Prove the equation ad+b+c=bc+a+d [closed]
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Let $b,cneq 1$. Prove that if $ac-a-c=b^2-2b$ and $bd-b-d=c^2-2c$ then $ad+b+c=bc+a+d$. Please help me. Thanky very much
algebra-precalculus
asked Dec 4 '18 at 18:18
brisk tutibrisk tuti
13
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closed as off-topic by Did, José Carlos Santos, Jyrki Lahtonen, Alexander Gruber♦ Jan 17 at 23:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, Jyrki Lahtonen, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $b,cneq 1$. Prove that if $ac-a-c=b^2-2b$ and $bd-b-d=c^2-2c$ then $ad+b+c=bc+a+d$. Please help me. Thanky very much
algebra-precalculus
asked Dec 4 '18 at 18:18
brisk tutibrisk tuti
13
$endgroup$
closed as off-topic by Did, José Carlos Santos, Jyrki Lahtonen, Alexander Gruber♦ Jan 17 at 23:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, Jyrki Lahtonen, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Please edit the question to show us what you tried. Hint: the condition $bc ne 1$ suggests that $b-1$ and $c-1$ might be useful.
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– Ethan Bolker
Dec 4 '18 at 18:21
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the example is thus given in book
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– brisk tuti
Dec 4 '18 at 18:25
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Well, try to manipulate the given expressions into something easier to work with. For instance, factoring gets a lot easier if you add $1$ to both sides, no? Try that.
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– lulu
Dec 4 '18 at 18:27
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i took it to find a for the first equation and b for the second equation, then i substitution in the end eqution but nothing gains, i also added 1 in the both sides in the fisrt and the second equation but nothing gainjs else
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– brisk tuti
Dec 4 '18 at 18:31
add a comment |
$begingroup$
Let $b,cneq 1$. Prove that if $ac-a-c=b^2-2b$ and $bd-b-d=c^2-2c$ then $ad+b+c=bc+a+d$. Please help me. Thanky very much
algebra-precalculus
asked Dec 4 '18 at 18:18
brisk tutibrisk tuti
13
$endgroup$
Let $b,cneq 1$. Prove that if $ac-a-c=b^2-2b$ and $bd-b-d=c^2-2c$ then $ad+b+c=bc+a+d$. Please help me. Thanky very much
algebra-precalculus
algebra-precalculus
asked Dec 4 '18 at 18:18
brisk tutibrisk tuti
13
asked Dec 4 '18 at 18:18
brisk tutibrisk tuti
13
asked Dec 4 '18 at 18:18
brisk tutibrisk tuti
13
asked Dec 4 '18 at 18:18
brisk tutibrisk tuti
13
asked Dec 4 '18 at 18:18
brisk tutibrisk tuti
13
13
closed as off-topic by Did, José Carlos Santos, Jyrki Lahtonen, Alexander Gruber♦ Jan 17 at 23:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, Jyrki Lahtonen, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, José Carlos Santos, Jyrki Lahtonen, Alexander Gruber♦ Jan 17 at 23:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, Jyrki Lahtonen, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Please edit the question to show us what you tried. Hint: the condition $bc ne 1$ suggests that $b-1$ and $c-1$ might be useful.
$endgroup$
– Ethan Bolker
Dec 4 '18 at 18:21
$begingroup$
the example is thus given in book
$endgroup$
– brisk tuti
Dec 4 '18 at 18:25
$begingroup$
Well, try to manipulate the given expressions into something easier to work with. For instance, factoring gets a lot easier if you add $1$ to both sides, no? Try that.
$endgroup$
– lulu
Dec 4 '18 at 18:27
$begingroup$
i took it to find a for the first equation and b for the second equation, then i substitution in the end eqution but nothing gains, i also added 1 in the both sides in the fisrt and the second equation but nothing gainjs else
$endgroup$
– brisk tuti
Dec 4 '18 at 18:31
add a comment |
$begingroup$
Please edit the question to show us what you tried. Hint: the condition $bc ne 1$ suggests that $b-1$ and $c-1$ might be useful.
$endgroup$
– Ethan Bolker
Dec 4 '18 at 18:21
$begingroup$
the example is thus given in book
$endgroup$
– brisk tuti
Dec 4 '18 at 18:25
$begingroup$
Well, try to manipulate the given expressions into something easier to work with. For instance, factoring gets a lot easier if you add $1$ to both sides, no? Try that.
$endgroup$
– lulu
Dec 4 '18 at 18:27
$begingroup$
i took it to find a for the first equation and b for the second equation, then i substitution in the end eqution but nothing gains, i also added 1 in the both sides in the fisrt and the second equation but nothing gainjs else
$endgroup$
– brisk tuti
Dec 4 '18 at 18:31
$begingroup$
Please edit the question to show us what you tried. Hint: the condition $bc ne 1$ suggests that $b-1$ and $c-1$ might be useful.
$endgroup$
– Ethan Bolker
Dec 4 '18 at 18:21
$begingroup$
Please edit the question to show us what you tried. Hint: the condition $bc ne 1$ suggests that $b-1$ and $c-1$ might be useful.
$endgroup$
– Ethan Bolker
Dec 4 '18 at 18:21
$begingroup$
the example is thus given in book
$endgroup$
– brisk tuti
Dec 4 '18 at 18:25
$begingroup$
the example is thus given in book
$endgroup$
– brisk tuti
Dec 4 '18 at 18:25
$begingroup$
Well, try to manipulate the given expressions into something easier to work with. For instance, factoring gets a lot easier if you add $1$ to both sides, no? Try that.
