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

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Solution for 11(z+4)-3(z-1)=3(z-4)+4(z-2) equation:


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

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

Reorder the terms in parentheses
+(+11z+44)-3*(z-1)=3*(z-4)+4*(z-2)

Remove unnecessary parentheses
+11z+44-3+*(+z-1+)=+3+*(+z-4+)+4+*(+z-2+)

Reorder the terms in parentheses
+11z+44+(-3z+3)=3*(z-4)+4*(z-2)

Remove unnecessary parentheses
+11z+44-3z+3=+3+*(+z-4+)+4+*(+z-2+)

Reorder the terms in parentheses
+11z+44-3z+3=+(+3z-12)+4*(z-2)

Remove unnecessary parentheses
+11z+44-3z+3=+3z-12+4+*(+z-2+)

Reorder the terms in parentheses
+11z+44-3z+3=+3z-12+(+4z-8)

Remove unnecessary parentheses
+11z+44-3z+3=+3z-12+4z-8

We move all terms containing z to the left and all other terms to the right.
+11z-3z-3z-4z=-12-8-44-3

We simplify left and right side of the equation.
+1z=-67

We divide both sides of the equation by 1 to get z.
z=-67

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