Answer:
[tex]x>\frac{9}{5}[/tex]
Step-by-step explanation:
To reduce any confusion later on in the problem the first step should getting the fractions to have an equal denominator, this can be done by finding the LCF, or Least Common Factor.
In this case the least common factor is [tex]15[/tex], [tex]3[/tex] × [tex]5 = 15[/tex], and [tex]5[/tex] × [tex]3 = 15[/tex]; whatever you do to the denominator has to be done to the numerator
[tex]\frac{2}{3}[/tex] × [tex]\frac{5}{5} = \frac{10}{15}[/tex] [tex]\frac{1}{5}[/tex] × [tex]\frac{3}{3} =\frac{3}{15}[/tex]
Now that the denominators are equal, lets input the values into the original equation:
[tex]\frac{10}{15} x-\frac{3}{15} >1[/tex]
add [tex]\frac{3}{15}[/tex] to both sides, to isolate the variable
[tex]\frac{10}{15}x > 1\frac{3}{15}[/tex] [tex]or \frac{18}{15}[/tex]
divide
both sides by [tex]\frac{10}{15}[/tex], when you divide fractions you multiply by the reciprocal of the divisor, then you multiply the numerators by the numerators and the denominators by the denominators. [tex]18[/tex] × [tex]15[/tex][tex]=270[/tex] , [tex]15[/tex] × [tex]10=150[/tex]
[tex]x>\frac{18}{15}[/tex] × [tex]\frac{15}{10}[/tex] ⇒ [tex]x>\frac{270}{150}[/tex]
[tex]270[/tex] and [tex]150[/tex] have a GCF, or Greatest Common Factor of [tex]30[/tex] which simplifies the answer to:
[tex]x>\frac{9}{5}[/tex]
