7-0.5(8y-3)=10-4/5(3-4y)

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Solution for 7-0.5(8y-3)=10-4/5(3-4y) equation: 



7-0.5(8y-3)=10-4/5(3-4y)

We move all terms to the left:

7-0.5(8y-3)-(10-4/5(3-4y))=0



Domain of the equation: 5(3-4y))!=0

y∈R



We add all the numbers together, and all the variables

-0.5(8y-3)-(10-4/5(-4y+3))+7=0

We multiply parentheses

-4y-(10-4/5(-4y+3))+1.5+7=0

We multiply all the terms by the denominator


-4y*5(-4y+3))-(10+(1.5)*5(-4y+3))+7*5(-4y-4+3))=0

We add all the numbers together, and all the variables

-4y*5(-4y+3))-(10+(1.5)*5(-4y+3))+7*5(-4y-1))=0

We add all the numbers together, and all the variables

-4y-4y*5(-4y+3))-(10+(1.5)*5(-4y+3))+7*5(=0

Wy multiply elements

-20y^2(-+(1.5)*5(-4y+3))-4y+7*5(=0

We add all the numbers together, and all the variables

-20y^2(-+(1.5)*5(-4y+3))-4y=0

We use the square of the difference formula

-20y^2(-(1.5)*5(-4y+3))-4y=0

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