How do you solve #5( 3r - 4) = 2( 4r - 3)#?

1 Answer
smendyka
Nov 24, 2016

#r = 2#

Explanation:

First, expand the terms in parenthesis on both sides of the equation:

#15r - 20 = 8r - 6#

Next isolate the terms containing #r# on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:

#15r - 20 + 20 - 8r = 8r - 6 + 20 - 8r#

#15r - 8r - 0 = 0 - 6 + 20#

#(15 - 8)r = 14#

#7r = 14#

Finally, divide by #7# on each side of the equation to solve for #r# while keeping the equation balanced:

#(7r)/7 = 14/7#

#(cancel(7)r)/cancel(7) = 2#

#r = 2#

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