How do you solve #3x+14/(5x)=31/5#?

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Answer 1 · 2018-03-17T17:15:43

#x=2/3, 7/5#

#(15x^2)/(5x)+14/(5x)=(31x)/(5x) rarr# Find the least common denominator so that the fractions can be eliminated

#15x^2+14=31x#

#15x^2-31x+14=0 rarr# Now multiply 15 and 14 together (it equals 210) and think of two numbers that add to -31 and multiply to 210

#15x^2-21x-10x+14=0#

#3x(5x-7)-2(5x-7)=0#

#(3x-2)(5x-7)=0#

#3x-2=0#

#3x=2#

#x=2/3#

#5x-7=0#

#5x=7#

#x=7/5#