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...
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)
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 + 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! ❤
Big fan
Thank you so much! ❤
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)
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Very interesting 😊
Glad you think so! ❤
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! ❤
a = 1/2 ; b = -1
تستطيع حلها عن طريق التمثيل الباتي في اقل من دقيقة
لطيف جدًا❤
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.
As soon as a case is disproven by one equation, you don't need to check the other equations.
Very nice! ❤