Question

Solve each of the following quadratic equations:
$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},x\ne 2,4$

Answer 1

As given,

$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}$

$\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}$

L.C.M. of $(x-2)and(x-4)=(x-2)(x-4)$

Then,

=> $\frac{\left[\right(x-1\left)\right(x-4)+(x-3\left)\right(x-2\left)\right]}{(x-2)(x-4)}=\frac{10}{3}$

=> $3(x^2-5x+4+x^2-5x+6)=10(x-2)(x-4)$

=> $3(2x^2–10x+10)=10(x^2–6x+8)$

=> $6x^2–30x+30=10x^2–60x+80$

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

=> $3x^2–15x+15=5x^2–30x+40$

=>$5x^2–3x^2–30x+15x+40–15=0$

=> $2x^2-15x+25=0$

Now, by the middle term splitting method,we have,

=> $2x^2-10x-5x+25=0$

=> $2x(x–5)-5(x-5)=0$

=> $(2x-5)(x-5)=0$

=> $2x-5=0$

So,$x=\frac{5}{2}$

Or, $x-5=0$

So, $x=5$

Hence, $x=5or\frac{5}{2}$