How do you solve #2=(5p-7)/4#?

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Answer 1 · 2016-07-12T00:28:17

p=3

Multiply 4 on both sides to get rid of the denominator:

#4xx2=(5p-7)/cancel4##xxcancel4#

Add 7 to both sides to get the constant by itself:

#8 = 5p-7#
#+7 color(white)(aaa) +7#

#15=5p#

Divide both sides by 5 to find p:
#15/5=(cancel5p)/cancel5#

#3=p#