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Question Number 109500 by john santu last updated on 24/Aug/20
Given { ((a^2 +ab+bc+ac=a+c)),((b^2 +ab+bc+ac=b+a)),((c^2 +ab+bc+ac=c+b)) :} find the value of a+b+c
$${Given}\:\begin{cases}{{a}^{\mathrm{2}} +{ab}+{bc}+{ac}={a}+{c}}\\{{b}^{\mathrm{2}} +{ab}+{bc}+{ac}={b}+{a}}\\{{c}^{\mathrm{2}} +{ab}+{bc}+{ac}={c}+{b}}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:{a}+{b}+{c}\: \\ $$
Answered by bemath last updated on 24/Aug/20
⇔ (a+b+c)^2 +ab+bc+ac=2(a+b+c) let a+b+c = p ⇒p^2 +ab+bc+ac = p ⇒(1/a)+(1/b)+(1/c)=((ab+ac+bc)/(abc))
$$\Leftrightarrow\:\left({a}+{b}+{c}\right)^{\mathrm{2}} +{ab}+{bc}+{ac}=\mathrm{2}\left({a}+{b}+{c}\right) \\ $$$${let}\:{a}+{b}+{c}\:=\:{p} \\ $$$$\Rightarrow{p}^{\mathrm{2}} +{ab}+{bc}+{ac}\:=\:{p} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{{ab}+{ac}+{bc}}{{abc}} \\ $$
Answered by mr W last updated on 24/Aug/20
due to symmetry: a=b=c ⇒4a^2 =2a ⇒a=0 or a=(1/2) ⇒a+b+c=0 or (3/2)
$${due}\:{to}\:{symmetry}: \\ $$$${a}={b}={c} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} =\mathrm{2}{a} \\ $$$$\Rightarrow{a}=\mathrm{0}\:{or}\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{0}\:{or}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by bemath last updated on 24/Aug/20
sir. how to know that equation due to symetry sir?
$${sir}.\:{how}\:{to}\:{know}\:{that}\:{equation}\:{due}\:{to} \\ $$$${symetry}\:{sir}? \\ $$
Commented by mr W last updated on 24/Aug/20
if you replace a with b or c, b with c or a, c with a or b, and the equations remain the same, then they are symmetric. a=b=c could be the solution, but mustn′t.
$${if}\:{you}\:{replace}\:{a}\:{with}\:{b}\:{or}\:{c},\:{b}\:{with}\:{c}\:{or} \\ $$$${a},\:{c}\:{with}\:{a}\:{or}\:{b},\:{and}\:{the}\:{equations} \\ $$$${remain}\:{the}\:{same},\:{then}\:{they}\:{are} \\ $$$${symmetric}.\:{a}={b}={c}\:{could}\:{be}\:{the} \\ $$$${solution},\:{but}\:{mustn}'{t}. \\ $$
Commented by bemath last updated on 24/Aug/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Aug/20
a+c−a^2 =b+a−b^2 =c+b−c^2 b+a−b^2 −a−c+a^2 =0 (a−b)(a+b−1)=0 × a=b ∣ a+b=1 Similarly b=c ∣ b+c=1 c=a ∣ c+a=1 a=b=c ∣ 2(a+b+c)=3 a+b+c=3k ∀k∈Z ∣ a+b+c=3/2
$${a}+{c}−{a}^{\mathrm{2}} ={b}+{a}−{b}^{\mathrm{2}} ={c}+{b}−{c}^{\mathrm{2}} \\ $$$${b}+{a}−{b}^{\mathrm{2}} −{a}−{c}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}−{b}\right)\left({a}+{b}−\mathrm{1}\right)=\mathrm{0}\:× \\ $$$${a}={b}\:\mid\:{a}+{b}=\mathrm{1} \\ $$$${Similarly} \\ $$$${b}={c}\:\mid\:{b}+{c}=\mathrm{1} \\ $$$${c}={a}\:\mid\:{c}+{a}=\mathrm{1} \\ $$$${a}={b}={c}\:\mid\:\mathrm{2}\left({a}+{b}+{c}\right)=\mathrm{3} \\ $$$${a}+{b}+{c}=\mathrm{3}{k}\:\forall{k}\in\mathbb{Z}\:\:\mid\:\:{a}+{b}+{c}=\mathrm{3}/\mathrm{2} \\ $$$$ \\ $$
Commented by mr W last updated on 24/Aug/20
i mean when a=b=c=k, then 4k^2 =2k ⇒k=0 or k=(1/2) there is no other solutions.
$${i}\:{mean}\:{when}\:{a}={b}={c}={k},\:{then} \\ $$$$\mathrm{4}{k}^{\mathrm{2}} =\mathrm{2}{k} \\ $$$$\Rightarrow{k}=\mathrm{0}\:{or}\:{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${there}\:{is}\:{no}\:{other}\:{solutions}. \\ $$
Commented by mr W last updated on 24/Aug/20
k=0 ?
$${k}=\mathrm{0}\:? \\ $$
Commented by Rasheed.Sindhi last updated on 24/Aug/20
Yes sir, I think a=b=c=0 also satisfy the given equations.
$${Yes}\:{sir},\:{I}\:{think}\:{a}={b}={c}=\mathrm{0}\:{also} \\ $$$${satisfy}\:{the}\:{given}\:{equations}. \\ $$
Commented by $@y@m last updated on 24/Aug/20
@ Mr. Rashid Sindhi! How c disappeared in 3^(rd) line?
$$@\:{Mr}.\:{Rashid}\:{Sindhi}! \\ $$$${How}\:\mathrm{c}\:{disappeared}\:{in}\:\mathrm{3}^{{rd}} \:{line}? \\ $$
Commented by $@y@m last updated on 24/Aug/20
@ Mr. W! Sorry, No doubt on your solution. This app does not specify comment over comment. All goes in a continuous chain.
$$@\:{Mr}.\:{W}! \\ $$$${Sorry}, \\ $$$${No}\:{doubt}\:{on}\:{your}\:{solution}. \\ $$$$\:{This}\:{app}\:{does}\:{not}\:{specify} \\ $$$${comment}\:{over}\:{comment}.\:{All}\:{goes} \\ $$$${in}\:{a}\:{continuous}\:{chain}. \\ $$
Commented by Rasheed.Sindhi last updated on 24/Aug/20
@ S@y@m b−b^2 −c+a^2 =0 a^2 −b^2 +b−c (a−b)(a+b)+b−c I thought c as a! This mistake made my calculation easy but you caught my mistake. hahaha.. and make my way difficult!
$$@\:{S}@{y}@{m} \\ $$$${b}−{b}^{\mathrm{2}} −{c}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{b}−{c} \\ $$$$\left({a}−{b}\right)\left({a}+{b}\right)+{b}−{c} \\ $$$${I}\:{thought}\:{c}\:{as}\:{a}!\:\mathcal{T}{his}\:{mistake} \\ $$$${made}\:{my}\:{calculation}\:{easy}\:{but} \\ $$$${you}\:{caught}\:{my}\:{mistake}.\:{hahaha}.. \\ $$$${and}\:{make}\:{my}\:{way}\:{difficult}! \\ $$
Commented by Rasheed.Sindhi last updated on 24/Aug/20
@ Mr W sir the solution is wrong.
$$@\:{Mr}\:{W} \\ $$$${sir}\:{the}\:{solution}\:{is}\:{wrong}. \\ $$

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