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

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


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

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

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

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

Reorder the terms in parentheses
+10z+30+(-3z+6)=4*(z-3)+2*(z-2)

Remove unnecessary parentheses
+10z+30-3z+6=+4+*(+z-3+)+2+*(+z-2+)

Reorder the terms in parentheses
+10z+30-3z+6=+(+4z-12)+2*(z-2)

Remove unnecessary parentheses
+10z+30-3z+6=+4z-12+2+*(+z-2+)

Reorder the terms in parentheses
+10z+30-3z+6=+4z-12+(+2z-4)

Remove unnecessary parentheses
+10z+30-3z+6=+4z-12+2z-4

We move all terms containing z to the left and all other terms to the right. 
+10z-3z-4z-2z=-12-4-30-6

We simplify left and right side of the equation. 
+1z=-52

We divide both sides of the equation by 1 to get z. 
z=-52

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