Answer:
Given, \[\frac{1}{2x-3}+\frac{1}{x-5}=1\frac{1}{9}\] \[\frac{x-5+2x-3}{(2x-3)(x-5)}=\frac{10}{9}\] \[\frac{3x-8}{2{{x}^{2}}-13x+15}=\frac{10}{9}\] \[9(3x-8)=10(2{{x}^{2}}-13x+15)\] \[27x-72=20{{x}^{2}}-130x+150\] \[20{{x}^{2}}-157x+222=0\] \[20{{x}^{2}}-120x-37x+222=0\] \[20x(x-6)-37(x-6)=0\] \[(x-6)(20x-37)=0\] Either \[x-6=0\] or \[20x-37=0\] \[\Rightarrow \] \[x=6\]\[x=\frac{37}{20}\]
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