Mixing
For the following system to be consistent, what must k not be equal to?
$begingroup$
$6x - 4y + 4z = 5$
$9x - 6y + kz = -4$
$12x - 8y = -10$
Originally I just multiplied the first row by (3/2) and subtracted it from the second, which gives you a value of 6 for the answer. However, this is not the correct answer. Any idea what I am doing wrong?
linear-algebra systems-of-equations
edited Sep 11 '14 at 14:44
amWhy
1
asked Sep 11 '14 at 14:23
MattMatt
111
$endgroup$
add a comment |
$begingroup$
$6x - 4y + 4z = 5$
$9x - 6y + kz = -4$
$12x - 8y = -10$
Originally I just multiplied the first row by (3/2) and subtracted it from the second, which gives you a value of 6 for the answer. However, this is not the correct answer. Any idea what I am doing wrong?
linear-algebra systems-of-equations
edited Sep 11 '14 at 14:44
amWhy
1
asked Sep 11 '14 at 14:23
MattMatt
111
$endgroup$
add a comment |
$begingroup$
$6x - 4y + 4z = 5$
$9x - 6y + kz = -4$
$12x - 8y = -10$
Originally I just multiplied the first row by (3/2) and subtracted it from the second, which gives you a value of 6 for the answer. However, this is not the correct answer. Any idea what I am doing wrong?
linear-algebra systems-of-equations
edited Sep 11 '14 at 14:44
amWhy
1
asked Sep 11 '14 at 14:23
MattMatt
111
$endgroup$
$6x - 4y + 4z = 5$
$9x - 6y + kz = -4$
$12x - 8y = -10$
Originally I just multiplied the first row by (3/2) and subtracted it from the second, which gives you a value of 6 for the answer. However, this is not the correct answer. Any idea what I am doing wrong?
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Sep 11 '14 at 14:44
amWhy
1
asked Sep 11 '14 at 14:23
MattMatt
111
edited Sep 11 '14 at 14:44
amWhy
1
asked Sep 11 '14 at 14:23
MattMatt
111
edited Sep 11 '14 at 14:44
amWhy
1
edited Sep 11 '14 at 14:44
amWhy
1
edited Sep 11 '14 at 14:44
amWhy
1
1
asked Sep 11 '14 at 14:23
MattMatt
111
asked Sep 11 '14 at 14:23
MattMatt
111
asked Sep 11 '14 at 14:23
MattMatt
111
111
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
you need to look at the third equation as well! How is that related to the second equation?
answered Sep 11 '14 at 14:29
almagestalmagest
12.1k1329
$endgroup$
add a comment |
$begingroup$
Use the first and third equations to find $z$. Then substitute that into the equation you got from the first two equations. There is only one value $k$ can be.
answered Sep 11 '14 at 14:58
Empy2Empy2
33.5k12261
$endgroup$
add a comment |
$begingroup$
Subtracting the first equation from the third twice yields $-8z=-20$, so $z=frac52$. Then multiplying the second equation by two and subtracting the first three times yields
$$-5k-30=-23.$$
This shows that $k=-frac75$.
answered Feb 24 '17 at 10:50
ServaesServaes
25.1k33994
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
you need to look at the third equation as well! How is that related to the second equation?
answered Sep 11 '14 at 14:29
almagestalmagest
12.1k1329
$endgroup$
add a comment |
$begingroup$
Hint:
you need to look at the third equation as well! How is that related to the second equation?
answered Sep 11 '14 at 14:29
almagestalmagest
12.1k1329
$endgroup$
add a comment |
$begingroup$
Hint:
you need to look at the third equation as well! How is that related to the second equation?
answered Sep 11 '14 at 14:29
almagestalmagest
12.1k1329
$endgroup$
Hint:
you need to look at the third equation as well! How is that related to the second equation?
answered Sep 11 '14 at 14:29
almagestalmagest
12.1k1329
answered Sep 11 '14 at 14:29
almagestalmagest
12.1k1329
answered Sep 11 '14 at 14:29
almagestalmagest
12.1k1329
answered Sep 11 '14 at 14:29
almagestalmagest
12.1k1329
12.1k1329
add a comment |
add a comment |
$begingroup$
Use the first and third equations to find $z$. Then substitute that into the equation you got from the first two equations. There is only one value $k$ can be.
answered Sep 11 '14 at 14:58
Empy2Empy2
33.5k12261
$endgroup$
add a comment |
$begingroup$
Use the first and third equations to find $z$. Then substitute that into the equation you got from the first two equations. There is only one value $k$ can be.
answered Sep 11 '14 at 14:58
Empy2Empy2
33.5k12261
$endgroup$
add a comment |
$begingroup$
Use the first and third equations to find $z$. Then substitute that into the equation you got from the first two equations. There is only one value $k$ can be.
answered Sep 11 '14 at 14:58
Empy2Empy2
33.5k12261
$endgroup$
Use the first and third equations to find $z$. Then substitute that into the equation you got from the first two equations. There is only one value $k$ can be.
answered Sep 11 '14 at 14:58
Empy2Empy2
33.5k12261
answered Sep 11 '14 at 14:58
Empy2Empy2
33.5k12261
answered Sep 11 '14 at 14:58
Empy2Empy2
33.5k12261
answered Sep 11 '14 at 14:58
Empy2Empy2
33.5k12261
33.5k12261
add a comment |
add a comment |
$begingroup$
Subtracting the first equation from the third twice yields $-8z=-20$, so $z=frac52$. Then multiplying the second equation by two and subtracting the first three times yields
$$-5k-30=-23.$$
This shows that $k=-frac75$.
answered Feb 24 '17 at 10:50
ServaesServaes
25.1k33994
$endgroup$
add a comment |
$begingroup$
Subtracting the first equation from the third twice yields $-8z=-20$, so $z=frac52$. Then multiplying the second equation by two and subtracting the first three times yields
$$-5k-30=-23.$$
This shows that $k=-frac75$.
answered Feb 24 '17 at 10:50
ServaesServaes
25.1k33994
$endgroup$
add a comment |
$begingroup$
Subtracting the first equation from the third twice yields $-8z=-20$, so $z=frac52$. Then multiplying the second equation by two and subtracting the first three times yields
$$-5k-30=-23.$$
This shows that $k=-frac75$.
answered Feb 24 '17 at 10:50
ServaesServaes
25.1k33994
$endgroup$
Subtracting the first equation from the third twice yields $-8z=-20$, so $z=frac52$. Then multiplying the second equation by two and subtracting the first three times yields
$$-5k-30=-23.$$
This shows that $k=-frac75$.
answered Feb 24 '17 at 10:50
ServaesServaes
25.1k33994
answered Feb 24 '17 at 10:50
ServaesServaes
25.1k33994
answered Feb 24 '17 at 10:50
ServaesServaes
25.1k33994
answered Feb 24 '17 at 10:50
ServaesServaes
25.1k33994
25.1k33994
add a comment |
add a comment |
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