6(z+2)-7z=2(-1/2z+1)+10

Solution for 6(z+2)-7z=2(-1/2z+1)+10 equation:

6(z+2)-7z=2(-1/2z+1)+10

We move all terms to the left:

6(z+2)-7z-(2(-1/2z+1)+10)=0


Domain of the equation: 2z+1)+10)!=0

z∈R


We add all the numbers together, and all the variables

-7z+6(z+2)-(2(-1/2z+1)+10)=0

We multiply parentheses

-7z+6z-(2(-1/2z+1)+10)+12=0

We multiply all the terms by the denominator


-7z*2z+1)+10)+6z*2z+1)+10)-(2(-1+12*2z+1)+10)=0

We add all the numbers together, and all the variables

-7z*2z+1)+10)+6z*2z+1)+10)-(2(+12*2z)+10)=0

Wy multiply elements

-14z^2+1)+10)+6z*2z+1)+10)-(2(+12*2z)+10)=0