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:

$\dfrac{5-2y}{12} > 1-6y$

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:

$y > \dfrac{1}{10}$

If, instead, you meant

$5-\dfrac{2y}{12} > 1-6y$

we proceed as follows: subtract 5 from both sides

$-\dfrac{2y}{12} > -4-6y$

Switch signs and inequality mark:

$\dfrac{2y}{12} < 4+6y$

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:

$y > -\dfrac{48}{70}$

Answer 2

Answer:

For all the values of y>0.1 the fraction $\frac{5-2y}{12}$ is greater than the binomial $1-6y$

Step-by-step explanation:

We have to find the values of y where $\frac{5-2y}{12}>1-6y$. The first step we are going to do is multiply by 12 in both sides of the expression.

$\frac{5-2y}{12}.12>(1-6y).12\\5-2y>(1-6y).12$

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

$70y>7\\\frac{70y}{70}>\frac{7}{70}\\y>\frac{1}{10}\\y>0.1$

Then for all the values of y>0.1 the fraction $\frac{5-2y}{12}$ is greater than the binomial $1-6y$