem is solved now?
$endgroup$
– Dr. Sonnhard Graubner
Oct 6 '18 at 13:55
$begingroup$
Yes. I thought so that I made a mistake
$endgroup$
– lincong wang
Oct 6 '18 at 13:57
$begingroup$
Yes. I thought so that I made a mistake
$endgroup$
– lincong wang
Oct 6 '18 at 13:57
$begingroup$
Sorry, I had the more longer answer
$endgroup$
– lincong wang
Jan 27 at 22:16
$begingroup$
Sorry, I had the more longer answer
$endgroup$
– lincong wang
Jan 27 at 22:16
add a comment |
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$begingroup$
Linear systems of equations are very tedious when you have to do them by hand. Just check you work. Again. Again. And you need a very steady hand too.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:31
$begingroup$
Or maybe my book is wrong.
$endgroup$
– lincong wang
Oct 6 '18 at 13:34
1
$begingroup$
No it's not. The given values satisfy the equations.
$endgroup$
– Parcly Taxel
Oct 6 '18 at 13:35