(4/7)b-8=(2/7)b+10

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Solution for (4/7)b-8=(2/7)b+10 equation: 



(4/7)b-8=(2/7)b+10

We move all terms to the left:

(4/7)b-8-((2/7)b+10)=0



Domain of the equation: 7)b!=0

b!=0/1

b!=0

b∈R





Domain of the equation: 7)b+10)!=0

b!=0/1

b!=0

b∈R



We add all the numbers together, and all the variables

(+4/7)b-((+2/7)b+10)-8=0

We multiply parentheses

4b^2-((+2/7)b+10)-8=0

We multiply all the terms by the denominator


4b^2*7)b+10)-((-8*7)b+10)+2=0

We add all the numbers together, and all the variables

4b^2*7)b+10)-((-56)b+10)+2=0

Wy multiply elements

28b^3-56)b+10)+2=0

We do not support ebpression: b^3

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