Vessel Transfer
A vessel contains $x$ gallons of wine and another contains $y$ gallons of water. From each vessel, $z$ gallons are taken out and transferred to the other. From the resulting mixture in each vessel, $z$ gallons are again taken out and transferred to the other. If after the second transfer, the quantity of wine in each vessel remains the same as it was after the first transfer, then show that $z(x+y)=xy$.
I have been trying to solve this question for many many days and would like a solid proof for it!
algebra-precalculus arithmetic
edited Dec 11 '18 at 22:32
platty
3,370320
asked Dec 11 '18 at 22:18
Pratik Apshinge
395
add a comment |
A vessel contains $x$ gallons of wine and another contains $y$ gallons of water. From each vessel, $z$ gallons are taken out and transferred to the other. From the resulting mixture in each vessel, $z$ gallons are again taken out and transferred to the other. If after the second transfer, the quantity of wine in each vessel remains the same as it was after the first transfer, then show that $z(x+y)=xy$.
I have been trying to solve this question for many many days and would like a solid proof for it!
algebra-precalculus arithmetic
edited Dec 11 '18 at 22:32
platty
3,370320
asked Dec 11 '18 at 22:18
Pratik Apshinge
395
add a comment |
A vessel contains $x$ gallons of wine and another contains $y$ gallons of water. From each vessel, $z$ gallons are taken out and transferred to the other. From the resulting mixture in each vessel, $z$ gallons are again taken out and transferred to the other. If after the second transfer, the quantity of wine in each vessel remains the same as it was after the first transfer, then show that $z(x+y)=xy$.
I have been trying to solve this question for many many days and would like a solid proof for it!
algebra-precalculus arithmetic
edited Dec 11 '18 at 22:32
platty
3,370320
asked Dec 11 '18 at 22:18
Pratik Apshinge
395
A vessel contains $x$ gallons of wine and another contains $y$ gallons of water. From each vessel, $z$ gallons are taken out and transferred to the other. From the resulting mixture in each vessel, $z$ gallons are again taken out and transferred to the other. If after the second transfer, the quantity of wine in each vessel remains the same as it was after the first transfer, then show that $z(x+y)=xy$.
I have been trying to solve this question for many many days and would like a solid proof for it!
algebra-precalculus arithmetic
algebra-precalculus arithmetic
edited Dec 11 '18 at 22:32
platty
3,370320
asked Dec 11 '18 at 22:18
Pratik Apshinge
395
edited Dec 11 '18 at 22:32
platty
3,370320
asked Dec 11 '18 at 22:18
Pratik Apshinge
395
edited Dec 11 '18 at 22:32
platty
3,370320
edited Dec 11 '18 at 22:32
platty
3,370320
edited Dec 11 '18 at 22:32
platty
3,370320
3,370320
asked Dec 11 '18 at 22:18
Pratik Apshinge
395
asked Dec 11 '18 at 22:18
Pratik Apshinge
395
asked Dec 11 '18 at 22:18
Pratik Apshinge
395
395
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
After the first transfer, the first vessel contains $x-z$ gallons of wine and $z$ gallons of water; the second contains $z$ gallons of wine and $y-z$ gallons of water.
The second transfer takes $z$ gallons of the first mixture and adds it to the second, and vice versa. We can write the resulting amount of wine in the first vessel as:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z)
$$
where the first term is the wine left from after the first transfer, and the second term is the new wine from the second transfer. From the problem, we may write:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z) = x-z
$$
From which we just need a bit of manipulation to produce:
$$
begin{align*}
(x-z)^2 y + z^2x &= (x-z)xy \
x^2y - 2xyz + yz^2 + xz^2 &= x^2y - xyz \
yz^2 + xz^2 &= xyz \
(x+y)(z)(z) &= (xy)z
end{align*}
$$
Which, assuming $z neq 0$, gives you the desired equality.
answered Dec 11 '18 at 22:31
platty
3,370320
Thank you so much!!
– Pratik Apshinge
Dec 11 '18 at 22:42
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
After the first transfer, the first vessel contains $x-z$ gallons of wine and $z$ gallons of water; the second contains $z$ gallons of wine and $y-z$ gallons of water.
