Solve the system of equations in the set of real numbers.
Solve the system of equations in the set of real numbers:
$$begin{cases}
frac1x + frac1{y+z} = frac13 \
frac1y + frac1{x+z} = frac15 \
frac1z + frac1{x+y} = frac17
end{cases}$$
I got:
$$begin{cases}
3(x+y+z)=x(y+z) \
5(x+y+z)=y(x+z) \
7(x+y+z)=z(x+y)
end{cases}$$
However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).
How can I solve this problem or how should I approach it?
algebra-precalculus systems-of-equations
asked Dec 18 '18 at 22:35
Pero
1307
add a comment |
Solve the system of equations in the set of real numbers:
$$begin{cases}
frac1x + frac1{y+z} = frac13 \
frac1y + frac1{x+z} = frac15 \
frac1z + frac1{x+y} = frac17
end{cases}$$
I got:
$$begin{cases}
3(x+y+z)=x(y+z) \
5(x+y+z)=y(x+z) \
7(x+y+z)=z(x+y)
end{cases}$$
However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).
How can I solve this problem or how should I approach it?
algebra-precalculus systems-of-equations
asked Dec 18 '18 at 22:35
Pero
1307
1
To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
– Card_Trick
Dec 18 '18 at 22:40
1
@Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
– Don Thousand
Dec 18 '18 at 22:41
1
I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
– Card_Trick
Dec 18 '18 at 22:42
@Card_Trick Pero is concluding that the $0$ solution is extraneous.
– The Great Duck
Dec 19 '18 at 5:09
add a comment |
Solve the system of equations in the set of real numbers:
$$begin{cases}
frac1x + frac1{y+z} = frac13 \
frac1y + frac1{x+z} = frac15 \
frac1z + frac1{x+y} = frac17
end{cases}$$
I got:
$$begin{cases}
3(x+y+z)=x(y+z) \
5(x+y+z)=y(x+z) \
7(x+y+z)=z(x+y)
end{cases}$$
However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).
How can I solve this problem or how should I approach it?
algebra-precalculus systems-of-equations
asked Dec 18 '18 at 22:35
Pero
1307
Solve the system of equations in the set of real numbers:
$$begin{cases}
frac1x + frac1{y+z} = frac13 \
frac1y + frac1{x+z} = frac15 \
frac1z + frac1{x+y} = frac17
end{cases}$$
I got:
$$begin{cases}
3(x+y+z)=x(y+z) \
5(x+y+z)=y(x+z) \
7(x+y+z)=z(x+y)
end{cases}$$
However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).
How can I solve this problem or how should I approach it?
algebra-precalculus systems-of-equations
algebra-precalculus systems-of-equations
asked Dec 18 '18 at 22:35
Pero
1307
asked Dec 18 '18 at 22:35
Pero
1307
asked Dec 18 '18 at 22:35
Pero
1307
asked Dec 18 '18 at 22:35
Pero
1307
asked Dec 18 '18 at 22:35
Pero
1307
1307
1
To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
– Card_Trick
Dec 18 '18 at 22:40
1
@Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
– Don Thousand
Dec 18 '18 at 22:41
1
I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
– Card_Trick
Dec 18 '18 at 22:42
@Card_Trick Pero is concluding that the $0$ solution is extraneous.
– The Great Duck
Dec 19 '18 at 5:09
add a comment |
1
To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
– Card_Trick
Dec 18 '18 at 22:40
1
@Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
– Don Thousand
Dec 18 '18 at 22:41
1
I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
– Card_Trick
Dec 18 '18 at 22:42
@Card_Trick Pero is concluding that the $0$ solution is extraneous.
– The Great Duck
Dec 19 '18 at 5:09
1
1
To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
– Card_Trick
Dec 18 '18 at 22:40
To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
– Card_Trick
Dec 18 '18 at 22:40
1
1
@Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
– Don Thousand
Dec 18 '18 at 22:41
@Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
– Don Thousand
Dec 18 '18 at 22:41
1
1
I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
– Card_Trick
Dec 18 '18 at 22:42
I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
– Card_Trick
Dec 18 '18 at 22:42
@Card_Trick Pero is concluding that the $0$ solution is extraneous.
– The Great Duck
Dec 19 '18 at 5:09
@Card_Trick Pero is concluding that the $0$ solution is extraneous.
– The Great Duck
Dec 19 '18 at 5:09
add a comment |
2 Answers
2
active
oldest
votes
Let $x+y+z=m$
Adding all the equations, we get,
$xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$
Subtracting each equation one by one from this, we get,
$xy=frac{m}{2}$
$yz=frac{9m}{2}$
$zx=frac{5m}{2}$
Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
$$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
$$Longrightarrow x:y:z=5:9:45$$
Now, let $x=5k, y=9k, z=45k$ and get the result.
