6(b+1)-7(3b+2)=10+1/2(-20b+6)

Solution for 6(b+1)-7(3b+2)=10+1/2(-20b+6) equation:

6(b+1)-7(3b+2)=10+1/2(-20b+6)

We move all terms to the left:

6(b+1)-7(3b+2)-(10+1/2(-20b+6))=0


Domain of the equation: 2(-20b+6))!=0

b∈R


We multiply parentheses

6b-21b-(10+1/2(-20b+6))+6-14=0

We multiply all the terms by the denominator


6b*2(-20b+6))-21b*2(-20b-20b+6))-14*2(-20b+6))+6))-(10+1+6*2(=0

We add all the numbers together, and all the variables

6b*2(-20b+6))-21b*2(-40b+6))-14*2(-20b+6))+6))-(10+1+6*2(=0

We add all the numbers together, and all the variables

-40b+6b*2(-20b+6))-21b*2(+6))-14*2(-20b+6))+6))-(10+1+6*2(=0

Wy multiply elements

24b^3(-+6))-14*2(-40b-20b+6))+6))-(10+1+6*2(=0

We do not support ebpression: b^3(