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
<u>Answer</u><u>:</u> Hence the value of x for the given problem is 7 or -2.
<u>VERIFICATION</u><u>:</u>
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)
<u>Hence</u><u>,</u><u> </u><u>verified</u><u>.</u>
Please let me know if you have any other questions.
