How do you solve #(8r+3)(5r-15)=0#?

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Answer 1 · 2017-06-02T20:01:26

See a solution process below:

To solve this problem equate each term in parenthesis to #0# and solve for #r#:

Solution 1)

#8r + 3 = 0#

#8r + 3 - color(red)(3) = 0 - color(red)(3)#

#8r + 0 = -3#

#8r = -3#

#(8r)/color(red)(8) = -3/color(red)(8)#

#(color(red)(cancel(color(black)(8)))r)/cancel(color(red)(8)) = -3/8#

#r = -3/8#

Solution 2)

#5r - 15 = 0#

#5r - 15 + color(red)(15) = 0 + color(red)(15)#

#5r + 0 = 15#

#5r = 15#

#(5r)/color(red)(5) = 15/color(red)(5)#

#(color(red)(cancel(color(black)(5)))r)/cancel(color(red)(5)) = 3#

#r = 3#

The solution is: #r = -3/8#; #r = 3#