How to solve these simultaneous equation?
$begingroup$
I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows
begin{array}{r l}
bc+3d+15a-db-15c-3b=60 & (1) \
d-15c=0 & (2) \
3a-b= -6 & (3) \
b-d-c+a=0 & (4)
end{array}
Just need the values and how to to them.. not case sensitive just in case
Thanks
algebra-precalculus systems-of-equations
edited Jan 15 at 15:57
Dylan
14.2k31127
asked Jan 15 at 14:33
user10276651user10276651
1
$endgroup$
add a comment |
$begingroup$
I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows
begin{array}{r l}
bc+3d+15a-db-15c-3b=60 & (1) \
d-15c=0 & (2) \
3a-b= -6 & (3) \
b-d-c+a=0 & (4)
end{array}
Just need the values and how to to them.. not case sensitive just in case
Thanks
algebra-precalculus systems-of-equations
edited Jan 15 at 15:57
Dylan
14.2k31127
asked Jan 15 at 14:33
user10276651user10276651
1
$endgroup$
$begingroup$
Where did you “stumble” upon this system of equations?
$endgroup$
– amd
Jan 15 at 20:44
add a comment |
$begingroup$
I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows
begin{array}{r l}
bc+3d+15a-db-15c-3b=60 & (1) \
d-15c=0 & (2) \
3a-b= -6 & (3) \
b-d-c+a=0 & (4)
end{array}
Just need the values and how to to them.. not case sensitive just in case
Thanks
algebra-precalculus systems-of-equations
edited Jan 15 at 15:57
Dylan
14.2k31127
asked Jan 15 at 14:33
user10276651user10276651
1
$endgroup$
I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows
begin{array}{r l}
bc+3d+15a-db-15c-3b=60 & (1) \
d-15c=0 & (2) \
3a-b= -6 & (3) \
b-d-c+a=0 & (4)
end{array}
Just need the values and how to to them.. not case sensitive just in case
Thanks
algebra-precalculus systems-of-equations
algebra-precalculus systems-of-equations
edited Jan 15 at 15:57
Dylan
14.2k31127
asked Jan 15 at 14:33
user10276651user10276651
1
edited Jan 15 at 15:57
Dylan
14.2k31127
asked Jan 15 at 14:33
user10276651user10276651
1
edited Jan 15 at 15:57
Dylan
14.2k31127
edited Jan 15 at 15:57
Dylan
14.2k31127
edited Jan 15 at 15:57
Dylan
14.2k31127
14.2k31127
asked Jan 15 at 14:33
user10276651user10276651
1
asked Jan 15 at 14:33
user10276651user10276651
1
asked Jan 15 at 14:33
user10276651user10276651
1
1
$begingroup$
Where did you “stumble” upon this system of equations?
$endgroup$
– amd
Jan 15 at 20:44
add a comment |
$begingroup$
Where did you “stumble” upon this system of equations?
$endgroup$
– amd
Jan 15 at 20:44
$begingroup$
Where did you “stumble” upon this system of equations?
$endgroup$
– amd
Jan 15 at 20:44
$begingroup$
Where did you “stumble” upon this system of equations?
$endgroup$
– amd
Jan 15 at 20:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
From $(2)$,
$$d=15c ldots(5)$$
Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
$$3a+6-15c-c+a=0$$
$$4a-16c+6=0$$
$$a = 4c-frac32 ldots(6)$$
So$$begin{align}
b&=3left(4c-frac32right)+6\
&=12c+frac32 ldots(7)
end{align}$$
Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.
Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.
answered Jan 15 at 15:47
Faiq IrfanFaiq Irfan
831317
$endgroup$
add a comment |
$begingroup$
Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.
answered Jan 15 at 14:42
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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$begingroup$
From $(2)$,
$$d=15c ldots(5)$$
Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
$$3a+6-15c-c+a=0$$
$$4a-16c+6=0$$
$$a = 4c-frac32 ldots(6)$$
So$$begin{align}
b&=3left(4c-frac32right)+6\
&=12c+frac32 ldots(7)
end{align}$$
Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.
Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.
answered Jan 15 at 15:47
Faiq IrfanFaiq Irfan
831317
$endgroup$
add a comment |
$begingroup$
From $(2)$,
$$d=15c ldots(5)$$
Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
$$3a+6-15c-c+a=0$$
$$4a-16c+6=0$$
$$a = 4c-frac32 ldots(6)$$
So$$begin{align}
b&=3left(4c-frac32right)+6\
&=12c+frac32 ldots(7)
end{align}$$
Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.
Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.
answered Jan 15 at 15:47
Faiq IrfanFaiq Irfan
831317
$endgroup$
add a comment |
$begingroup$
From $(2)$,
$$d=15c ldots(5)$$
Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
$$3a+6-15c-c+a=0$$
$$4a-16c+6=0$$
$$a = 4c-frac32 ldots(6)$$
So$$begin{align}
b&=3left(4c-frac32right)+6\
&=12c+frac32 ldots(7)
end{align}$$
Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.
Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.
answered Jan 15 at 15:47
Faiq IrfanFaiq Irfan
831317
$endgroup$
From $(2)$,
$$d=15c ldots(5)$$
Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
$$3a+6-15c-c+a=0$$
$$4a-16c+6=0$$
$$a = 4c-frac32 ldots(6)$$
So$$begin{align}
b&=3left(4c-frac32right)+6\
&=12c+frac32 ldots(7)
end{align}$$
Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.
Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.
answered Jan 15 at 15:47
Faiq IrfanFaiq Irfan
831317
answered Jan 15 at 15:47
Faiq IrfanFaiq Irfan
831317
answered Jan 15 at 15:47
Faiq IrfanFaiq Irfan
831317
answered Jan 15 at 15:47
Faiq IrfanFaiq Irfan
831317
831317
add a comment |
add a comment |
$begingroup$
Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.
answered Jan 15 at 14:42
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
$begingroup$
Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.
answered Jan 15 at 14:42
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
$begingroup$
Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.
answered Jan 15 at 14:42
Robert IsraelRobert Israel
330k23218473
$endgroup$
Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.
answered Jan 15 at 14:42
Robert IsraelRobert Israel
330k23218473
answered Jan 15 at 14:42
Robert IsraelRobert Israel
330k23218473
answered Jan 15 at 14:42
Robert IsraelRobert Israel
330k23218473
answered Jan 15 at 14:42
Robert IsraelRobert Israel
330k23218473
330k23218473
add a comment |
add a comment |
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$begingroup$
Where did you “stumble” upon this system of equations?
$endgroup$
– amd
Jan 15 at 20:44