solve for x and y , x-1 / x-2 + x-3 / x-4 = 10/3.
Answers
\rm{ \dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = \dfrac{10}{3} }
x−2
x−1
+
x−4
x−3
=
3
10
Required Answer:-
For solving for x, we have to take the LCM i.e. (x - 2)(x -4). Then,
⇒ \rm{ \dfrac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \dfrac{10}{3} }
(x−2)(x−4)
(x−1)(x−4)+(x−3)(x−2)
=
3
10
Cross multiplying for easier calculation:
⇒ \rm{3 \{(x - 1)(x - 4) + (x - 3)(x - 2) \}= 10(x - 2)( - 4)}3{(x−1)(x−4)+(x−3)(x−2)}=10(x−2)(−4)
Multiplying the terms inside the curly brackets:
⇒ \rm3 {\{ {x}^{2} - 5x + 4 + {x}^{2} - 5x + 6 \} = 10( {x}^{2} - 6x + 8}3{x
2
−5x+4+x
2
−5x+6}=10(x
2
−6x+8
Opening the parentheses,
⇒ \rm{3(2 {x}^{2} - 10x + 10) = 10{x}^{2} - 60x + 80}3(2x
2
−10x+10)=10x
2
−60x+80
⇒ \rm{6 {x}^{2} - 30x + 30 = 10 {x}^{2} - 60x + 80}6x
2
−30x+30=10x
2
−60x+80
Taking 2 as a common from both sides & cancelling it, we have,
⇒ \rm{3{x}^{2} - 15x+15 = 5{x}^2 - 30x+40}3x
2
−15x+15=5x
2
−30x+40
Now shifting the terms to one side of the equation:
⇒ \rm{5{x}^2 - 3{x}^2 - 30x+15x+40 - 15=0}5x
2
−3x
2
−30x+15x+40−15=0
⇒ \rm{2x^2 -15x+25=0}2x
2
−15x+25=0
Now find the possible values of x by Middle term factorisation,
⇒ \rm{2x^2 -10x-5x+25=0}2x
2
−10x−5x+25=0
⇒ \rm{2x(x - 5) -5(x-5) =0}2x(x−5)−5(x−5)=0
⇒ \rm{(2x-5)(x-5)=0}(2x−5)(x−5)=0
Equating to 0, We get
⇒ \rm{x = \dfrac{5}{2} \: or \: 5}x=
2
5
or5
Hence:-
The value of x after solving: 5/2 or 5
Answer:
refer to the attachment
