Mixing
Solving system of equations with three unknowns, but equations are missing components?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
The equations are:
$ y * z = 8$
$x + z = 8$
$x + y = 6$
Solving for these gets me to:
$z = frac167, y = frac72$ and $x = 8 - frac167$.
This is incorrect however. Putting it into a solver yields no solution.
Why can't this be solved?
algebra-precalculus systems-of-equations
edited Jul 24 at 14:14
Jneven
479218
asked Jul 24 at 12:10
user9995331
1114
add a comment |Â
up vote
0
down vote
favorite
The equations are:
$ y * z = 8$
$x + z = 8$
$x + y = 6$
Solving for these gets me to:
$z = frac167, y = frac72$ and $x = 8 - frac167$.
This is incorrect however. Putting it into a solver yields no solution.
Why can't this be solved?
algebra-precalculus systems-of-equations
edited Jul 24 at 14:14
Jneven
479218
asked Jul 24 at 12:10
user9995331
1114
btw, which solver do you use?
– Siong Thye Goh
Jul 24 at 12:19
They’re not missing components. The coefficients of the “missing†terms are zero.
– amd
Jul 24 at 19:03
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The equations are:
$ y * z = 8$
$x + z = 8$
$x + y = 6$
Solving for these gets me to:
$z = frac167, y = frac72$ and $x = 8 - frac167$.
This is incorrect however. Putting it into a solver yields no solution.
Why can't this be solved?
algebra-precalculus systems-of-equations
edited Jul 24 at 14:14
Jneven
479218
asked Jul 24 at 12:10
user9995331
1114
The equations are:
$ y * z = 8$
$x + z = 8$
$x + y = 6$
Solving for these gets me to:
$z = frac167, y = frac72$ and $x = 8 - frac167$.
This is incorrect however. Putting it into a solver yields no solution.
Why can't this be solved?
algebra-precalculus systems-of-equations
edited Jul 24 at 14:14
Jneven
479218
asked Jul 24 at 12:10
user9995331
1114
edited Jul 24 at 14:14
Jneven
479218
edited Jul 24 at 14:14
Jneven
479218
edited Jul 24 at 14:14
Jneven
479218
479218
asked Jul 24 at 12:10
user9995331
1114
asked Jul 24 at 12:10
user9995331
1114
asked Jul 24 at 12:10
user9995331
1114
1114
btw, which solver do you use?
– Siong Thye Goh
Jul 24 at 12:19
They’re not missing components. The coefficients of the “missing†terms are zero.
– amd
Jul 24 at 19:03
add a comment |Â
btw, which solver do you use?
– Siong Thye Goh
Jul 24 at 12:19
They’re not missing components. The coefficients of the “missing†terms are zero.
– amd
Jul 24 at 19:03
btw, which solver do you use?
– Siong Thye Goh
Jul 24 at 12:19
btw, which solver do you use?
– Siong Thye Goh
Jul 24 at 12:19
They’re not missing components. The coefficients of the “missing†terms are zero.
– amd
Jul 24 at 19:03
They’re not missing components. The coefficients of the “missing†terms are zero.
– amd
Jul 24 at 19:03
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
$$yz=8tag1$$
$$x+z=8 implies z=8-xtag2$$
$$x+y=6 implies y = 6-xtag3$$
substitute $(2)$ and $(3)$ into $(1)$, we get
$$(6-x)(8-x)=8$$
which is a quadratic equation, now you can compute the discriminant to check if there is such real $x$.
- If the discriminant is positive, there are two real solutions for $x$.
- If the dscriminant is zero, there is one real solution for $x$.
- If the dscriminant is negative, there is no solution for $x$.
answered Jul 24 at 12:15
Siong Thye Goh
77.4k134795
add a comment |Â
up vote
4
down vote
$z = 8 - x$ and $y = 6 - x$, then $$yz = (6-x)(8-x) = 8$$
which is
$$x^2 - 14x + 40 = 0$$
The roots of the above equation are:
$$x_1 = 10$$
and
$$x_2 = 4$$
For $x_1 = 10$, we get $y_1 = 6-x_1 = -4$ and $z_1 = 8-10 = -2$.
For $x_2 = 4$, we get $y_2 = 6-x_2 = 2$ and $z_2 = 8-4 = 4$.
Therefore two solutions:
$x_1 = 10$, $y_1 = -4$, and $z_1 = -2$.
$x_2 = 4$, $y_2 = 2$ and $z_2 = 4$.
answered Jul 24 at 12:15
Ahmad Bazzi
2,6271417
@ThePhenotype yes typo mistake .. thank you!!
– Ahmad Bazzi
Jul 24 at 12:27
add a comment |Â
up vote
0
down vote
With $(2)$ and $(3)$ one can deduce that $z$ is two greater than $y$ [e.g. $z = y+2$].
