Determinati numerele naturale x si y stiind ca :
a.sunt invers proportionale cu 2 si 3 ,iar x-y=10;
b.sunt invers proportionale cu 3 si 4,iar x-y=63;
c.sunt invers proportionele cu 2 si 3,iar [tex] x^{2} [/tex] + [tex]y 2[/tex] =208
d.sunt invers proportionale cu 3 si 4,iar [tex] x^{2} [/tex] + [tex]y ^{2} [/tex] = 175
e.sunt invers proportionale cu  3 si 5,iar xy=135

Răspuns :

a)x-y=10                     {x,y}ip{2,3}⇒2x=3y=k
2x=k⇒x=k/2
3y=k⇒y=k/3
k/2-k/3=10⇒    (3k-2k)/6=10
k/6=10⇒    k=6·10⇒  k=60
x=60/2=30
y=60/3=20
b) x-y=63
3x=4y=k⇒  
 3x=k⇒x=k/3  si 4y=k⇒y=k/4
k/3-k/4=63⇒    (4k-3k)/12=63
k=63·12=756
x=756/3⇒x=253
y=756/4⇒y=189
c)x²+y²=208
2x=3y=k
x=k/2 si y=k/3
k²/4+k²/9=208
(9k²+4k²)/36=208
13k²/36=208
k²=208·36/13
k²=576
k=√576⇒k=24
x=24/2⇒x=12
y=24/3⇒y=8
d) x²+y²=175
3x=4y=k
3x=k⇒x=k/3
4y=k⇒y=k/4
k²/9+k²/16=175
(16k²+9k²)144=175
25k²/144=175
k²=175·144/25
k²=1008
k=√1008⇒k=31,75
x=31,75/3⇒x=10,58
y=31,75/4⇒y=7,93
e) xy=135
3x=5y=k
3x=k⇒x=k/3
5y=k⇒y=k/5
k/3·k/5=135
k²/15=135
k²=135·15
k²=2025
k=√2025⇒k=45
x=45/3⇒x=15
y=45/5⇒y=9