(2z+5)/5=(3z+1)/2+(7-z)2

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Solution for (2z+5)/5=(3z+1)/2+(7-z)2 equation:



(2z+5)/5=(3z+1)/2+(7-z)2
We move all terms to the left:
(2z+5)/5-((3z+1)/2+(7-z)2)=0
Domain of the equation: 2+(7-z)2)!=0
We move all terms containing z to the left, all other terms to the right
(7-z)2)!=-2
z∈R
We add all the numbers together, and all the variables
(2z+5)/5-((3z+1)/2+(-1z+7)2)=0
We calculate fractions
(4z+(-1z+7)2)+10)/((-1z+7)2)+5*2)+(-((3z+1)*5)/((-1z+7)2)+5*2)=0
We calculate terms in parentheses: +(4z+(-1z+7)2), so:
4z+(-1z+7)2
We multiply parentheses
4z-2z+14
We add all the numbers together, and all the variables
2z+14
Back to the equation:
+(2z+14)
We can not solve this equation

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