Respuesta :
Answer:
[tex]x + y < 2x[/tex]
[tex]\frac{x}{y} > \frac{y}{x}[/tex]
[tex] \frac{x+y}{y} > 2[/tex]
Step-by-step explanation:
For solving the question first we need to revise the properties of inequality,
a < b ⇒ a ± c < b ± c ∀ a, b, c ∈ R,
a < b ⇒ ca < cb if c > 0,
a < b ⇒ ca > cb if c < 0
Part 1 :
Given,
x and y are two numbers ( must be positive real numbers because they are showing the heights )
Such that, x > y,
Adding x on both sides of the inequality,
x + x > y + x
2x > y + x
⇒ x + y < 2x
Thus, SECOND option is correct.
Part 2 :
∵ x > 0 ⇒ 1/x > 0,
x > y
⇒ [tex]\frac{x}{x} > \frac{y}{x}[/tex]
[tex]\implies \frac{y}{x} < 1-----(X)[/tex]
Similarly, y > 0 ⇒ 1/y > 0
[tex]\frac{x}{y}> \frac{y}{y}[/tex]
[tex]\implies \frac{x}{y} > 1----(Y)[/tex]
From equation (X) and (Y),
[tex]\frac{x}{y} > \frac{y}{x}[/tex]
Thus, FIRST option is correct.
Part 3 :
Now, x > y
x + y > 2y
Since, y > 0 ⇒ [tex]\frac{1}{y}> 0[/tex]
[tex]\implies \frac{x+y}{y} > 2[/tex]
Thus, FIRST option is correct.
