(4y+3)+(y-2=)

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Solution for (4y+3)+(y-2=) equation: 



(4y+3)+(y-2=)

We move all terms to the left:

(4y+3)+(y-2-())=0

We get rid of parentheses

4y+(y-2-())+3=0



We calculate terms in parentheses: +(y-2-()), so:

y-2-()

We add all the numbers together, and all the variables

y

Back to the equation:

+(y)



We add all the numbers together, and all the variables

5y+3=0

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

5y=-3

y=-3/5

y=-3/5

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