How do you solve #y+ \frac { 3} { 4} = 7#?

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Answer 1 · 2017-03-20T11:54:45

#color(red)(y=6 1/4#

#y+3/4=7#

#:.y/1+3/4=7/1#

#:.(4y+3=28)/4#

multiply L.H.S and R.H.S. by 4

#:.4y+3=28#

#:.4y=28-3#

#:.4y=25#

#:.y=25/4#

#:.color(red)(y=6 1/4#

check:

substitute#color(red)(y=6 1/4#

#:.(color(red)(6 1/4))+3/4=7#

#:.6+cancel4^color(red)1/cancel4^color(red)1=7#

#:.6+1=7#

#:.7=7#