Answer:
For the values of y greater than [tex]\frac{3}{17}[/tex]
Step-by-step explanation:
Suppose 5y-1 greater than the value of the fraction [tex]\frac{3y-1}{4}[/tex]
That is,
[tex]5y - 1 > \frac{3y-1}{4}[/tex]
[tex]4(5y - 1) > 3y - 1[/tex]
[tex]20y - 4 > 3y - 1[/tex]
[tex]20y - 3y > -1 + 4[/tex]
[tex]17y > 3[/tex]
[tex]\implies y>\frac{3}{17}[/tex] ( a > b ⇒ [tex]\frac{a}{d} > \frac{b}{d}[/tex] where, d > 0 )
Hence, for the values of y greater than [tex]\frac{3}{17}[/tex] the value of the binomial 5y-1 greater than the value of the fraction [tex]\frac{3y-1}{4}[/tex].
