15(z+1)-4(z-3)=4(z-1)+6(z-3)

Solution for 15(z+1)-4(z-3)=4(z-1)+6(z-3) equation:

15(z+1)-4(z-3)=4(z-1)+6(z-3)

We move all terms to the left:

15(z+1)-4(z-3)-(4(z-1)+6(z-3))=0

We multiply parentheses

15z-4z-(4(z-1)+6(z-3))+15+12=0


We calculate terms in parentheses: -(4(z-1)+6(z-3)), so:

4(z-1)+6(z-3)

We multiply parentheses

4z+6z-4-18

We add all the numbers together, and all the variables

10z-22

Back to the equation:

-(10z-22)


We add all the numbers together, and all the variables

11z-(10z-22)+27=0

We get rid of parentheses

11z-10z+22+27=0

We add all the numbers together, and all the variables

z+49=0

We move all terms containing z to the left, all other terms to the right

z=-49