Question

solve for x and y , x-1 / x-2 + x-3 / x-4 = 10/3.​

Answer 1

$\huge\mathtt\red{ꪖꪀs᭙ᴇʀ}$

To Solve:-

$\rm{ \dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = \dfrac{10}{3} }x−2x−1+x−4x−3=310$

Required Answer:-

For solving for x, we have to take the LCM i.e. (x - 2)(x -4). Then,

$⇒ \rm{ \dfrac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \dfrac{10}{3} }(x−2)(x−4)(x−1)(x−4)+(x−3)(x−2)=310$

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:

$⇒ \rm3 {\{ {x}^{2} - 5x + 4 + {x}^{2} - 5x + 6 \} = 10( {x}^{2} - 6x + 8}3{x2−5x+4+x2−5x+6}=10(x2−6x+8$

Opening the parentheses,

$⇒ \rm{3(2 {x}^{2} - 10x + 10) = 10{x}^{2} - 60x + 80}3(2x2−10x+10)=10x2−60x+80$

$⇒ \rm{6 {x}^{2} - 30x + 30 = 10 {x}^{2} - 60x + 80}6x2−30x+30=10x2−60x+80$

Taking 2 as a common from both sides & cancelling it, we have,

$⇒ \rm{3{x}^{2} - 15x+15 = 5{x}^2 - 30x+40}3x2−15x+15=5x2−30x+40$

Now shifting the terms to one side of the equation:

$⇒ \rm{5{x}^2 - 3{x}^2 - 30x+15x+40 - 15=0}5x2−3x2−30x+15x+40−15=0$

$⇒ \rm{2x^2 -15x+25=0}2x2−15x+25=0$

Now find the possible values of x by Middle term factorisation,

$⇒ \rm{2x^2 -10x-5x+25=0}2x2−10x−5x+25=0$

$⇒ \rm{2x(x - 5) -5(x-5) =0}2x(x−5)−5(x−5)=0$

$⇒ \rm{(2x-5)(x-5)=0}(2x−5)(x−5)=0$

Equating to 0, We get

$⇒ \rm{x = \dfrac{5}{2} \: or \: 5}x=25 or 5$

Hence The value of x after solving: 5/2 or 5.

$\huge\mathtt\red{ᴛʜᴇᴍøøɴᘜɪʀꪶ}$

Answer 2

Answer:

To Solve:-

$x−2</p><p>x−1</p><p> </p><p> + </p><p>x−4</p><p>x−3</p><p> </p><p> = </p><p>3</p><p>10</p><p>$

Required Answer:-

For solving for x, we have to take the LCM i.e.

$(x - 2)(x -4)$

Then,

$(x−2)(x−4)(x−1)(x−4)+(x−3)(x−2)= 310$

Cross multiplying for easier calculation:

$3{(x−1)(x−4)+(x−3)(x−2)} \\ =10(x−2)(−4)$

Multiplying the terms inside the curly brackets:

$3 {\{ {x}^{2} - 5x + 4 + {x}^{2} - 5x + 6 \}= 10( {x}^{2} - 6x + 8}3{x </p><p>2</p><p> −5x+4+x </p><p>2</p><p> −5x+6}= \\ 10(x </p><p>2</p><p> −6x+8</p><p>$

Opening the parentheses,

⇒

$\rm{3(2 {x}^{2} - 10x + 10) 10{x}^{2} - 60x + 80}3(2x </p><p>2</p><p> −10x+10)=10x </p><p>2</p><p> −60x+80$

$\rm{6 {x}^{2} - 30x + 30 </p><p>= 10 {x}^{2} - 60x + 80}6x 2−30x+30</p><p>=10x 2−60x+80</p><p>$

Taking 2 as a common from both sides & cancelling it, we have,

$\rm{3{x}^{2} - 15x+15 = 5{x}^2 - 30x+40}3x </p><p>2</p><p> −15x+15=5x </p><p>2</p><p> −30x+40$

Now shifting the terms to one side of the equation:

$\rm{5{x}^2 - 3{x}^2 - 30x+15x+40 - 15=0}5x </p><p>2</p><p> −3x </p><p>2</p><p> −30x+15x+40−15=0$

$\rm{2x^2 -15x+25=0}2x </p><p>2</p><p> −15x+25=0$

Now find the possible values of x by Middle term factorisation,

$\rm{2x^2 -10x-5x+25=0}2x </p><p>2</p><p> −10x−5x+25=0$

$\rm{2x(x - 5) -5(x-5) =0}2x(x−5)−5(x−5)=0$

$\rm{(2x-5)(x-5)=0}(2x−5)(x−5)=0$

Equating to 0, We get

$\rm{x = \dfrac{5}{2} \: or \: 5}x= </p><p>2</p><p>5</p><p> </p><p> or5</p><p>$

Hence:-

The value of x after solving: 5/2 or 5