Mixing
If $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y} $ find the ratio of $x$, $y$ and $ z$
$begingroup$
I have this question from higher algebra by Hall and Knight:
if $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $frac x4 =frac y2 =frac z3$ and the second is $frac x1 =frac y{-1} =frac z0 $. Now I solve this question in the following manner:
Adding numerator and denominator gives $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y}= 2frac{x+y}{x+y} =2$ (when x+y is not zero).This gives the first answer.
When (x+y ) is zero => $ y= -x $ that is $frac xy = -1$ and now there are two things
(i)If I put these values in the original expression I get $ frac{y}{x-z} =frac 0z= -1$ this implies 0= 1 where am I making the mistake?
(ii)Also from the original expression I have$ frac{x+y}{z} = frac{x}{y}$ multiplying by z I get $ x+y =zfrac xy$. This implies that z=0 and x:y:z =x:-x:0 = 1:-1:0 and this gives the second answer.But the problem is how z can be zero when it appears in the denominator in the expression
I think these things are very basic but still I am stuck.Could anyone please help me in knowing where the mistake is?
algebra-precalculus ratio
asked Nov 14 '15 at 5:49
Sanjeev VermaSanjeev Verma
63
$endgroup$
add a comment |
$begingroup$
I have this question from higher algebra by Hall and Knight:
if $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $frac x4 =frac y2 =frac z3$ and the second is $frac x1 =frac y{-1} =frac z0 $. Now I solve this question in the following manner:
Adding numerator and denominator gives $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y}= 2frac{x+y}{x+y} =2$ (when x+y is not zero).This gives the first answer.
When (x+y ) is zero => $ y= -x $ that is $frac xy = -1$ and now there are two things
(i)If I put these values in the original expression I get $ frac{y}{x-z} =frac 0z= -1$ this implies 0= 1 where am I making the mistake?
(ii)Also from the original expression I have$ frac{x+y}{z} = frac{x}{y}$ multiplying by z I get $ x+y =zfrac xy$. This implies that z=0 and x:y:z =x:-x:0 = 1:-1:0 and this gives the second answer.But the problem is how z can be zero when it appears in the denominator in the expression
I think these things are very basic but still I am stuck.Could anyone please help me in knowing where the mistake is?
algebra-precalculus ratio
asked Nov 14 '15 at 5:49
Sanjeev VermaSanjeev Verma
63
$endgroup$
add a comment |
$begingroup$
I have this question from higher algebra by Hall and Knight:
if $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $frac x4 =frac y2 =frac z3$ and the second is $frac x1 =frac y{-1} =frac z0 $. Now I solve this question in the following manner:
Adding numerator and denominator gives $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y}= 2frac{x+y}{x+y} =2$ (when x+y is not zero).This gives the first answer.
When (x+y ) is zero => $ y= -x $ that is $frac xy = -1$ and now there are two things
(i)If I put these values in the original expression I get $ frac{y}{x-z} =frac 0z= -1$ this implies 0= 1 where am I making the mistake?
(ii)Also from the original expression I have$ frac{x+y}{z} = frac{x}{y}$ multiplying by z I get $ x+y =zfrac xy$. This implies that z=0 and x:y:z =x:-x:0 = 1:-1:0 and this gives the second answer.But the problem is how z can be zero when it appears in the denominator in the expression
I think these things are very basic but still I am stuck.Could anyone please help me in knowing where the mistake is?
algebra-precalculus ratio
asked Nov 14 '15 at 5:49
Sanjeev VermaSanjeev Verma
63
$endgroup$
I have this question from higher algebra by Hall and Knight:
if $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $frac x4 =frac y2 =frac z3$ and the second is $frac x1 =frac y{-1} =frac z0 $. Now I solve this question in the following manner:
Adding numerator and denominator gives $ frac{y}{x-z} = frac{x+y}{z} = frac{x}{y}= 2frac{x+y}{x+y} =2$ (when x+y is not zero).This gives the first answer.
When (x+y ) is zero => $ y= -x $ that is $frac xy = -1$ and now there are two things
(i)If I put these values in the original expression I get $ frac{y}{x-z} =frac 0z= -1$ this implies 0= 1 where am I making the mistake?
(ii)Also from the original expression I have$ frac{x+y}{z} = frac{x}{y}$ multiplying by z I get $ x+y =zfrac xy$. This implies that z=0 and x:y:z =x:-x:0 = 1:-1:0 and this gives the second answer.But the problem is how z can be zero when it appears in the denominator in the expression
I think these things are very basic but still I am stuck.Could anyone please help me in knowing where the mistake is?
algebra-precalculus ratio
algebra-precalculus ratio
asked Nov 14 '15 at 5:49
Sanjeev VermaSanjeev Verma
63
asked Nov 14 '15 at 5:49
Sanjeev VermaSanjeev Verma
63
edited Nov 14 '15 at 5:51
user147263
asked Nov 14 '15 at 5:49
Sanjeev VermaSanjeev Verma
63
asked Nov 14 '15 at 5:49
Sanjeev VermaSanjeev Verma
63
asked Nov 14 '15 at 5:49
Sanjeev VermaSanjeev Verma
63
63
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I too am concerned about an answer of $frac{z}{0}$ being listed. That is a poor way of writing that $z$ is not related to $x$ and $y$ (if that is actually what it means).
