simple proof algebra question on' or'
$begingroup$
If $x^2+5y=y^2+5x$ then $x=y$ or $x+y=5$, where $x$ and $y$ are real numbers . Prove this statement.
Can someone help me with this problem or how to approach it? I can get x=y Does this mean i have proved the statement because it is an 'or'? Thanks
proof-writing
edited Dec 4 '18 at 17:31
user3482749
4,271919
asked Dec 4 '18 at 17:28
HarryHarry
253
$endgroup$
add a comment |
$begingroup$
If $x^2+5y=y^2+5x$ then $x=y$ or $x+y=5$, where $x$ and $y$ are real numbers . Prove this statement.
Can someone help me with this problem or how to approach it? I can get x=y Does this mean i have proved the statement because it is an 'or'? Thanks
proof-writing
edited Dec 4 '18 at 17:31
user3482749
4,271919
asked Dec 4 '18 at 17:28
HarryHarry
253
$endgroup$
$begingroup$
No, you need to prove the complete "or" condition.
$endgroup$
– Yves Daoust
Dec 4 '18 at 17:31
$begingroup$
It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
$endgroup$
– user3482749
Dec 4 '18 at 17:32
add a comment |
$begingroup$
If $x^2+5y=y^2+5x$ then $x=y$ or $x+y=5$, where $x$ and $y$ are real numbers . Prove this statement.
Can someone help me with this problem or how to approach it? I can get x=y Does this mean i have proved the statement because it is an 'or'? Thanks
proof-writing
edited Dec 4 '18 at 17:31
user3482749
4,271919
asked Dec 4 '18 at 17:28
HarryHarry
253
$endgroup$
If $x^2+5y=y^2+5x$ then $x=y$ or $x+y=5$, where $x$ and $y$ are real numbers . Prove this statement.
Can someone help me with this problem or how to approach it? I can get x=y Does this mean i have proved the statement because it is an 'or'? Thanks
proof-writing
proof-writing
edited Dec 4 '18 at 17:31
user3482749
4,271919
asked Dec 4 '18 at 17:28
HarryHarry
253
edited Dec 4 '18 at 17:31
user3482749
4,271919
asked Dec 4 '18 at 17:28
HarryHarry
253
edited Dec 4 '18 at 17:31
user3482749
4,271919
edited Dec 4 '18 at 17:31
user3482749
4,271919
edited Dec 4 '18 at 17:31
user3482749
4,271919
4,271919
asked Dec 4 '18 at 17:28
HarryHarry
253
asked Dec 4 '18 at 17:28
HarryHarry
253
asked Dec 4 '18 at 17:28
HarryHarry
253
253
$begingroup$
No, you need to prove the complete "or" condition.
$endgroup$
– Yves Daoust
Dec 4 '18 at 17:31
$begingroup$
It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
$endgroup$
– user3482749
Dec 4 '18 at 17:32
add a comment |
$begingroup$
No, you need to prove the complete "or" condition.
$endgroup$
– Yves Daoust
Dec 4 '18 at 17:31
$begingroup$
It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
$endgroup$
– user3482749
Dec 4 '18 at 17:32
$begingroup$
No, you need to prove the complete "or" condition.
$endgroup$
– Yves Daoust
Dec 4 '18 at 17:31
$begingroup$
No, you need to prove the complete "or" condition.
$endgroup$
– Yves Daoust
Dec 4 '18 at 17:31
$begingroup$
It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
$endgroup$
– user3482749
Dec 4 '18 at 17:32
$begingroup$
It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
$endgroup$
– user3482749
Dec 4 '18 at 17:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
\iff(x+y-5)(x-y)=0.$$
Now a product is zero when either factor is zero, that means
$$x+y-5=0text{ or }x-y=0.$$
answered Dec 4 '18 at 17:33
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
$begingroup$
Hint
$$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$
answered Dec 4 '18 at 17:32
mflmfl
26.3k12141
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
$$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
\iff(x+y-5)(x-y)=0.$$
Now a product is zero when either factor is zero, that means
$$x+y-5=0text{ or }x-y=0.$$
answered Dec 4 '18 at 17:33
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
$begingroup$
$$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
\iff(x+y-5)(x-y)=0.$$
Now a product is zero when either factor is zero, that means
$$x+y-5=0text{ or }x-y=0.$$
answered Dec 4 '18 at 17:33
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
$begingroup$
$$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
\iff(x+y-5)(x-y)=0.$$
Now a product is zero when either factor is zero, that means
$$x+y-5=0text{ or }x-y=0.$$
answered Dec 4 '18 at 17:33
Yves DaoustYves Daoust
126k672226
$endgroup$
$$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
\iff(x+y-5)(x-y)=0.$$
Now a product is zero when either factor is zero, that means
$$x+y-5=0text{ or }x-y=0.$$
answered Dec 4 '18 at 17:33
Yves DaoustYves Daoust
126k672226
edited Dec 4 '18 at 18:07
answered Dec 4 '18 at 17:33
Yves DaoustYves Daoust
126k672226
answered Dec 4 '18 at 17:33
Yves DaoustYves Daoust
126k672226
answered Dec 4 '18 at 17:33
Yves DaoustYves Daoust
126k672226
126k672226
add a comment |
add a comment |
$begingroup$
Hint
$$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$
answered Dec 4 '18 at 17:32
mflmfl
26.3k12141
$endgroup$
add a comment |
$begingroup$
Hint
$$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$
answered Dec 4 '18 at 17:32
mflmfl
26.3k12141
$endgroup$
add a comment |
$begingroup$
Hint
$$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$
answered Dec 4 '18 at 17:32
mflmfl
26.3k12141
$endgroup$
Hint
$$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$
answered Dec 4 '18 at 17:32
mflmfl
26.3k12141
answered Dec 4 '18 at 17:32
mflmfl
26.3k12141
answered Dec 4 '18 at 17:32
mflmfl
26.3k12141
answered Dec 4 '18 at 17:32
mflmfl
26.3k12141
26.3k12141
add a comment |
add a comment |
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$begingroup$
No, you need to prove the complete "or" condition.
$endgroup$
– Yves Daoust
Dec 4 '18 at 17:31
$begingroup$
It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
$endgroup$
– user3482749
Dec 4 '18 at 17:32