a(b + c) = 300 & b(a + c) = 400 & c(a + b) = 500 ---(1) adding 3 equations => 2(ab + bc + ca) = 1200 => ab + bc + ca = 600 ---(2) from (1) and (2) => bc = 300 & ca = 200 & ab = 100 ---(3) multiplying 3 equations, (abc)^2 = 6*10^6 => abc = ±√6*10^3 ---(4) from (3) and (4) , a = ±10*√6/3 & b = ±10*√6/2 & c = ±10*√6 => a + b + c = ±55/3*√6
at first i can see ab=100, ac=200, bc=300 eliminating ab from the first two equations gives c(b-a)=100 subs into eqn 3 100(a+b)=500b -500a 6a=4b etc so c=3a=2b, so c( a+b)=500 becomes c(c/3+c/2)=500 5/6 c^2=500, c^2=600, c =10 sqrt(6) a+b+c =c+c/3+c/2 = (11/6)c =110sqrt(6)/6
a(b + c) = 300 & b(a + c) = 400 & c(a + b) = 500 ---(1)
adding 3 equations => 2(ab + bc + ca) = 1200 => ab + bc + ca = 600 ---(2)
from (1) and (2) => bc = 300 & ca = 200 & ab = 100 ---(3)
multiplying 3 equations, (abc)^2 = 6*10^6 => abc = ±√6*10^3 ---(4)
from (3) and (4) , a = ±10*√6/3 & b = ±10*√6/2 & c = ±10*√6
=> a + b + c = ±55/3*√6
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^=read as to the power
*=read as square root
As per question
a(b+c)=300...eqn1
b(a+c)=400....eqn2
C(a+b)=500...eqn3
Eqn1 -eqn3
ab+ac-ac-bc=300-500
ab-bc=-200
b(a-c)=-200....eqn4
Eqn2/eqn4
b(a+c)/{b(a-c)}=400/(-200)
(a+c)/(a-c)=(-2)
(a+c)=-2(a-c)
a+c=-2a+2c
a+2a=2c-c
C=3a
a=c/3......eqn5
Eqn2 - eqn3
ab+bc-ac-bc=400-500
ab-ac=-100
a(b-c)=-100......eqn6
Eqn1/eqn6
a(b+c)/{a(b-c)}=300/-100
(b+c)/(b-c)=-3
b+c=-3b+3c
b+3b=3c-c
4b=2c
b=2c/4=c/2.....eqn7
Put eqn1 & eqn2 in eqn3
C(a+b)=500
C{(c/3)+(c/2)}=500
C(5c/6)=500
5c^2=500×6=3000
C^2=3000/5=600
*(c^2)=*600
C=10.*6
a=c/3
=(10.*6)/3
b=c/2
=(10.*6)/2
=5.*6
a+b+c={(10.*6)/3}+(5.*6)+(10*6)
={10.*6)+(15.*6)+30.*6)}/3
=(55.*6)/3
a(b+c)=300 b(a+c)=400 c(a+b)=500 a+b+c=(±55Sqrt[6])/3=±18.3 recurring Sqrt[6]=(18 1/3) Sqrt[6]
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There is much easier way of solving it, it doesn't involve quadratic equation.
Add all->ab+bc+ca= 600; subtract each from 600 to get ab, bc, ca. From these a , b, c
at first i can see ab=100, ac=200, bc=300
eliminating ab from the first two equations gives c(b-a)=100
subs into eqn 3
100(a+b)=500b -500a 6a=4b etc
so c=3a=2b, so c( a+b)=500 becomes c(c/3+c/2)=500
5/6 c^2=500, c^2=600, c =10 sqrt(6) a+b+c =c+c/3+c/2 = (11/6)c =110sqrt(6)/6
В конце запутались. Проще можно: 10sq(2/3)+10sq(3/2)+10sq(6)
10 => ±10
I square up the 3 equation first
a(b+c) =300 ...(1)
b(a+c) =400 ...(2)
c(a+b) =500 ...(3)
since, 300^2 +400^2=500^2 , get the equation
[a(b+c)]^2 +[b(a+c)]^2 = [c(a+b)]^2
a^2(b+c)^2 +b^2(a+c)^2 = c^2(a+b)^2
a^2(b^2+c^2+2bc)+ b^2(a^2+c^2+2ac )^2 = c^2(a^2+b^2 +2ab)
(a^2b^2)+(a^2c^2)+2(a^2bc)+ (a^2b^2)+(b^2c^2)+(2ab^2c ) = (a^2c^2)+(b^2c^2) +(2abc^2)
(a^2b^2)+2(a^2bc)+ (a^2b^2)+(2ab^2c ) = (2abc^2)
2(a^2b^2)+2(a^2bc)+(2ab^2c ) = (2abc^2)
(a^2b^2)+(a^2bc)+(ab^2c ) = (abc^2)
(ab)+(ac)+(bc ) = (c^2) ...(4)
summing up ...(1)...(2)...(3)
a(b+c) b(a+c) c(a+b) =300 +400 +500
2[(ab)+(ac)+(bc ) ] =1200
[(ab)+(ac)+(bc ) ] =600
...(4)=...(5)
=> (c^2) =600
c = sqrt(600)
= 10sqrt(6)
...(1)+...(2)
a(b+c) + b(a+c) =300 +400
2ab+ac + bc =700
2ab+(a + b) [10sqrt(6)]=700 ...(5)
from (3)
c(a+b) =500
(a+b) =50 /sqrt(6) ...(6)
sub into ...(5)
2ab+(a + b) [10sqrt(6)]=700
2ab+[50 /sqrt(6)] [10sqrt(6)]=700
ab=100...(7)
sub into ...(1)
a(b+c) =300
ab+ac =300
100 +a(10sqrt(6)) =300
a(sqrt(6)) =20
a =20/(sqrt(6))
sub into ...(7)
ab=100
b[20/(sqrt(6))] =100
b =100/[20/(sqrt(6))]
=5(sqrt(6))
a +b+ c =20/(sqrt(6)) +5(sqrt(6)) +10sqrt(6)
=20/(sqrt(6)) +15(sqrt(6))
=110/(sqrt(6))
X=AB， Y=BC， Z=CA， X+Z=300， X+Y=400， Y+Z=500， X+Y+Z=600， X=100， Y=300，Z=200，XYZ=6*10^6=（ABC）^2，ABC=-+1000Sqrt（6）
Vimp
另a+b+c=x a(x-a)=300 b(x-b)=400 c(x-c)=500 全加起來 ax-aa+bx-bb+cx-cc=1200 x(a+b+c)-aa-bb-cc=1200 xx=1200+aa+bb+cc xx=1200+200/3+150+600=6050/3