Adding 3 equations => 2(ab + bc + ca) = 1200 => ab + bc + ca = 600 => bc = 300 , ca = 200 , ab = 100 Multiplying 3 equations => (abc)^2 = 6*10^6 => abc = ±1000√6 => a = ±10√6/3, b = ±10√6/2, c = ±10√6 => a + b + c = ±55√6/3
First, forget the 100's and solve a*(b+c) = 3 .... (1); b*(c+a) = 4 .... (2); c*(a+b) = 5 .... (3). Then we can scale up a, b and c each by a factor of 10 to get the final answer. With cyclic equations like these, it often helps to subtract two equations and use the result together with the third. So we try b+c = 3/a and c+a = 4/b and subtracting, we get b-a = 3/a - 4/b. This is not terribly helpful so we try another way: ab + ac - bc - ba = 3-4 --> c*(a-b) = -1 --> a-b = -1/c while the last equation can be written as a+b = 5/c. Now we can easily solve a = 2/c and b = 3/c. .... (4). Then substitute these in either ..(1) or ..(2) to get (2/c) * (3/c + c) = 3 --> 2*(3+c^2) = 3c^2 --> c^2 = 6 and c = plusorminus sqrt(6). If we take c = sqrt(6); a = 2/sqrt(6) = sqrt(6)/3 and b = sqrt(6)/2. Finally, scaling up by 10, Final answer: a = 10*sqrt(6)/3; b = 5*sqrt(6) and c = 10*sqrt(6) and a+b+c = (55/3) * sqrt(6). If we take c = -sqrt(6), a and b will also be negative but the products will all be positive. Nothing wrong there so another answer is a+b+c= -(55/3) * sqrt(6).
Yes, this is similar to the route I took. I simply expressed 'a' and 'b' in terms of 'c' and substituted in 'c(a+b)=500' to get 'c(c/3+c/2)=500' and a solution for 'c'. Then of course, 'c+c/3+c/2=a+b+c'.
Is an answer in the form a+b+ c a solution. I got the answers a=100/(150^0.5) b=(150^0.5) and c= 300/(150^0.5) and second answer a=-100/(150^0.5) b=-(150^0.5) and c= -300/(150^0.5)
From (1) a.(b + c) = 300 ab + ac = 300 ← equation (4) a.(b + c) = 300 b + c = 300/a ← equation (8) From (2) b.(a + c) = 400 ab + bc = 400 ← equation (5) From (3) c.(a + b) = 500 ac + bc = 500 ← equation (6) c.(a + b) = 500 (a + b) = 500/c ← equation (7) (4) - (5) (ab + ac) - (ab + bc) = 300 - 400 ab + ac - ab - bc = - 100 ac - bc = - 100 c.(a - b) = - 100 a - b = - 100/c → recall (7): a + b = 500/c a + b = 500/c --------------------------------------------------------sum 2a = - (100/c) + (500/c) 2a = 400/c a = 200/c ← equation (9) (5) - (6) (ab + bc) - (ac + bc) = 400 - 500 ab + bc - ac - bc = - 100 ab - ac = - 100 a.(b - c) = - 100 b - c = - 100/a → recall (8): b + c = 300/a b + c = 300/a -------------------------------------------------------sum 2b = - (100/a) + (300/a) 2b = 200/a b = 100/a → recall (9): a = 200/c b = 100/(200/c) b = 100c/200 b = c/2 ← equation (10) Restart from (7) (a + b) = 500/c → recall (9): a = 200/c (200/c) + b = 500/c → recall (10): b = c/2 (200/c) + (c/2) = 500/c c/2 = (500/c) - (200/c) c/2 = 300/c c² = 600 c = ± 10√6 First case: c = 10√6 Recall (10): b = c/2 → b = 5√6 Recall (9): a = 200/c → a = 200/(10√6) = 20/√6 = (20√6)/6 = (10√6)/3 Second case: c = - 10√6 Recall (10): b = c/2 → b = - 5√6 Recall (9): a = 200/c → a = 200/(- 10√6) = - 20/√6 = - (20√6)/6 = - (10√6)/3 Solution: [a ; b ; c] [(10√6)/3 ; 5√6 ; 10√6] [- (10√6)/3 ; - 5√6 ; - 10√6]
a b + b c + c a = (300 +400 + 500)/2 = 600 c a = ( a b + b c + c a) - ( a b + b c) = 200 b c = 600 - 300 = 300 a b = 600 - 500 = 100 a : b : c = (1/3 ) : (1/ 2) : 1 = 2 : 3 : 6 a b c = 1000 √ 6, -1000 √ 6, a = 10 √6 /3, b = 10 √6 /2, c = 10 √6 Or a = -10 √6 /3, b = -10 √6 /2, c = -10 √6
sum all
2(ab+bc+ca)=1200, ab+bc+ca=600
bc=300
ca=200
ab=100
(abc)^2=6000000
abc=1000sqrt(6), -1000sqrt(6)
a=+-10/3sqrt(6)
b=+-5sqrt(6)
c=+-10sqrt(6)
Sres. Reciban un cordial saludo, gracias por este bonito ejercicio de aplicación de productos notables. Éxitos.
