Question 897709
Here is one way to solve for x. First I combined the fractions on the left side (by getting the LCD first). From there I cross multiplied and then isolated x.


{{{5/x+2/5=7/x}}}


{{{(5*5)/(5*x)+2/5=7/x}}}


{{{25/(5x)+2/5=7/x}}}


{{{25/(5x)+(x*2)/(x*5)=7/x}}}


{{{25/(5x)+(2x)/(5x)=7/x}}}


{{{(25+2x)/(5x)=7/x}}}


{{{(25+2x)*x=5x*7}}}


{{{x(25+2x)=5x*7}}}


{{{x(25)+x(2x)=5x*7}}}


{{{25x+2x^2=35x}}}


{{{25x+2x^2-35x=0}}}


{{{2x^2+25x-35x=0}}}


{{{2x^2-10x=0}}}


{{{2x(x-5)=0}}}


{{{2x=0}}} or {{{x-5=0}}}


{{{x=0/2}}} or {{{x-5=0}}}


{{{x=0}}} or {{{x-5=0}}}


{{{x=0}}} or {{{x=0+5}}}


{{{x=0}}} or {{{x=5}}}


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The two *possible* solutions are {{{x=0}}} or {{{x=5}}}


However, only {{{x=5}}} works because plugging {{{x=0}}} back into the original equation gives you a division by zero error. 


So the final answer is {{{x = 5}}}