How do you solve #\frac { 5x + 7} { 4x - 1} = 2#?

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Answer 1 · 2016-11-18T16:50:51

#x = 3#

#(5x+7)/(4x-1) = 2/1#

There is one fraction on each side of an equal sign, so you can cross-multiply. This will get rid of the fractions

#2(4x-1) = 5x+7#

#8x-2 = 5x+7#

#8x -5x = 7+2" "larr# re-arrange the equation

#3x =9#

#x = 3#