Question

1/(x-1)(x-2) + 1/(x-2)(x-3) + 1/(x-3)(x-4) = 1/6​

Answer 1

Answer:

see below

Step-by-step explanation:

See attached screenshot (too time consuming to type out using latex, so used MS word).

Answer 2

Answer:

The value of x for the given equation is 7 or -2.

Step-by-step explanation:

Given equation is [1/{(x-1)(x-2)} + 1/{(x-2)(x-3)} + 1/{(x-3)(x-4)}] = 1/6

⇛[{(x-3)(x-4) + (x-1)(x-4) + (x-1)(x-2)}/{(x-1)(x-2)(x-3)(x-4)}] = 1/6

⇛[{x(x-4)-3(x-4) + x(x-4)-1(x-4) + x(x-2)-1(x-2)}/{(x-1)(x-2)(x-3)(x-4)}] = 1/6

⇛[{x²-4x-3x+12 + x²-4x-x+4 + x²-2x-x+2}/{(x-1)(x-2)(x-3)(x-4)}] = 1/6

⇛[{x²-7x+12 + x²-5x+4 + x²-3x+2}/{(x-1)(x-2)(x-3)(x-4)}] = 1/6

⇛[{x²+x²+x² - 7x-5x-3x + 12+4+2}/{(x-1)(x-2)(x-3)(x-4)}] = 1/6

⇛[{2x²+x² - 12x-3x + 16+2}/{(x-1)(x-2)(x-3)(x-4)}] = 1/6

⇛[{3x² - 15x + 18}/{(x-1)(x-2)(x-3)(x-4)}] = 1/6

On applying cross multiplication then

⇛6(3x² - 15x + 18) = 1{(x-1)(x-2)(x-3)(x-4)}

Multiply the number outside of the bracket with the numbers in the bracket, on both LHS and RHS.

⇛18x² - 90x + 108 = (x-1)(x-2)(x-3)(x-4)

⇛18x²(x²-5x+6) = (x-1)(x-2)(x-3)(x-4)

⇛18(x-2)(x-3) = (x-1)(x-2)(x-3)(x-4)

Now, cancel value {(x-2)(x-3)} on both LHS and RHS.

⇛18 = (x-1)(x-4)

⇛18 = x(x-4)-1(x-4)

⇛18 = x² - 4x - x + 4

⇛18 = x² - 5x + 4

⇛x² - 5x + 4 - 18 = 0

⇛x² - 5x - 14 = 0

By splitting the middle term, we get

⇛x² - 7x + 2x - 14 = 0

⇛x(x-7) + 2(x-7) = 0

⇛(x-7)(x+2) = 0

⇛x-7 = 0 | x+2 = 0

Therefore, x = 7 or x = -2

Answer: Hence the value of x for the given problem is 7 or -2.

VERIFICATION:

If x = 7 then, the equation is

[1/{(x-1)(x-2)} + 1/{(x-2)(x-3)} + 1/{(x-3)(x-4)}]

Substitute the value of x = 7 in equation then

= [1/{(7-1)(7-2)} + 1/{(7-2)(7-3)} + 1/{(7-3)(7-4)}]

= [1/{(6)(5)} + 1/{(5)(4)} + 1/{(4)(3)}]

= {1/(6*5) + 1/(5*4) + 1/(4*3)}

= {(1/30) + (1/20) + (1/12)}

Take the LCM of the denominator 12, 20 and 60 is 60.

= {(1*2 + 1*3 + 1*5)/60}

= {(2 + 3 + 5)/60}

= {(5 + 5)/60}

= (10/60)

= (1/6)

If x = -2, then the equation is

[1/{(x-1)(x-2)} + 1/{(x-2)(x-3)} + 1/{(x-3)(x-4)}]

Substitute the value of x = -2 in equation then

= [1/{(-2-1)(-2-2)} + 1/{(-2-2)(-2-3)} + 1/{(-2-3)(-2-4)}]

= [1/{(-3)(-4)} + 1/{(-4)(-5)} + 1/{(-5)(-6)}]

= {1/(-3*-4) + 1/(-4*-5) + 1/(-5*-6)}

= {(1/12) + (1/20) + (1/30)}

Take the LCM of the denominator 12, 20 and 60 is 60.

= {(5*1 + 3*1 + 2*1)/60}

= {(5 + 3 + 2)/60}

= {(5 + 5)/60}

= (10/60)

= (1/6)

Hence, verified.

Please let me know if you have any other questions.