With $(1)$ we have
$$beginalign
yz &= 8\
yz -8&= 0\
y(y+2) -8&= 0\
y^2 +2y -8&= 0\
(y+4)(y-2)&=0\
endalign
$$
which gives a solution of
$$(x,y,z)in lbrace (4, 2, 4), (10, -4, -2)rbrace $$
answered Jul 28 at 6:13
John Joy
5,88511526
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
$$yz=8tag1$$
$$x+z=8 implies z=8-xtag2$$
$$x+y=6 implies y = 6-xtag3$$
substitute $(2)$ and $(3)$ into $(1)$, we get
$$(6-x)(8-x)=8$$
which is a quadratic equation, now you can compute the discriminant to check if there is such real $x$.
- If the discriminant is positive, there are two real solutions for $x$.
- If the dscriminant is zero, there is one real solution for $x$.
- If the dscriminant is negative, there is no solution for $x$.
answered Jul 24 at 12:15
Siong Thye Goh
77.4k134795
add a comment |Â
up vote
6
down vote
$$yz=8tag1$$
$$x+z=8 implies z=8-xtag2$$
$$x+y=6 implies y = 6-xtag3$$
substitute $(2)$ and $(3)$ into $(1)$, we get
$$(6-x)(8-x)=8$$
which is a quadratic equation, now you can compute the discriminant to check if there is such real $x$.
- If the discriminant is positive, there are two real solutions for $x$.
- If the dscriminant is zero, there is one real solution for $x$.
- If the dscriminant is negative, there is no solution for $x$.
answered Jul 24 at 12:15
Siong Thye Goh
77.4k134795
add a comment |Â
up vote
6
down vote
up vote
6
down vote
$$yz=8tag1$$
$$x+z=8 implies z=8-xtag2$$
$$x+y=6 implies y = 6-xtag3$$
substitute $(2)$ and $(3)$ into $(1)$, we get
$$(6-x)(8-x)=8$$
which is a quadratic equation, now you can compute the discriminant to check if there is such real $x$.
- If the discriminant is positive, there are two real solutions for $x$.
- If the dscriminant is zero, there is one real solution for $x$.
- If the dscriminant is negative, there is no solution for $x$.
answered Jul 24 at 12:15
Siong Thye Goh
77.4k134795
$$yz=8tag1$$
$$x+z=8 implies z=8-xtag2$$
$$x+y=6 implies y = 6-xtag3$$
substitute $(2)$ and $(3)$ into $(1)$, we get
$$(6-x)(8-x)=8$$
which is a quadratic equation, now you can compute the discriminant to check if there is such real $x$.
- If the discriminant is positive, there are two real solutions for $x$.
- If the dscriminant is zero, there is one real solution for $x$.
- If the dscriminant is negative, there is no solution for $x$.
answered Jul 24 at 12:15
Siong Thye Goh
77.4k134795
edited Jul 24 at 12:21
answered Jul 24 at 12:15
Siong Thye Goh
77.4k134795
answered Jul 24 at 12:15
Siong Thye Goh
77.4k134795
answered Jul 24 at 12:15
Siong Thye Goh
77.4k134795
77.4k134795
add a comment |Â
add a comment |Â
up vote
4
down vote
$z = 8 - x$ and $y = 6 - x$, then $$yz = (6-x)(8-x) = 8$$
which is
$$x^2 - 14x + 40 = 0$$
The roots of the above equation are:
$$x_1 = 10$$
and
$$x_2 = 4$$
For $x_1 = 10$, we get $y_1 = 6-x_1 = -4$ and $z_1 = 8-10 = -2$.
For $x_2 = 4$, we get $y_2 = 6-x_2 = 2$ and $z_2 = 8-4 = 4$.
Therefore two solutions:
$x_1 = 10$, $y_1 = -4$, and $z_1 = -2$.
$x_2 = 4$, $y_2 = 2$ and $z_2 = 4$.
answered Jul 24 at 12:15
Ahmad Bazzi
2,6271417
@ThePhenotype yes typo mistake .. thank you!!
– Ahmad Bazzi
Jul 24 at 12:27
add a comment |Â
up vote
4
down vote
$z = 8 - x$ and $y = 6 - x$, then $$yz = (6-x)(8-x) = 8$$
which is
$$x^2 - 14x + 40 = 0$$
The roots of the above equation are:
$$x_1 = 10$$
and
$$x_2 = 4$$
For $x_1 = 10$, we get $y_1 = 6-x_1 = -4$ and $z_1 = 8-10 = -2$.
For $x_2 = 4$, we get $y_2 = 6-x_2 = 2$ and $z_2 = 8-4 = 4$.
Therefore two solutions:
$x_1 = 10$, $y_1 = -4$, and $z_1 = -2$.
$x_2 = 4$, $y_2 = 2$ and $z_2 = 4$.
answered Jul 24 at 12:15
Ahmad Bazzi
2,6271417
@ThePhenotype yes typo mistake .. thank you!!
