Question

For what values of y is the value of the fraction 5−2y 12 greater than the value of the binomial 1−6y?

Answer 1

The question is ambiguous, since it's not clear how we should consider the fraction. Anyway, I see two alternatives:

`(5-2y)/(12) > 1-6y`

Multiply both sides by 12:

`5-2y > 12-72y`

Add 72y to both sides:

`5+70y > 12`

Subtract 5 from both sides:

`70y > 7`

Divide both sides by 70:

`y > (1)/(10)`

If, instead, you meant

`5-(2y)/(12) > 1-6y`

we proceed as follows: subtract 5 from both sides

`-(2y)/(12) > -4-6y`

Switch signs and inequality mark:

`(2y)/(12) < 4+6y`

Multiply both sides by 12:

`2y < 48+72y`

Subtract 2y from both sides:

`0 < 48+70y`

subtract 48 from both sides:

`-48 < 70 y`

Divide both sides by 70:

`y > -(48)/(70)`

Answer 2

For all the values of y>0.1 the fraction
`(5-2y)/(12)` is greater than the binomial
`1-6y`

Explanation:

We have to find the values of y where
`(5-2y)/(12)>1-6y`. The first step we are going to do is multiply by 12 in both sides of the expression.

`(5-2y)/(12).12>(1-6y).12\\5-2y>(1-6y).12`

Now we are going to use distributive property on the right side,

Observation: distributive property: (b+c)a=ba+ca

`5-2y>(1-6y).12\\\\5-2y>1.12-6y.12\\5-2y>12-72y`

Then we have to add 72y in both sides.

`5-2y>12-72y\\5-2y+72y>12-72y+72y\\5+70y>12`

Now subtract 5 from both sides,

`5+70y>12\\5+70y-5>12-5\\70y>7`

Divide both sides in 70

`70y>7\\(70y)/(70)>(7)/(70)\\y>(1)/(10)\\y>0.1`

Then for all the values of y>0.1 the fraction
`(5-2y)/(12)` is greater than the binomial
`1-6y`