Y1=3(2x-5)=Y2=2(3x-1)+x

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Solution for Y1=3(2x-5)=Y2=2(3x-1)+x equation: 



1=3(2Y-5)=2=2(3Y-1)+Y

We move all terms to the left:

1-(3(2Y-5))=0



We calculate terms in parentheses: -(3(2Y-5)), so:

3(2Y-5)

We multiply parentheses

6Y-15

Back to the equation:

-(6Y-15)



We get rid of parentheses

-6Y+15+1=0

We add all the numbers together, and all the variables

-6Y+16=0

We move all terms containing Y to the left, all other terms to the right

-6Y=-16

Y=-16/-6

Y=2+2/3

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