Question

For what values of y:

Answer 1

Answer:

$y > \frac{3}{17}$

Step-by-step explanation:

To solve this problem, simply set the binomial, $5y - 1$ as greater than $\frac{3y - 1}{4}$, and solve for the value of y as you would solve an equation.

Thus:

$5y - 1 > \frac{3y - 1}{4}$

Multiply both sides by 4

$4(5y - 1) > \frac{3y - 1}{4}*4$

$4*5y - 4*1 > 3y - 1$

$20y - 4 > 3y - 1$

Subtract 3y from both sides

$20y - 4 - 3y > 3y - 1 - 3y$

$17y - 4 > -1$

Add 4 to both sides

$17y - 4 + 4 > -1 + 4$

$17y > 3$

Divide both sides by 17

$\frac{17y}{17} > \frac{3}{17}$

$y > \frac{3}{17}$

Therefore, the value of y that will make the binomial $5y - 1$ greater than $\frac{3y - 1}{4}$ is $y > \frac{3}{17}$.