You're welcome.💕🔥🥰🤩✅ Thanks too.
ab+ac=300;ab+bc=400;ac+bc=500;bc-ac=100;2ac=400;ac=200;ab=100;c=2b; bc=300;ac=200;b/a=1.5;b=1.5a;c=3a;1.5 a^2 +3a^2 = 300=>a= (10/3)*sqrt(6);b=1.5 a; c=3a
Adding 3 equations => 2(ab + bc + ca) = 1200 => ab + bc + ca = 600
=> bc = 300 , ca = 200 , ab = 100
Multiplying 3 equations => (abc)^2 = 6*10^6 => abc = ±1000√6
=> a = ±10√6/3, b = ±10√6/2, c = ±10√6 => a + b + c = ±55√6/3
First, forget the 100's and solve a*(b+c) = 3 .... (1); b*(c+a) = 4 .... (2); c*(a+b) = 5 .... (3). Then we can scale up a, b and c each by a factor of 10 to get the final answer.
With cyclic equations like these, it often helps to subtract two equations and use the result together with the third. So we try
b+c = 3/a and c+a = 4/b and subtracting, we get b-a = 3/a - 4/b. This is not terribly helpful so we try another way:
ab + ac - bc - ba = 3-4 --> c*(a-b) = -1 --> a-b = -1/c while the last equation can be written as a+b = 5/c. Now we can easily solve a = 2/c and b = 3/c. .... (4).
Then substitute these in either ..(1) or ..(2) to get
(2/c) * (3/c + c) = 3 --> 2*(3+c^2) = 3c^2 --> c^2 = 6 and c = plusorminus sqrt(6).
If we take c = sqrt(6); a = 2/sqrt(6) = sqrt(6)/3 and b = sqrt(6)/2. Finally, scaling up by 10,
Final answer: a = 10*sqrt(6)/3; b = 5*sqrt(6) and c = 10*sqrt(6) and a+b+c = (55/3) * sqrt(6).
If we take c = -sqrt(6), a and b will also be negative but the products will all be positive. Nothing wrong there so another answer is a+b+c= -(55/3) * sqrt(6).
Yes, this is similar to the route I took. I simply expressed 'a' and 'b' in terms of 'c' and substituted in 'c(a+b)=500' to get 'c(c/3+c/2)=500' and a solution for 'c'. Then of course, 'c+c/3+c/2=a+b+c'.
Is an answer in the form a+b+ c a solution. I got the answers a=100/(150^0.5) b=(150^0.5) and c= 300/(150^0.5) and second answer a=-100/(150^0.5) b=-(150^0.5) and c= -300/(150^0.5)
From (1)
a.(b + c) = 300
ab + ac = 300 ← equation (4)
a.(b + c) = 300
b + c = 300/a ← equation (8)
From (2)
b.(a + c) = 400
ab + bc = 400 ← equation (5)
From (3)
c.(a + b) = 500
ac + bc = 500 ← equation (6)
c.(a + b) = 500
(a + b) = 500/c ← equation (7)
(4) - (5)
(ab + ac) - (ab + bc) = 300 - 400
ab + ac - ab - bc = - 100
ac - bc = - 100
c.(a - b) = - 100
a - b = - 100/c → recall (7): a + b = 500/c
a + b = 500/c
--------------------------------------------------------sum
2a = - (100/c) + (500/c)
2a = 400/c
a = 200/c ← equation (9)
(5) - (6)
(ab + bc) - (ac + bc) = 400 - 500
ab + bc - ac - bc = - 100
ab - ac = - 100
a.(b - c) = - 100
b - c = - 100/a → recall (8): b + c = 300/a
b + c = 300/a
-------------------------------------------------------sum
2b = - (100/a) + (300/a)
2b = 200/a
b = 100/a → recall (9): a = 200/c
b = 100/(200/c)
b = 100c/200
b = c/2 ← equation (10)
Restart from (7)
(a + b) = 500/c → recall (9): a = 200/c
(200/c) + b = 500/c → recall (10): b = c/2
(200/c) + (c/2) = 500/c
c/2 = (500/c) - (200/c)
c/2 = 300/c
c² = 600
c = ± 10√6
First case: c = 10√6
Recall (10): b = c/2 → b = 5√6
Recall (9): a = 200/c → a = 200/(10√6) = 20/√6 = (20√6)/6 = (10√6)/3
Second case: c = - 10√6
Recall (10): b = c/2 → b = - 5√6
Recall (9): a = 200/c → a = 200/(- 10√6) = - 20/√6 = - (20√6)/6 = - (10√6)/3
Solution: [a ; b ; c]
[(10√6)/3 ; 5√6 ; 10√6]
[- (10√6)/3 ; - 5√6 ; - 10√6]
Thanks for detailed and resourceful explanation 🙏🤩🤩💕🥰. But you're required to find the sum of a, b and c.
@@superacademy247 sorry for this missing
a + b + c = ± [(10√6)/3 + 5√6 + 10√6]
a + b + c = ± (10√6 + 15√6 + 30√6)/3
a + b + c = ± (55√6)/3
a b + b c + c a
= (300 +400 + 500)/2 = 600
c a = ( a b + b c + c a) - ( a b + b c)
= 200
b c = 600 - 300 = 300
a b = 600 - 500 = 100
a : b : c = (1/3 ) : (1/ 2) : 1
= 2 : 3 : 6
a b c = 1000 √ 6, -1000 √ 6,
a = 10 √6 /3,
b = 10 √6 /2,
c = 10 √6
Or
a = -10 √6 /3,
b = -10 √6 /2,
c = -10 √6
Write with spidol, I can't see
Thanks for your tips. I've already begun writing with spidol
@@superacademy247 thanks for your attention sir