Question

Answer kindly please

$(\frac{1}{x} +\frac{6}{y} )(x+y)=14$ is true for positive integer x and y.

Then, what is the value of $(4x+y)(\frac{1}{x} +\frac{1}{y} )$?

Answer 1

Answer:$\frac{35}{3}$ or $10$

Step-by-step explanation:

If $(\frac{1}{x} +\frac{6}{y} )(x+y)=14$,

then $1+\frac{y}{x} +\frac{6x}{y} +6=14$

Let's take $t=\frac{x}{y}$

If we substitute, we get $6t-7+\frac{1}{t} =0$

First take $t\neq 0$, then multiply $t$

The quadratic equation $6t^2-7t+1=0$ has two solution $t=\frac{1}{6}$, $t=1$

$(4x+y)(\frac{1}{x} +\frac{1}{y} )$ is expanded.

We have $4+\frac{4x}{y} +\frac{y}{x}+1$, or $4t+\frac{1}{t} +5$

If we consider $t=\frac{1}{6}$

The value will be $\frac{35}{3}$

If we consider $t=1$

The value will be $10$

Answer 2

Answer:

\frac{35}{3}335 or 1010

Step-by-step explanation:

If (\frac{1}{x} +\frac{6}{y} )(x+y)=14(x1+y6)(x+y)=14 ,

then 1+\frac{y}{x} +\frac{6x}{y} +6=141+xy+y6x+6=14

Let's take t=\frac{x}{y}t=yx

If we substitute, we get 6t-7+\frac{1}{t} =06t−7+t1=0

First take t\neq 0t=0 , then multiply tt

The quadratic equation 6t^2-7t+1=06t2−7t+1=0 has two solution t=\frac{1}{6}t=61 , t=1t=1

(4x+y)(\frac{1}{x} +\frac{1}{y} )(4x+y)(x1+y1) is expanded.

We have 4+\frac{4x}{y} +\frac{y}{x}+14+y4x+xy+1 , or 4t+\frac{1}{t} +54t+t1+5

If we consider t=\frac{1}{6}t=61

The value will be \frac{35}{3}335

If we consider t=1t=1

The value will be 1010

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