Advertisements
Advertisements
Question
Solve the following equation by factorization
`a/(ax - 1) + b/(bx - 1) = a + b, a + b ≠ 0, ab ≠ 0`
Advertisements
Solution
`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)`.
APPEARS IN
RELATED QUESTIONS
Solve for x :
`2/(x+1)+3/(2(x-2))=23/(5x), x!=0,-1,2`
Solve the following quadratic equations by factorization:
3x2 = -11x - 10
Solve the following quadratic equations by factorization:
`a/(x - a) + b/(x - b) = (2c)/(x - c)`
Solve the following quadratic equations by factorization:
`4(2x – 3)^2 – (2x – 3) – 14 = 0`
Find two consecutive multiples of 3 whose product is 648.
Solve the following quadratic equation for x:
`4sqrt3x^3+5x-2sqrt3=0`
Solve the following quadratic equation by factorisation.
2y2 + 27y + 13 = 0
Solve the following quadratic equation by factorization method : `"x"^2 - 5"x" - 36 = 0`
Solve the following equation by factorization
`(x^2 - 5x)/(2)` = 0
Solve the following equation by factorization
`(1)/(7)(3x – 5)^2`= 28