$endgroup$
– lulu
Dec 4 '18 at 18:27
$begingroup$
Well, try to manipulate the given expressions into something easier to work with. For instance, factoring gets a lot easier if you add $1$ to both sides, no? Try that.
$endgroup$
– lulu
Dec 4 '18 at 18:27
$begingroup$
i took it to find a for the first equation and b for the second equation, then i substitution in the end eqution but nothing gains, i also added 1 in the both sides in the fisrt and the second equation but nothing gainjs else
$endgroup$
– brisk tuti
Dec 4 '18 at 18:31
$begingroup$
i took it to find a for the first equation and b for the second equation, then i substitution in the end eqution but nothing gains, i also added 1 in the both sides in the fisrt and the second equation but nothing gainjs else
$endgroup$
– brisk tuti
Dec 4 '18 at 18:31
add a comment |
1 Answer
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$ac-a-c=b^2-2b$ then $a(c-1)-c+1=b^2-2b+1$ then $(c-1)(a-1)=(b-1)^2$ similar
$(d-1)(b-1)=(c-1)^2$ now by this relation we have $frac{d-1}{c-1}=frac{c-1}{b-1}=frac{b-1}{a-1}$. Then $(d-1)(a-1)=(b-1)(c-1)$.
$(d-1)(a-1)=(c-1)(b-1)$ then $a(d-1)-d+1=b(c-1)-c+1$ then $ad+b+c=bc+a+d$
answered Dec 4 '18 at 19:01
yavaryavar
993
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$ac-a-c=b^2-2b$ then $a(c-1)-c+1=b^2-2b+1$ then $(c-1)(a-1)=(b-1)^2$ similar
$(d-1)(b-1)=(c-1)^2$ now by this relation we have $frac{d-1}{c-1}=frac{c-1}{b-1}=frac{b-1}{a-1}$. Then $(d-1)(a-1)=(b-1)(c-1)$.
$(d-1)(a-1)=(c-1)(b-1)$ then $a(d-1)-d+1=b(c-1)-c+1$ then $ad+b+c=bc+a+d$
answered Dec 4 '18 at 19:01
yavaryavar
993
$endgroup$
add a comment |
$begingroup$
$ac-a-c=b^2-2b$ then $a(c-1)-c+1=b^2-2b+1$ then $(c-1)(a-1)=(b-1)^2$ similar
$(d-1)(b-1)=(c-1)^2$ now by this relation we have $frac{d-1}{c-1}=frac{c-1}{b-1}=frac{b-1}{a-1}$. Then $(d-1)(a-1)=(b-1)(c-1)$.
$(d-1)(a-1)=(c-1)(b-1)$ then $a(d-1)-d+1=b(c-1)-c+1$ then $ad+b+c=bc+a+d$
answered Dec 4 '18 at 19:01
yavaryavar
993
$endgroup$
add a comment |
$begingroup$
$ac-a-c=b^2-2b$ then $a(c-1)-c+1=b^2-2b+1$ then $(c-1)(a-1)=(b-1)^2$ similar
$(d-1)(b-1)=(c-1)^2$ now by this relation we have $frac{d-1}{c-1}=frac{c-1}{b-1}=frac{b-1}{a-1}$. Then $(d-1)(a-1)=(b-1)(c-1)$.
$(d-1)(a-1)=(c-1)(b-1)$ then $a(d-1)-d+1=b(c-1)-c+1$ then $ad+b+c=bc+a+d$
answered Dec 4 '18 at 19:01
yavaryavar
993
$endgroup$
$ac-a-c=b^2-2b$ then $a(c-1)-c+1=b^2-2b+1$ then $(c-1)(a-1)=(b-1)^2$ similar
$(d-1)(b-1)=(c-1)^2$ now by this relation we have $frac{d-1}{c-1}=frac{c-1}{b-1}=frac{b-1}{a-1}$. Then $(d-1)(a-1)=(b-1)(c-1)$.
$(d-1)(a-1)=(c-1)(b-1)$ then $a(d-1)-d+1=b(c-1)-c+1$ then $ad+b+c=bc+a+d$
answered Dec 4 '18 at 19:01
yavaryavar
993
answered Dec 4 '18 at 19:01
yavaryavar
993
answered Dec 4 '18 at 19:01
yavaryavar
993
answered Dec 4 '18 at 19:01
yavaryavar
993
993
add a comment |
add a comment |
$begingroup$
Please edit the question to show us what you tried. Hint: the condition $bc ne 1$ suggests that $b-1$ and $c-1$ might be useful.
$endgroup$
– Ethan Bolker
Dec 4 '18 at 18:21
$begingroup$
the example is thus given in book
$endgroup$
– brisk tuti
Dec 4 '18 at 18:25
$begingroup$
Well, try to manipulate the given expressions into something easier to work with. For instance, factoring gets a lot easier if you add $1$ to both sides, no? Try that.
$endgroup$
– lulu
Dec 4 '18 at 18:27
$begingroup$
i took it to find a for the first equation and b for the second equation, then i substitution in the end eqution but nothing gains, i also added 1 in the both sides in the fisrt and the second equation but nothing gainjs else
$endgroup$
– brisk tuti
Dec 4 '18 at 18:31