a + b = ab = a/b (•°•) a + b = ab ab = a/b a + b = a/b By the equation: a + b = a/b Multiply both sides by b..... b(a + b) = b • a/b => ab + b² = a - (1) Substitute the value of a by (1) in the other equation: a + b = ab => (ab + b²) + b = ab => ab + b² + b - ab = 0 [Cancel → ab from both sides] => b² + b = 0 => b(b + 1) = 0 So either..... b = 0 or b + 1 = 0 since b is in the denominator of (a/b) it is a non zero so first case → [rejected] b + 1 = 0 => b = -1 When b = -1, a + b = ab => a - 1 = -a [Add → +a on both sides of LHS and RHS] => a - 1 + a = 0 => 2a - 1 = 0 => 2a = 1 => a = 1/2 Verifying this value: a = 1/2, b = -1 a + b = ab 1/2 - 1 = 1/2 => 1/2 = 1/2 [Verified] Since all the equations are same and equal... verification in one is true for all.... hence, a = 1/2 ; b = -1 is the solution that satisfies all equation...
Just note ab=a/b therefore b=1/b therefore b squared =1 therefore b =+ or-1. If +1 then a+1=a so we discard +1 And so a= -1 and as a+b = a/b then a-1 = -a/1 Therefore 2a=1 and so a=1/2. Job done. Well once you stick the values in all 3 equations as a check.
You've already showed that a=0 implies: b=0, which is impossible at first place. This fact excludes the solution: a=0. Therefore, it is logically pointless to put afterwards the value : a=0 into the other equations.
ab = a/b → where: b ≠ 0 ab - (a/b) = 0 (ab² - a)/b = 0 ab² - a = 0 a.(b² - 1) = 0 First case: a = 0 Given: a + b = a/b → when: a = 0 0 + b = 0/b b = 0 → rejected because the condition: b ≠ 0 Second case: (b² - 1) = 0 b² = 1 b = ± 1 First solution: b = 1 Given: a + b = a/b → when: b = 1 a + 1 = a/1 a + 1 = a 1 = 0 ← false, solution to be rejected Second solution: b = - 1 Given: a + b = a/b → where: b = - 1 a - 1 = a/- 1 a - 1 = - a 2a = 1 a = 1/2 Unique solution (a ; b) → (1/2 ; - 1)
let a/b = k => a = bk, ab = k => kb^2 = k =>k(b^2 - 1) = 0 => k = 0 or b = 1, -1 (case k = 0) => a = 0, b = 0 => since b ≠ 0, no solution. (case b = 1) => a + 1 = a = a/1 => no solution. (case b = -1) => a - 1 = -a = a/-1 => a =1/2 Answer (a,b) = (1/2, -1)
I have a doubt in the solution why can't "a" tend to postive infinity And b belongs to (0+,+infinity) Such that ab also tends to postive infinity and a/b also tends to postive infinity
a + b = ab = a/b
(•°•) a + b = ab
ab = a/b
a + b = a/b
By the equation: a + b = a/b
Multiply both sides by b.....
b(a + b) = b • a/b
=> ab + b² = a - (1)
Substitute the value of a by (1) in the other equation:
a + b = ab
=> (ab + b²) + b = ab
=> ab + b² + b - ab = 0 [Cancel → ab from both sides]
=> b² + b = 0
=> b(b + 1) = 0
So either.....
b = 0 or b + 1 = 0
since b is in the denominator of (a/b) it is a non zero so first case → [rejected]
b + 1 = 0
=> b = -1
When b = -1, a + b = ab
=> a - 1 = -a
[Add → +a on both sides of LHS and RHS]
=> a - 1 + a = 0
=> 2a - 1 = 0
=> 2a = 1
=> a = 1/2
Verifying this value:
a = 1/2, b = -1
a + b = ab
1/2 - 1 = 1/2
=> 1/2 = 1/2 [Verified]
Since all the equations are same and equal... verification in one is true for all....
hence, a = 1/2 ; b = -1 is the solution that satisfies all equation...
