How do you solve #17=5y-3#?

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Answer 1 · 2018-05-28T20:38:35

#y = 4#

#17 = 5y - 3#

To solve for the variable #y#, we have to make it by itself.

First, add #color(blue)3# to both sides of the equation:
#17 quadcolor(blue)(+quad3) = 5y - 3 quadcolor(blue)(+quad3)#

#20 = 5y#

Now divide both sides by #color(blue)5#:
#20/color(blue)5 = (5y)/color(blue)5#

#4 = y#

Therefore, #y = 4#.

Hope this helps!