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| 1. |
x-4/x-5+x-4/x-5=10/3 solve the quadratic equation by factorisation |
| Answer» {tex}\\frac{x-4}{x-5}+\\frac{x-6}{x-7}=\\frac{10}{3}{/tex}, x\xa0{tex}\\neq{/tex}\xa05, 7{tex}\\Rightarrow \\frac{(x-4)(x-7)+(x-6)(x-5)}{(x-5)(x-7)}=\\frac{10}{3}{/tex}{tex}=\\frac{x^{2}-11 x+28+x^{2}-11 x+30}{x^{2}-12 x+35}=\\frac{10}{3}{/tex}{tex}\\Rightarrow{/tex}\xa03[2x2 - 22x + 58] = 10[x2 - 12x + 35]{tex}\\Rightarrow{/tex}\xa06x2 - 66x + 174 = 10x2 - 120x + 350{tex}\\Rightarrow{/tex}\xa04x2 - 54x + 176 = 0{tex}\\Rightarrow{/tex}\xa02x2 - 27x + 88 = 0{tex}\\Rightarrow{/tex}\xa02x2 - 16x - 11x + 88 = 0{tex}\\Rightarrow{/tex}\xa02x(x - 8)\xa0-11(x - 8)\xa0= 0{tex}\\Rightarrow{/tex}\xa0(2x - 11) (x - 8) = 0{tex}\\Rightarrow{/tex}\xa0x =\xa0{tex}\\frac{{11}}{2}{/tex},\xa08 | |