10(z+3)-3(z-1)=2(z-4)+4(z-4)

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Solution for 10(z+3)-3(z-1)=2(z-4)+4(z-4) equation:


10(z+3)-3(z-1)=2(z-4)+4(z-4)

We simplify the equation to the form, which is simple to understand 
10(z+3)-3(z-1)=2(z-4)+4(z-4)

Reorder the terms in parentheses
+(+10z+30)-3*(z-1)=2*(z-4)+4*(z-4)

Remove unnecessary parentheses
+10z+30-3+*(+z-1+)=+2+*(+z-4+)+4+*(+z-4+)

Reorder the terms in parentheses
+10z+30+(-3z+3)=2*(z-4)+4*(z-4)

Remove unnecessary parentheses
+10z+30-3z+3=+2+*(+z-4+)+4+*(+z-4+)

Reorder the terms in parentheses
+10z+30-3z+3=+(+2z-8)+4*(z-4)

Remove unnecessary parentheses
+10z+30-3z+3=+2z-8+4+*(+z-4+)

Reorder the terms in parentheses
+10z+30-3z+3=+2z-8+(+4z-16)

Remove unnecessary parentheses
+10z+30-3z+3=+2z-8+4z-16

We move all terms containing z to the left and all other terms to the right. 
+10z-3z-2z-4z=-8-16-30-3

We simplify left and right side of the equation. 
+1z=-57

We divide both sides of the equation by 1 to get z. 
z=-57

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