(8/9)z+2=(1/3)(4z-6)

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Solution for (8/9)z+2=(1/3)(4z-6) equation: 



(8/9)z+2=(1/3)(4z-6)

We move all terms to the left:

(8/9)z+2-((1/3)(4z-6))=0



Domain of the equation: 9)z!=0

z!=0/1

z!=0

z∈R





Domain of the equation: 3)(4z-6))!=0

z∈R



We add all the numbers together, and all the variables

(+8/9)z-((+1/3)(4z-6))+2=0

We multiply parentheses

8z^2-((+1/3)(4z-6))+2=0

We multiply parentheses ..

8z^2-((+4z^2+1/3*-6))+2=0

We multiply all the terms by the denominator


4z^2+8z^2*3*+1-6))-((+2*3*-6))=0

We add all the numbers together, and all the variables

4z^2+8z^2*3*+1-6))-((-36))=0

We add all the numbers together, and all the variables

4z^2+8z^2*3*=0

Wy multiply elements

4z^2+24z^3=0

We do not support ezpression: z^3

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