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(3-x)/(x-5)=(2)/(5-x)
We move all terms to the left:
(3-x)/(x-5)-((2)/(5-x))=0
Domain of the equation: (x-5)!=0
We move all terms containing x to the left, all other terms to the right
x!=5
x∈R
Domain of the equation: (5-x))!=0We add all the numbers together, and all the variables
We move all terms containing x to the left, all other terms to the right
-x)!=-5
x!=-5/1
x!=-5
x∈R
(-1x+3)/(x-5)-(2/(-1x+5))=0
We calculate fractions
((-1x+3)*(-1x+5)))/((x-5)*(-1x+5)))+(-(2*(x-5))/((x-5)*(-1x+5)))=0
We calculate terms in parentheses: +((-1x+3)*(-1x+5)))/((x-5)*(-1x+5)))+(-(2*(x-5))/((x-5)*(-1x+5))), so:
(-1x+3)*(-1x+5)))/((x-5)*(-1x+5)))+(-(2*(x-5))/((x-5)*(-1x+5))
determiningTheFunctionDomain (-1x+3)*(-1x+5)))/((x-1x+5)))+(-(2*(x-5))/((x-1x-5)*(-5)*(+5))
We add all the numbers together, and all the variables
(-1x+3)*(-1x+5)))/(5))+(-(2*(x-5))/((-5)*(-5)*5)
We add all the numbers together, and all the variables
(-1x+3)*(-1x+5)))/5)+(-(2*(x-5))/((
We calculate fractions
((-1x+3)*(-1x+5)))*(()/(-(2*(x*(5+()+(-5))*5)+(-(2*x/(-(2*(x*(5+()
We calculate terms in parentheses: +((-1x+3)*(-1x+5)))*(()/(-(2*(x*(5+()+(-5))*5)+(-(2*x/(-(2*(x*(5+(), so:
(-1x+3)*(-1x+5)))*(()/(-(2*(x*(5+()+(-5))*5)+(-(2*x/(-(2*(x*(5+(
We calculate fractions
((-1x+3)*(-1x+5)))*(()*(-(2*(x*(5+()/((*(-(2*(x*(5+()+(-(2*(x*(5+()+(-5))*5)+(-(2*x*()/((*(-(2*(x*(5+()
We calculate terms in parentheses: +((-1x+3)*(-1x+5)))*(()*(-(2*(x*(5+()/((*(-(2*(x*(5+()+(-(2*(x*(5+()+(-5))*5)+(-(2*x*()/((*(-(2*(x*(5+(), so:
(-1x+3)*(-1x+5)))*(()*(-(2*(x*(5+()/((*(-(2*(x*(5+()+(-(2*(x*(5+()+(-5))*5)+(-(2*x*()/((*(-(2*(x*(5+(
We can not solve this equation
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