What the second answer should say is "If both $x$ and $y$ are zero then $z$ can be any value." Asking this as a ratio question makes it very awkward.
answered Nov 14 '15 at 5:58
Ian MillerIan Miller
10.5k11438
$endgroup$
$begingroup$
Yes Miller i agree but i am worried because it is from a very reputed book
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:05
$begingroup$
I think answer should be "if x+y is zero then z can be any value"
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:36
$begingroup$
The more I think about it the more I feel that second answer is nonsense. If $x+y=0$ from the first equality then $y=0$ and by the second $x=0$ but then that makes the third fraction undefined.
$endgroup$
– Ian Miller
Nov 14 '15 at 8:27
add a comment |
$begingroup$
y/(x-z)=(y+x)/z=x/y = be k
y= kx - kz ..... 1
y+x = kz ...... 2
x=yk ................3
Add 1 & 2 , 2y+ x = kx , substitute 3 , 2y+ yk = k *yk
2 +k = k² , k²- k - 2 = 0 , k= 2 or k = -1
for k = 2 , x = 2y , substitute in 2 , 3y = 2z
x:y:z = 2y:y:3y/2 = 4:2: 3
answered Jun 16 '17 at 6:10
user455479user455479
1
$endgroup$
$begingroup$
Please use MathJax in the future when posting answers. As it stands this looks very confusing and difficult to comprehend.
$endgroup$
– Rumplestillskin
Jun 16 '17 at 6:42
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1528191%2fif-fracyx-z-fracxyz-fracxy-find-the-ratio-of-x-y-an%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I too am concerned about an answer of $frac{z}{0}$ being listed. That is a poor way of writing that $z$ is not related to $x$ and $y$ (if that is actually what it means).
What the second answer should say is "If both $x$ and $y$ are zero then $z$ can be any value." Asking this as a ratio question makes it very awkward.
answered Nov 14 '15 at 5:58
Ian MillerIan Miller
10.5k11438
$endgroup$
$begingroup$
Yes Miller i agree but i am worried because it is from a very reputed book
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:05
$begingroup$
I think answer should be "if x+y is zero then z can be any value"
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:36
$begingroup$
The more I think about it the more I feel that second answer is nonsense. If $x+y=0$ from the first equality then $y=0$ and by the second $x=0$ but then that makes the third fraction undefined.
$endgroup$
– Ian Miller
Nov 14 '15 at 8:27
add a comment |
$begingroup$
I too am concerned about an answer of $frac{z}{0}$ being listed. That is a poor way of writing that $z$ is not related to $x$ and $y$ (if that is actually what it means).
What the second answer should say is "If both $x$ and $y$ are zero then $z$ can be any value." Asking this as a ratio question makes it very awkward.
answered Nov 14 '15 at 5:58
Ian MillerIan Miller
10.5k11438
$endgroup$
$begingroup$
Yes Miller i agree but i am worried because it is from a very reputed book
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:05
$begingroup$
I think answer should be "if x+y is zero then z can be any value"
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:36
$begingroup$
The more I think about it the more I feel that second answer is nonsense. If $x+y=0$ from the first equality then $y=0$ and by the second $x=0$ but then that makes the third fraction undefined.
$endgroup$
– Ian Miller
Nov 14 '15 at 8:27
add a comment |
$begingroup$
I too am concerned about an answer of $frac{z}{0}$ being listed. That is a poor way of writing that $z$ is not related to $x$ and $y$ (if that is actually what it means).
What the second answer should say is "If both $x$ and $y$ are zero then $z$ can be any value." Asking this as a ratio question makes it very awkward.
answered Nov 14 '15 at 5:58
Ian MillerIan Miller
10.5k11438
$endgroup$
I too am concerned about an answer of $frac{z}{0}$ being listed. That is a poor way of writing that $z$ is not related to $x$ and $y$ (if that is actually what it means).
What the second answer should say is "If both $x$ and $y$ are zero then $z$ can be any value." Asking this as a ratio question makes it very awkward.
answered Nov 14 '15 at 5:58
Ian MillerIan Miller
10.5k11438
answered Nov 14 '15 at 5:58
Ian MillerIan Miller
10.5k11438
answered Nov 14 '15 at 5:58
Ian MillerIan Miller
10.5k11438
answered Nov 14 '15 at 5:58
Ian MillerIan Miller
10.5k11438
10.5k11438
$begingroup$
Yes Miller i agree but i am worried because it is from a very reputed book
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:05
$begingroup$
I think answer should be "if x+y is zero then z can be any value"
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:36
$begingroup$
The more I think about it the more I feel that second answer is nonsense. If $x+y=0$ from the first equality then $y=0$ and by the second $x=0$ but then that makes the third fraction undefined.