Very nice! ❤
Just note ab=a/b therefore b=1/b therefore b squared =1 therefore b =+ or-1. If +1 then a+1=a so we discard +1 And so a= -1 and as a+b = a/b then a-1 = -a/1 Therefore 2a=1 and so a=1/2. Job done. Well once you stick the values in all 3 equations as a check.
Very nice! ❤
a = 1/2 ; b = -1
Yes, you are right! ❤
a=ab-b and a=ab^2 or b=±1
But b=1 is wrong in a+b=ab because 1=0 is wrong.
So b=-1
a-1=-a
2a=1
a=0,5
Very nice! ❤
You've already showed that a=0 implies: b=0, which is impossible at first place. This fact excludes the solution: a=0. Therefore, it is logically pointless to put afterwards the value : a=0 into the other equations.
a=0 is not the solution! ❤
Very interesting 😊
Glad you think so! ❤
ab = a/b → where: b ≠ 0
ab - (a/b) = 0
(ab² - a)/b = 0
ab² - a = 0
a.(b² - 1) = 0
First case: a = 0
Given: a + b = a/b → when: a = 0
0 + b = 0/b
b = 0 → rejected because the condition: b ≠ 0
Second case: (b² - 1) = 0
b² = 1
b = ± 1
First solution: b = 1
Given: a + b = a/b → when: b = 1
a + 1 = a/1
a + 1 = a
1 = 0 ← false, solution to be rejected
Second solution: b = - 1
Given: a + b = a/b → where: b = - 1
a - 1 = a/- 1
a - 1 = - a
2a = 1
a = 1/2
Unique solution (a ; b) → (1/2 ; - 1)
Very nice! ❤
let a/b = k => a = bk, ab = k => kb^2 = k =>k(b^2 - 1) = 0 => k = 0 or b = 1, -1
(case k = 0) => a = 0, b = 0 => since b ≠ 0, no solution.
(case b = 1) => a + 1 = a = a/1 => no solution.
(case b = -1) => a - 1 = -a = a/-1 => a =1/2 Answer (a,b) = (1/2, -1)
Very nice! ❤
ab=a/b , ab^2=a , a(b^2-1)=0
a=0 or b=1. or b=-1
case1 a=0. a+b=ab , 0+b=0×b
b=0 unacceptable
case2. b=1. a+1=a , 1=0 impossible case3. b=-1
a-1=-a. , a=1/2. acceptable
Very nice! ❤
a+b=a/b⇒ab+b²=a. a+b=ab⇒ -a- +b+b²= -a- . b²+b=0. b(b+1)=0 при b≠0. b+1=0⇒b=-1. a-1=-a⇒2a=1⇒a=½. Проверка: ½-1=-½. ½×(1-)=-½. ½÷(-1)=-½.
Very nice! ❤
Big fan
Thank you so much! ❤
a=0.5 , b=-1.
Very nice! ❤
@@SALogics Thank You!
تستطيع حلها عن طريق التمثيل الباتي في اقل من دقيقة
لطيف جدًا❤
As soon as a case is disproven by one equation, you don't need to check the other equations.
Very nice! ❤
I have a doubt in the solution why can't
"a" tend to postive infinity
And b belongs to (0+,+infinity)
Such that ab also tends to postive infinity and a/b also tends to postive infinity
The Solution are verified in the end! ❤
Sir another answer is
a) (√5 -1)/2
b) 1
•(√5-1)/2+1 = √5-1/2
•(√5-1)/2 ×1 = "
•(√5-1/2)÷1 = "
Very nice! ❤
Err, doesn't (√5-1)/2 + 1 = (√5)/2 + 1/2, while (√5-1)/2 ×1 = (√5-1/2)÷1 = (√5)/2 - 1/2 ?