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प्रश्न
Solve the following equation by factorization
`a/(ax - 1) + b/(bx - 1) = a + b, a + b ≠ 0, ab ≠ 0`
उत्तर
`a/(ax - 1) + b/(bx - 1) = a + b, a + b ≠ 0, ab ≠ 0`
⇒ `(a/(ax - 1) - b) + (b/(bx - 1) - a)` = 0
⇒ `(a - abx + b)/((ax - 1)) + (b - abx + a)/((bx - 1))` = 0
⇒ `(a + b - abx) [1/(ax - 1) + 1/(bx - 1)]` = 0
⇒ `(a + b - abx) [(bx - 1 + ax - 1)/((ax - 1)(bx - 1))]` = 0
⇒ `((a + b - abx)(ax + bx - 2))/((ax - 1)(bx - 1)` = 0
⇒ `(a + b - abx) (ax + bx - 2)` = 0
⇒ Either a + b - abx = 0,
then a+ b = abx
x = `(a + b)/(ab)`
or
ax + bx - 2 = 0,
then x(a + b) = 2
x = `(2)/(a + b)`
Hence x = `(a + b)/(ab), (2)/(a + b)`.
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