Question

x-1/x-2 + x-3/x-4 =3 1/3​

Answer 1

Answer:

No solution

Step-by-step explanation:

$\frac{x-1}{x-2} +\frac{x-3}{x-4} =3\frac{1}{3} \\\\Taking LCM,\\\\\frac{(x-4)(x-1)+(x-2)(x-3)}{(x-2)(x-4)} =\frac{10}{3} \\\\\frac{(x^{2} -x-4x+4)+(x^{2} -3x-2x+6)}{(x-2)(x-4)} =\frac{10}{3}\\\\\frac{x^{2} -x-4x+4+x^{2} -3x-2x+6}{(x-2)(x-4)} =\frac{10}{3}\\\\\frac{2x^{2} -10x+10}{(x-2)(x-3)} =\frac{10}{3} \\\\2(2x^{2} -10x+10)=10(x-2)(x-3)\\\\4x^{2} -20x+20=(10x-20)(x-3)\\\\4x^{2} -20x+20=10x^{2} -30x-20x+60\\\\4x^{2} -20x+20=10x^{2} -50x+60\\\\10x^{2} -50x+60-(4x^{2} -20x+20)=0$

$10x^{2} -50x+60-4x^{2} +20x-20=0\\\\6x^{2} -30x+40=0$

Using Quadratic formula,

$2(3x^{2} -15x+20)=0\\\\3x^{2} -15x+20=0$

x = [-b ± √b² - 4ac ] /2a

x = [-(-15) ± √(-15²) - 4(3)(20)] / 2(3)

x = [15 ± √225 - 240] / 6

x = [15 ± √-15] / 6

→The square root of a negative number is not a real number.

∴ No solution