(7/8)b-32=(3/4)b+22

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Solution for (7/8)b-32=(3/4)b+22 equation:



(7/8)b-32=(3/4)b+22
We move all terms to the left:
(7/8)b-32-((3/4)b+22)=0
Domain of the equation: 8)b!=0
b!=0/1
b!=0
b∈R
Domain of the equation: 4)b+22)!=0
b!=0/1
b!=0
b∈R
We add all the numbers together, and all the variables
(+7/8)b-((+3/4)b+22)-32=0
We multiply parentheses
7b^2-((+3/4)b+22)-32=0
We multiply all the terms by the denominator
7b^2*4)b+22)-((-32*4)b+22)+3=0
We add all the numbers together, and all the variables
7b^2*4)b+22)-((-128)b+22)+3=0
Wy multiply elements
28b^3-128)b+22)+3=0
We do not support ebpression: b^3

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