$endgroup$
– Ian Miller
Nov 14 '15 at 8:27
add a comment |
$begingroup$
Yes Miller i agree but i am worried because it is from a very reputed book
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:05
$begingroup$
I think answer should be "if x+y is zero then z can be any value"
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:36
$begingroup$
The more I think about it the more I feel that second answer is nonsense. If $x+y=0$ from the first equality then $y=0$ and by the second $x=0$ but then that makes the third fraction undefined.
$endgroup$
– Ian Miller
Nov 14 '15 at 8:27
$begingroup$
Yes Miller i agree but i am worried because it is from a very reputed book
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:05
$begingroup$
Yes Miller i agree but i am worried because it is from a very reputed book
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:05
$begingroup$
I think answer should be "if x+y is zero then z can be any value"
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:36
$begingroup$
I think answer should be "if x+y is zero then z can be any value"
$endgroup$
– Sanjeev Verma
Nov 14 '15 at 6:36
$begingroup$
The more I think about it the more I feel that second answer is nonsense. If $x+y=0$ from the first equality then $y=0$ and by the second $x=0$ but then that makes the third fraction undefined.
$endgroup$
– Ian Miller
Nov 14 '15 at 8:27
$begingroup$
The more I think about it the more I feel that second answer is nonsense. If $x+y=0$ from the first equality then $y=0$ and by the second $x=0$ but then that makes the third fraction undefined.
$endgroup$
– Ian Miller
Nov 14 '15 at 8:27
add a comment |
$begingroup$
y/(x-z)=(y+x)/z=x/y = be k
y= kx - kz ..... 1
y+x = kz ...... 2
x=yk ................3
Add 1 & 2 , 2y+ x = kx , substitute 3 , 2y+ yk = k *yk
2 +k = k² , k²- k - 2 = 0 , k= 2 or k = -1
for k = 2 , x = 2y , substitute in 2 , 3y = 2z
x:y:z = 2y:y:3y/2 = 4:2: 3
answered Jun 16 '17 at 6:10
user455479user455479
1
$endgroup$
$begingroup$
Please use MathJax in the future when posting answers. As it stands this looks very confusing and difficult to comprehend.
$endgroup$
– Rumplestillskin
Jun 16 '17 at 6:42
add a comment |
$begingroup$
y/(x-z)=(y+x)/z=x/y = be k
y= kx - kz ..... 1
y+x = kz ...... 2
x=yk ................3
Add 1 & 2 , 2y+ x = kx , substitute 3 , 2y+ yk = k *yk
2 +k = k² , k²- k - 2 = 0 , k= 2 or k = -1
for k = 2 , x = 2y , substitute in 2 , 3y = 2z
x:y:z = 2y:y:3y/2 = 4:2: 3
answered Jun 16 '17 at 6:10
user455479user455479
1
$endgroup$
$begingroup$
Please use MathJax in the future when posting answers. As it stands this looks very confusing and difficult to comprehend.
$endgroup$
– Rumplestillskin
Jun 16 '17 at 6:42
add a comment |
$begingroup$
y/(x-z)=(y+x)/z=x/y = be k
y= kx - kz ..... 1
y+x = kz ...... 2
x=yk ................3
Add 1 & 2 , 2y+ x = kx , substitute 3 , 2y+ yk = k *yk
2 +k = k² , k²- k - 2 = 0 , k= 2 or k = -1
for k = 2 , x = 2y , substitute in 2 , 3y = 2z
x:y:z = 2y:y:3y/2 = 4:2: 3
answered Jun 16 '17 at 6:10
user455479user455479
1
$endgroup$
y/(x-z)=(y+x)/z=x/y = be k
y= kx - kz ..... 1
y+x = kz ...... 2
x=yk ................3
Add 1 & 2 , 2y+ x = kx , substitute 3 , 2y+ yk = k *yk
2 +k = k² , k²- k - 2 = 0 , k= 2 or k = -1
for k = 2 , x = 2y , substitute in 2 , 3y = 2z
x:y:z = 2y:y:3y/2 = 4:2: 3
answered Jun 16 '17 at 6:10
user455479user455479
1
answered Jun 16 '17 at 6:10
user455479user455479
1
answered Jun 16 '17 at 6:10
user455479user455479
1
answered Jun 16 '17 at 6:10
user455479user455479
1
1
$begingroup$
Please use MathJax in the future when posting answers. As it stands this looks very confusing and difficult to comprehend.
$endgroup$
– Rumplestillskin
Jun 16 '17 at 6:42
add a comment |
$begingroup$
Please use MathJax in the future when posting answers. As it stands this looks very confusing and difficult to comprehend.
$endgroup$
– Rumplestillskin
Jun 16 '17 at 6:42
$begingroup$
Please use MathJax in the future when posting answers. As it stands this looks very confusing and difficult to comprehend.
$endgroup$
– Rumplestillskin
Jun 16 '17 at 6:42
$begingroup$
Please use MathJax in the future when posting answers. As it stands this looks very confusing and difficult to comprehend.
$endgroup$
– Rumplestillskin
Jun 16 '17 at 6:42
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1528191%2fif-fracyx-z-fracxyz-fracxy-find-the-ratio-of-x-y-an%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