The second transfer takes $z$ gallons of the first mixture and adds it to the second, and vice versa. We can write the resulting amount of wine in the first vessel as:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z)
$$
where the first term is the wine left from after the first transfer, and the second term is the new wine from the second transfer. From the problem, we may write:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z) = x-z
$$
From which we just need a bit of manipulation to produce:
$$
begin{align*}
(x-z)^2 y + z^2x &= (x-z)xy \
x^2y - 2xyz + yz^2 + xz^2 &= x^2y - xyz \
yz^2 + xz^2 &= xyz \
(x+y)(z)(z) &= (xy)z
end{align*}
$$
Which, assuming $z neq 0$, gives you the desired equality.
answered Dec 11 '18 at 22:31
platty
3,370320
Thank you so much!!
– Pratik Apshinge
Dec 11 '18 at 22:42
add a comment |
After the first transfer, the first vessel contains $x-z$ gallons of wine and $z$ gallons of water; the second contains $z$ gallons of wine and $y-z$ gallons of water.
The second transfer takes $z$ gallons of the first mixture and adds it to the second, and vice versa. We can write the resulting amount of wine in the first vessel as:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z)
$$
where the first term is the wine left from after the first transfer, and the second term is the new wine from the second transfer. From the problem, we may write:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z) = x-z
$$
From which we just need a bit of manipulation to produce:
$$
begin{align*}
(x-z)^2 y + z^2x &= (x-z)xy \
x^2y - 2xyz + yz^2 + xz^2 &= x^2y - xyz \
yz^2 + xz^2 &= xyz \
(x+y)(z)(z) &= (xy)z
end{align*}
$$
Which, assuming $z neq 0$, gives you the desired equality.
answered Dec 11 '18 at 22:31
platty
3,370320
Thank you so much!!
– Pratik Apshinge
Dec 11 '18 at 22:42
add a comment |
After the first transfer, the first vessel contains $x-z$ gallons of wine and $z$ gallons of water; the second contains $z$ gallons of wine and $y-z$ gallons of water.
The second transfer takes $z$ gallons of the first mixture and adds it to the second, and vice versa. We can write the resulting amount of wine in the first vessel as:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z)
$$
where the first term is the wine left from after the first transfer, and the second term is the new wine from the second transfer. From the problem, we may write:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z) = x-z
$$
From which we just need a bit of manipulation to produce:
$$
begin{align*}
(x-z)^2 y + z^2x &= (x-z)xy \
x^2y - 2xyz + yz^2 + xz^2 &= x^2y - xyz \
yz^2 + xz^2 &= xyz \
(x+y)(z)(z) &= (xy)z
end{align*}
$$
Which, assuming $z neq 0$, gives you the desired equality.
answered Dec 11 '18 at 22:31
platty
3,370320
After the first transfer, the first vessel contains $x-z$ gallons of wine and $z$ gallons of water; the second contains $z$ gallons of wine and $y-z$ gallons of water.
The second transfer takes $z$ gallons of the first mixture and adds it to the second, and vice versa. We can write the resulting amount of wine in the first vessel as:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z)
$$
where the first term is the wine left from after the first transfer, and the second term is the new wine from the second transfer. From the problem, we may write:
$$
frac{x-z}{x} (x-z) + frac{z}{y}(z) = x-z
$$
From which we just need a bit of manipulation to produce:
$$
begin{align*}
(x-z)^2 y + z^2x &= (x-z)xy \
x^2y - 2xyz + yz^2 + xz^2 &= x^2y - xyz \
yz^2 + xz^2 &= xyz \
(x+y)(z)(z) &= (xy)z
end{align*}
$$
Which, assuming $z neq 0$, gives you the desired equality.
answered Dec 11 '18 at 22:31
platty
3,370320
answered Dec 11 '18 at 22:31
platty
3,370320
answered Dec 11 '18 at 22:31
platty
3,370320
answered Dec 11 '18 at 22:31
platty
3,370320
3,370320
Thank you so much!!
– Pratik Apshinge
Dec 11 '18 at 22:42
add a comment |
Thank you so much!!
– Pratik Apshinge
Dec 11 '18 at 22:42
Thank you so much!!
– Pratik Apshinge
Dec 11 '18 at 22:42
Thank you so much!!
– Pratik Apshinge
Dec 11 '18 at 22:42
add a comment |
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