-1/2(z+12)=2/5(9-3z)

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Solution for -1/2(z+12)=2/5(9-3z) equation: 



-1/2(z+12)=2/5(9-3z)

We move all terms to the left:

-1/2(z+12)-(2/5(9-3z))=0



Domain of the equation: 2(z+12)!=0

z∈R





Domain of the equation: 5(9-3z))!=0

z∈R



We add all the numbers together, and all the variables

-1/2(z+12)-(2/5(-3z+9))=0

We calculate fractions


(-5z(-)/(2(z+12)*5(-3z+9)))+(-4zz/(2(z+12)*5(-3z+9)))=0



We calculate terms in parentheses: +(-5z(-)/(2(z+12)*5(-3z+9))), so:

-5z(-)/(2(z+12)*5(-3z+9))

We add all the numbers together, and all the variables

-5z0/(2(z+12)*5(-3z+9))

We multiply all the terms by the denominator


-5z0

We add all the numbers together, and all the variables

-5z

Back to the equation:

+(-5z)





We calculate terms in parentheses: +(-4zz/(2(z+12)*5(-3z+9))), so:

-4zz/(2(z+12)*5(-3z+9))

We multiply all the terms by the denominator


-4zz

Back to the equation:

+(-4zz)



We get rid of parentheses

-5z-4zz=0

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