Hope it is helpful
answered Dec 18 '18 at 22:59
Martund
1,405212
add a comment |
We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!
answered Dec 18 '18 at 22:44
Don Thousand
4,277734
How do we get $z=9x, z=5y$?
– Pero
Dec 18 '18 at 22:58
Divide the three equations I've provided from each other.
– Don Thousand
Dec 18 '18 at 23:00
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $x+y+z=m$
Adding all the equations, we get,
$xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$
Subtracting each equation one by one from this, we get,
$xy=frac{m}{2}$
$yz=frac{9m}{2}$
$zx=frac{5m}{2}$
Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
$$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
$$Longrightarrow x:y:z=5:9:45$$
Now, let $x=5k, y=9k, z=45k$ and get the result.
Hope it is helpful
answered Dec 18 '18 at 22:59
Martund
1,405212
add a comment |
Let $x+y+z=m$
Adding all the equations, we get,
$xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$
Subtracting each equation one by one from this, we get,
$xy=frac{m}{2}$
$yz=frac{9m}{2}$
$zx=frac{5m}{2}$
Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
$$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
$$Longrightarrow x:y:z=5:9:45$$
Now, let $x=5k, y=9k, z=45k$ and get the result.
Hope it is helpful
answered Dec 18 '18 at 22:59
Martund
1,405212
add a comment |
Let $x+y+z=m$
Adding all the equations, we get,
$xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$
Subtracting each equation one by one from this, we get,
$xy=frac{m}{2}$
$yz=frac{9m}{2}$
$zx=frac{5m}{2}$
Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
$$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
$$Longrightarrow x:y:z=5:9:45$$
Now, let $x=5k, y=9k, z=45k$ and get the result.
Hope it is helpful
answered Dec 18 '18 at 22:59
Martund
1,405212
Let $x+y+z=m$
Adding all the equations, we get,
$xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$
Subtracting each equation one by one from this, we get,
$xy=frac{m}{2}$
$yz=frac{9m}{2}$
$zx=frac{5m}{2}$
Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
$$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
$$Longrightarrow x:y:z=5:9:45$$
Now, let $x=5k, y=9k, z=45k$ and get the result.
Hope it is helpful
answered Dec 18 '18 at 22:59
Martund
1,405212
answered Dec 18 '18 at 22:59
Martund
1,405212
answered Dec 18 '18 at 22:59
Martund
1,405212
answered Dec 18 '18 at 22:59
Martund
1,405212
1,405212
add a comment |
add a comment |
We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!
answered Dec 18 '18 at 22:44
Don Thousand
4,277734
How do we get $z=9x, z=5y$?
– Pero
Dec 18 '18 at 22:58
Divide the three equations I've provided from each other.
– Don Thousand
Dec 18 '18 at 23:00
add a comment |
We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!
answered Dec 18 '18 at 22:44
Don Thousand
4,277734
How do we get $z=9x, z=5y$?
– Pero
Dec 18 '18 at 22:58
Divide the three equations I've provided from each other.
– Don Thousand
Dec 18 '18 at 23:00
add a comment |
We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!
answered Dec 18 '18 at 22:44
Don Thousand
4,277734
We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!
answered Dec 18 '18 at 22:44
Don Thousand
4,277734
answered Dec 18 '18 at 22:44
Don Thousand
4,277734
answered Dec 18 '18 at 22:44
Don Thousand
4,277734
answered Dec 18 '18 at 22:44
Don Thousand
4,277734
4,277734
How do we get $z=9x, z=5y$?
– Pero
Dec 18 '18 at 22:58
Divide the three equations I've provided from each other.
– Don Thousand
Dec 18 '18 at 23:00
add a comment |
How do we get $z=9x, z=5y$?
– Pero
Dec 18 '18 at 22:58
Divide the three equations I've provided from each other.
– Don Thousand
Dec 18 '18 at 23:00
How do we get $z=9x, z=5y$?
– Pero
Dec 18 '18 at 22:58
How do we get $z=9x, z=5y$?
– Pero
Dec 18 '18 at 22:58
Divide the three equations I've provided from each other.
– Don Thousand
Dec 18 '18 at 23:00
Divide the three equations I've provided from each other.
– Don Thousand
Dec 18 '18 at 23:00
add a comment |
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1
To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
– Card_Trick
Dec 18 '18 at 22:40
1
@Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
– Don Thousand
Dec 18 '18 at 22:41
1
I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
– Card_Trick
Dec 18 '18 at 22:42
@Card_Trick Pero is concluding that the $0$ solution is extraneous.
– The Great Duck
Dec 19 '18 at 5:09