3(z+2)=-2(2z+5)

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



3(z+2)=-2(2z+5)
We move all terms to the left:
3(z+2)-(-2(2z+5))=0
We multiply parentheses
3z-(-2(2z+5))+6=0
We calculate terms in parentheses: -(-2(2z+5)), so:
-2(2z+5)
We multiply parentheses
-4z-10
Back to the equation:
-(-4z-10)
We get rid of parentheses
3z+4z+10+6=0
We add all the numbers together, and all the variables
7z+16=0
We move all terms containing z to the left, all other terms to the right
7z=-16
z=-16/7
z=-2+2/7

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