x-4/x-5+x-4/x-5=10/3 solve the quadratic equation by …
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Sia ? 6 years, 3 months ago
{tex}\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}{/tex}, x {tex}\neq{/tex} 5, 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} 3[2x2 - 22x + 58] = 10[x2 - 12x + 35]
{tex}\Rightarrow{/tex} 6x2 - 66x + 174 = 10x2 - 120x + 350
{tex}\Rightarrow{/tex} 4x2 - 54x + 176 = 0
{tex}\Rightarrow{/tex} 2x2 - 27x + 88 = 0
{tex}\Rightarrow{/tex} 2x2 - 16x - 11x + 88 = 0
{tex}\Rightarrow{/tex} 2x(x - 8) -11(x - 8) = 0
{tex}\Rightarrow{/tex} (2x - 11) (x - 8) = 0
{tex}\Rightarrow{/tex} x = {tex}\frac{{11}}{2}{/tex}, 8
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