12(z+2)-4(z-3)=3(z-2)+4(z-4)

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Solution for 12(z+2)-4(z-3)=3(z-2)+4(z-4) equation:


12(z+2)-4(z-3)=3(z-2)+4(z-4)

We simplify the equation to the form, which is simple to understand 
12(z+2)-4(z-3)=3(z-2)+4(z-4)

Reorder the terms in parentheses
+(+12z+24)-4*(z-3)=3*(z-2)+4*(z-4)

Remove unnecessary parentheses
+12z+24-4+*(+z-3+)=+3+*(+z-2+)+4+*(+z-4+)

Reorder the terms in parentheses
+12z+24+(-4z+12)=3*(z-2)+4*(z-4)

Remove unnecessary parentheses
+12z+24-4z+12=+3+*(+z-2+)+4+*(+z-4+)

Reorder the terms in parentheses
+12z+24-4z+12=+(+3z-6)+4*(z-4)

Remove unnecessary parentheses
+12z+24-4z+12=+3z-6+4+*(+z-4+)

Reorder the terms in parentheses
+12z+24-4z+12=+3z-6+(+4z-16)

Remove unnecessary parentheses
+12z+24-4z+12=+3z-6+4z-16

We move all terms containing z to the left and all other terms to the right. 
+12z-4z-3z-4z=-6-16-24-12

We simplify left and right side of the equation. 
+1z=-58

We divide both sides of the equation by 1 to get z. 
z=-58

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