Answer:
The sum of the squares of the numbers is 7.
Step-by-step explanation:
Let's denote the two numbers as [tex]x[/tex] and [tex]y[/tex]. According to the given conditions:
The sum of the two numbers is 3:
[tex] x + y = 3 [/tex]
The sum of their reciprocals is also 3:
[tex] \dfrac{1}{x} + \dfrac{1}{y} = 3 [/tex]
We need to find the sum of the squares of the numbers, [tex]x^2 + y^2[/tex].
We can start by squaring the first equation [tex]x + y = 3[/tex]:
[tex] (x + y)^2 = 3^2 [/tex]
[tex] x^2 + 2xy + y^2 = 9 [/tex]
Next, we can rewrite the second equation ([tex] \dfrac{1}{x} + \dfrac{1}{y} = 3 [/tex]) with a common denominator and then simplify:
[tex] \dfrac{y + x}{xy} = 3 [/tex]
[tex] \dfrac{3}{xy} = 3 [/tex]
[tex] 3 = 3xy [/tex]
[tex] xy = 1 [/tex]
Now, we have [tex]x + y = 3[/tex] and [tex]xy = 1[/tex]. We can use these to find [tex]x^2 + y^2[/tex].
We know that:
[tex] (x + y)^2 = x^2 + 2xy + y^2 = 9 [/tex]
[tex] x^2 + 2xy + y^2 = 9 [/tex]
[tex] x^2 + 2(1) + y^2 = 9 [/tex]
[tex] x^2 + 2 + y^2 = 9 [/tex]
[tex] x^2 + y^2 = 9 - 2 [/tex]
[tex] x^2 + y^2 = 7 [/tex]
So, the sum of the squares of the numbers is 7.