– Ahmad Bazzi
Jul 24 at 12:27
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$z = 8 - x$ and $y = 6 - x$, then $$yz = (6-x)(8-x) = 8$$
which is
$$x^2 - 14x + 40 = 0$$
The roots of the above equation are:
$$x_1 = 10$$
and
$$x_2 = 4$$
For $x_1 = 10$, we get $y_1 = 6-x_1 = -4$ and $z_1 = 8-10 = -2$.
For $x_2 = 4$, we get $y_2 = 6-x_2 = 2$ and $z_2 = 8-4 = 4$.
Therefore two solutions:
$x_1 = 10$, $y_1 = -4$, and $z_1 = -2$.
$x_2 = 4$, $y_2 = 2$ and $z_2 = 4$.
answered Jul 24 at 12:15
Ahmad Bazzi
2,6271417
$z = 8 - x$ and $y = 6 - x$, then $$yz = (6-x)(8-x) = 8$$
which is
$$x^2 - 14x + 40 = 0$$
The roots of the above equation are:
$$x_1 = 10$$
and
$$x_2 = 4$$
For $x_1 = 10$, we get $y_1 = 6-x_1 = -4$ and $z_1 = 8-10 = -2$.
For $x_2 = 4$, we get $y_2 = 6-x_2 = 2$ and $z_2 = 8-4 = 4$.
Therefore two solutions:
$x_1 = 10$, $y_1 = -4$, and $z_1 = -2$.
$x_2 = 4$, $y_2 = 2$ and $z_2 = 4$.
answered Jul 24 at 12:15
Ahmad Bazzi
2,6271417
edited Jul 24 at 12:24
answered Jul 24 at 12:15
Ahmad Bazzi
2,6271417
answered Jul 24 at 12:15
Ahmad Bazzi
2,6271417
answered Jul 24 at 12:15
Ahmad Bazzi
2,6271417
2,6271417
@ThePhenotype yes typo mistake .. thank you!!
– Ahmad Bazzi
Jul 24 at 12:27
add a comment |Â
@ThePhenotype yes typo mistake .. thank you!!
– Ahmad Bazzi
Jul 24 at 12:27
@ThePhenotype yes typo mistake .. thank you!!
– Ahmad Bazzi
Jul 24 at 12:27
@ThePhenotype yes typo mistake .. thank you!!
– Ahmad Bazzi
Jul 24 at 12:27
add a comment |Â
up vote
0
down vote
With $(2)$ and $(3)$ one can deduce that $z$ is two greater than $y$ [e.g. $z = y+2$].
With $(1)$ we have
$$beginalign
yz &= 8\
yz -8&= 0\
y(y+2) -8&= 0\
y^2 +2y -8&= 0\
(y+4)(y-2)&=0\
endalign
$$
which gives a solution of
$$(x,y,z)in lbrace (4, 2, 4), (10, -4, -2)rbrace $$
answered Jul 28 at 6:13
John Joy
5,88511526
add a comment |Â
up vote
0
down vote
With $(2)$ and $(3)$ one can deduce that $z$ is two greater than $y$ [e.g. $z = y+2$].
With $(1)$ we have
$$beginalign
yz &= 8\
yz -8&= 0\
y(y+2) -8&= 0\
y^2 +2y -8&= 0\
(y+4)(y-2)&=0\
endalign
$$
which gives a solution of
$$(x,y,z)in lbrace (4, 2, 4), (10, -4, -2)rbrace $$
answered Jul 28 at 6:13
John Joy
5,88511526
add a comment |Â
up vote
0
down vote
up vote
0
down vote
With $(2)$ and $(3)$ one can deduce that $z$ is two greater than $y$ [e.g. $z = y+2$].
With $(1)$ we have
$$beginalign
yz &= 8\
yz -8&= 0\
y(y+2) -8&= 0\
y^2 +2y -8&= 0\
(y+4)(y-2)&=0\
endalign
$$
which gives a solution of
$$(x,y,z)in lbrace (4, 2, 4), (10, -4, -2)rbrace $$
answered Jul 28 at 6:13
John Joy
5,88511526
With $(2)$ and $(3)$ one can deduce that $z$ is two greater than $y$ [e.g. $z = y+2$].
With $(1)$ we have
$$beginalign
yz &= 8\
yz -8&= 0\
y(y+2) -8&= 0\
y^2 +2y -8&= 0\
(y+4)(y-2)&=0\
endalign
$$
which gives a solution of
$$(x,y,z)in lbrace (4, 2, 4), (10, -4, -2)rbrace $$
answered Jul 28 at 6:13
John Joy
5,88511526
answered Jul 28 at 6:13
John Joy
5,88511526
answered Jul 28 at 6:13
John Joy
5,88511526
answered Jul 28 at 6:13
John Joy
5,88511526
5,88511526
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861264%2fsolving-system-of-equations-with-three-unknowns-but-equations-are-missing-compo%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
btw, which solver do you use?
– Siong Thye Goh
Jul 24 at 12:19
They’re not missing components. The coefficients of the “missing†terms are zero.
– amd
Jul 24 at 19:03