3/4(16y-32)=3/7(56-28y)

Solution for 3/4(16y-32)=3/7(56-28y) equation:

3/4(16y-32)=3/7(56-28y)

We move all terms to the left:

3/4(16y-32)-(3/7(56-28y))=0


Domain of the equation: 4(16y-32)!=0

y∈R



Domain of the equation: 7(56-28y))!=0

y∈R


We add all the numbers together, and all the variables

3/4(16y-32)-(3/7(-28y+56))=0

We calculate fractions


(21y(-)/(4(16y-32)*7(-28y+56)))+(-12y1/(4(16y-32)*7(-28y+56)))=0


We calculate terms in parentheses: +(21y(-)/(4(16y-32)*7(-28y+56))), so:

21y(-)/(4(16y-32)*7(-28y+56))

We add all the numbers together, and all the variables

21y0/(4(16y-32)*7(-28y+56))

We multiply all the terms by the denominator


21y0

We add all the numbers together, and all the variables

21y

Back to the equation:

+(21y)



We calculate terms in parentheses: +(-12y1/(4(16y-32)*7(-28y+56))), so:

-12y1/(4(16y-32)*7(-28y+56))

We multiply all the terms by the denominator


-12y1

We add all the numbers together, and all the variables

-12y

Back to the equation:

+(-12y)


We get rid of parentheses

21y-12y=0

We add all the numbers together, and all the variables

9y=0

y=0/9

y=0