Question

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

Answer 1

$\underline{\boxed{\gray{\sf Given \: : }}}$

$\underline{\boxed{\purple{\sf\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3} }}}$

$\underline{\boxed{\pink{\sf\implies \frac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)}=\frac{10}{3} }}}$

$\underline{\boxed{\purple{\sf \implies \frac{x^{2}-4x-x+4+x^{2}-2x-3x+6}{x^{2}-4x-2x+8}=\frac{10}{3}}}}$

$\underline{\boxed{\pink{\sf \implies \frac{2x^{2}-10x+10}{x^{2}-6x+8}=\frac{10}{3} }}}$

$\underline{\boxed{\purple{\sf \implies 3(2x^{2}-10x+10)=10(x^{2}-6x+8)}}}$

$\underline{\boxed{\pink{\sf \implies 6x^{2}-30x+30=10x^{2}-60x+80}}}$

$\underline{\boxed{\purple{\sf\implies 0= 10x^{2}-60x+80-6x^{2}+30x-30 }}}$

$\underline{\boxed{\pink{\sf \implies 4x^{2}-30x+50=0 }}}$

$\large\mapsto\boxed{\sf\green{Divide \: each \: term \: by \: 2 ,we \: get}}$

$\underline{\boxed{\red{\sf \implies 2x^{2}-15x+25=0}}}$

$\large\mapsto\boxed{\sf\orange{Splitting \: the \: middle \: term, \: we \: get:}}$

$\underline{\boxed{\green{\sf \implies 2x^{2}-10x-5x+25=0}}}$

$\underline{\boxed{\red{\sf \implies 2x(x-5)-5(x-5)=0 }}}$

$\underline{\boxed{\green{\sf \implies (x-5)(2x-5)=0 }}}$

$\underline{\boxed{\red{\sf\implies x-5 = 0 \: Or \: 2x-5 = 0 }}}$

$\underline{\boxed{\green{\sf \implies x = 5 \: Or \: x=\frac{5}{2} }}}$

$\huge\pink{\boxed{therefore,}}$

$\underline{\boxed{\blue{\sf x = 5 Or x = \frac{5}{2}25</p><p> }}}</p><p>$

#thanks:)

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