Answer:
[tex]x + y = \frac{5}{6} [/tex]
Step-by-step explanation:
Let:
[tex]u = \frac{1}{x} \\ v = \frac{1}{y} [/tex]
This allows us to manipulate the equations like we normally would.
[tex]2u + 3v = 13 \\ 5u - 4v = - 2 \\ \\ + 8u + 12v = 52 \\ + 15u - 12v = - 6 \\ 23u = 46 \\ u = 2 \\ \\ 2(2) + 3v = 13 \\ 4 + 3v = 13 \\ 3v = 9 \\ v = 3[/tex]
Then, we return the values where they belong.
[tex]u = \frac{1}{x} \\ 2 = \frac{1}{x} \\ x = \frac{1}{2} \\ \\ v = \frac{1}{y} \\ 3 = \frac{1}{y} \\ y = \frac{1}{3} [/tex]
Finally, we add:
[tex] \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} [/tex]
Given that
(2/x)+(3/y) = 13--------(1)
(5/x)-(4/y) = -2 -------(2)
Put 1/x = a and 1/y = b then
2a + 3b = 13 ----------(3)
On multiplying with 5 then
10a +15 b = 65 -------(4)
and
5a -4b= -2 ----------(5)
On multiplying with 2 then
10 a - 8b = -4 -------(6)
On Subtracting (6) from (4) then
10a + 15b = 65
10a - 8b = -4
(-)
_____________
0 + 23 b = 69
______________
⇛ 23b = 69
⇛ b = 69/23
⇛ b =3
On Substituting the value of b in (5)
5a -4b= -2
⇛ 5a -4(3) = -2
⇛ 5a -12 = -2
⇛ 5a = -2+12
⇛ 5a = 10
⇛ a = 10/5
⇛ a = 2
Now we have
a = 2
⇛1/x = 2
⇛ x = 1/2
and
b = 3
⇛1/y = 3
⇛ y = 1/3
Answer :-The solution for the given problem is (1/2,1/3)
Check: If x = 1/2 and y = 1/3 then
LHS = (2/x)+(3/y)
= 2/(1/2)+3/(1/3)
= (2×2)+(3×3)
= 4+9
= 13
= RHS
LHS=RHS is true
and
LHS=(5/x)-(4/y)
⇛ 5/(1/2)- 4/(1/3)
⇛(5×2)-(4×3)
⇛ 10-12
⇛ -2
⇛RHS
LHS = RHS is true
Now,
X+Y
= (1/2) + (1/3)
Take the LCM of 2 and 3 is 6.
= (3+2)/6
= 5/6.
