1/2x + 1/3y = 2 ;
1/3x + 1/2y = 13/6

Given that:

(1/2x)+(1/3y) = 2 (1)

(1/3x)+(1/2y) = 13/6(2)

Put 1/x = a and 1/y = b then

(1) becomes

(a/2 )+( b/3) = 2

⇛ (3a+2b)/6 = 2

⇛ 3a+2b = 2×6

⇛ 3a+2b = 12 (3)

(2) becomes

(a/3)+(b/2)=13/6

⇛ (2a+3b)/6 = 13/6

⇛ 2a+3b=13(4)

On adding (3)&(4) then

3a+2b=12

2a+3b=13

(+)

5a+5b = 25

⇛ 5a+5b = 25

⇛ 5(a+b)=25

⇛ a+b=25/5

⇛ a+b=5 (5)

On subtracting (3) from (4)

2a+3b=13

3a+2b=12

(-)

-a+b = 1

⇛ -a+b = 1

⇛ b = 1+a (6)

On Substituting the value of b in (5) then

a+1+a = 5

⇛2a+1 = 5

⇛ 2a = 5-1

⇛ 2a = 4

⇛ a = 4/2

⇛ a = 2

On Substituting the value of a in (6) then

b = 1+2

b = 3

Now,

a = 2

⇛ 1/x=2

⇛ “x” = 1/2

and

b= 3

⇛ 1/y = 3

⇛“y” = 1/3

Answer:- Hence, the value of x and y with be 1/2 and 1/3 respectively.

The solution for the given problem is (1/2,1/3)

Check:-

If x = 1/2 and y = 1/3 then

LHS = (1/2x)+(1/3y)

= 1/2(1/2)+1/3(1/3)

= 1/(2/2)+1/(3/3)

= (1/1)+(1/1)

= 1+1

=2

= RHS

LHS=RHS is true

and

LHS=(1/3x)+(1/2y)

⇛ 1/3(1/2)+ 1/2(1/3)

⇛ 1/(3/2)+1/(2/3)

⇛ (2/3)+(3/2)

⇛ (4+9)/6

⇛ 13/6

⇛ RHS

LHS = RHS is true

also read similar questions: Simultaneous equations: 1. 2x+2y=10 X+2y=6 2. 3x+y=18 2x+y=13 3. X+y=1 X-y=5 4. 3x+4y=29 X-4y=-17 5. 4c+3y=11 2x+y=7

the axial of symmetry

step-by-step explanation:

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step-by-step explanation: