System of three linear equations with unknown constant
$begingroup$
I have a linear system with three equations:
$$begin{align}
x + y + z &= 3a\
x + 2y + (a+2)z &= a\
x - (a+1)y - z &= 0
end{align}$$
I want to find values for a where the system is inconsistent in addition to the values of $x, y, z$ and $a$ for where the system is consistent.
How should one approach a problem like this?
Thanks in advance.
linear-algebra systems-of-equations
edited Dec 16 '18 at 15:14
mrtaurho
5,74551540
asked Dec 16 '18 at 14:45
Olof AlmqvistOlof Almqvist
82
$endgroup$
add a comment |
$begingroup$
I have a linear system with three equations:
$$begin{align}
x + y + z &= 3a\
x + 2y + (a+2)z &= a\
x - (a+1)y - z &= 0
end{align}$$
I want to find values for a where the system is inconsistent in addition to the values of $x, y, z$ and $a$ for where the system is consistent.
How should one approach a problem like this?
Thanks in advance.
linear-algebra systems-of-equations
edited Dec 16 '18 at 15:14
mrtaurho
5,74551540
asked Dec 16 '18 at 14:45
Olof AlmqvistOlof Almqvist
82
$endgroup$
add a comment |
$begingroup$
I have a linear system with three equations:
$$begin{align}
x + y + z &= 3a\
x + 2y + (a+2)z &= a\
x - (a+1)y - z &= 0
end{align}$$
I want to find values for a where the system is inconsistent in addition to the values of $x, y, z$ and $a$ for where the system is consistent.
How should one approach a problem like this?
Thanks in advance.
linear-algebra systems-of-equations
edited Dec 16 '18 at 15:14
mrtaurho
5,74551540
asked Dec 16 '18 at 14:45
Olof AlmqvistOlof Almqvist
82
$endgroup$
I have a linear system with three equations:
$$begin{align}
x + y + z &= 3a\
x + 2y + (a+2)z &= a\
x - (a+1)y - z &= 0
end{align}$$
I want to find values for a where the system is inconsistent in addition to the values of $x, y, z$ and $a$ for where the system is consistent.
How should one approach a problem like this?
Thanks in advance.
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Dec 16 '18 at 15:14
mrtaurho
5,74551540
asked Dec 16 '18 at 14:45
Olof AlmqvistOlof Almqvist
82
edited Dec 16 '18 at 15:14
mrtaurho
5,74551540
asked Dec 16 '18 at 14:45
Olof AlmqvistOlof Almqvist
82
edited Dec 16 '18 at 15:14
mrtaurho
5,74551540
edited Dec 16 '18 at 15:14
mrtaurho
5,74551540
edited Dec 16 '18 at 15:14
mrtaurho
5,74551540
5,74551540
asked Dec 16 '18 at 14:45
Olof AlmqvistOlof Almqvist
82
asked Dec 16 '18 at 14:45
Olof AlmqvistOlof Almqvist
82
asked Dec 16 '18 at 14:45
Olof AlmqvistOlof Almqvist
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With $$z=x-(a+1)y$$ we get
$$2x+y-(a+1)y=3a$$
and
$$2x-ay=3a$$
eliminating $x$ we obtain
$$-frac{3a(a+3)}{2}y=-frac{3a(a+3)}{2}$$
Now you can do the rest!
answered Dec 16 '18 at 14:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
$begingroup$
Thanks for your help Dr. Graubner! :)
$endgroup$
– Olof Almqvist
Dec 16 '18 at 17:34
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With $$z=x-(a+1)y$$ we get
$$2x+y-(a+1)y=3a$$
and
$$2x-ay=3a$$
eliminating $x$ we obtain
$$-frac{3a(a+3)}{2}y=-frac{3a(a+3)}{2}$$
Now you can do the rest!
answered Dec 16 '18 at 14:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
$begingroup$
Thanks for your help Dr. Graubner! :)
$endgroup$
– Olof Almqvist
Dec 16 '18 at 17:34
add a comment |
$begingroup$
With $$z=x-(a+1)y$$ we get
$$2x+y-(a+1)y=3a$$
and
$$2x-ay=3a$$
eliminating $x$ we obtain
$$-frac{3a(a+3)}{2}y=-frac{3a(a+3)}{2}$$
Now you can do the rest!
answered Dec 16 '18 at 14:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
$begingroup$
Thanks for your help Dr. Graubner! :)
$endgroup$
– Olof Almqvist
Dec 16 '18 at 17:34
add a comment |
$begingroup$
With $$z=x-(a+1)y$$ we get
$$2x+y-(a+1)y=3a$$
and
$$2x-ay=3a$$
eliminating $x$ we obtain
$$-frac{3a(a+3)}{2}y=-frac{3a(a+3)}{2}$$
Now you can do the rest!
answered Dec 16 '18 at 14:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
With $$z=x-(a+1)y$$ we get
$$2x+y-(a+1)y=3a$$
and
$$2x-ay=3a$$
eliminating $x$ we obtain
$$-frac{3a(a+3)}{2}y=-frac{3a(a+3)}{2}$$
Now you can do the rest!
answered Dec 16 '18 at 14:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
answered Dec 16 '18 at 14:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
answered Dec 16 '18 at 14:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
answered Dec 16 '18 at 14:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
77.2k42866
$begingroup$
Thanks for your help Dr. Graubner! :)
$endgroup$
– Olof Almqvist
Dec 16 '18 at 17:34
add a comment |
$begingroup$
Thanks for your help Dr. Graubner! :)
$endgroup$
– Olof Almqvist
Dec 16 '18 at 17:34
$begingroup$
Thanks for your help Dr. Graubner! :)
$endgroup$
– Olof Almqvist
Dec 16 '18 at 17:34
$begingroup$
Thanks for your help Dr. Graubner! :)
$endgroup$
– Olof Almqvist
Dec 16 '18 at 17:34
add a comment |
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