Question

2/y + 5/x = 19 , 5/y - 3/x = 1 using substitution method​

Answer 1

Answer:

x = 1/3, y = 1/2

Step-by-step explanation:

Let 1/x = u and 1/y = v

So the equations become

2v + 5u = 19 and 5v - 3u = 1

2v + 5u = 19

v = (19 - 5u)/2

Substituting the value of v in the second equation, we get,

5(19 - 5u)/2 - 3u = 1

95 - 25u - 6u = 2 [multiplying by 2 on both sides]

95 - 31u = 2

93 = 31u

93/31 = u

3 = u

Thus, v = {(19 - 5(3)}/2 = 4/2 = 2

u = 1/x = 3 => x = 1/3

v = 1/y = 2 => y = 1/2

I hope you find the solution helpful :)

Answer 2

Step-by-step explanation:

Given :-

2/y + 5/x = 19 ,

5/y - 3/x = 1

To find :-

Solve by using substitution method ?

Solution :-

Given equations are :

2/y + 5/x = 19

= > (5/x) +(2/y) = 19

=> 5(1/x) +2(1/y) = 19 -------(1)

and

5/y - 3/x = 1

=> (-3/x) +(5/y) = 1

=> -3(1/x) +5(1/y) = 1

=> 3(1/x)-5(1/y) = -1 --------(2)

Put 1/x = a and 1/y = b then (1) and (2) becomes

5a +2b=19----------(3)

and

3a -5b = -1 -----------(4)

=> 3 a = 5b-1

=> a = (5b-1)/3 --------(5)

On Substituting the value of a in (3)

=> 5[(5b-1)/3] +2b = 19

=> [(25b-5)/3]+2b = 19

=> (25b-5+6b)/3 = 19

=> (31b-5 )/3 = 19

=> 31b-5 = 19×3

=> 31b -5 = 57

=> 31b = 57+5

=> 31b = 62

=> b = 62/31

=> b = 2

On Substituting the value of b in (5)

a = [5(2)-1]/3

=> a = [10-1]/3

=> a =9/3

=> a = 3

Now we have,

a = 3

=> 1/x = 3

=> x = 1/3

and

b = 2

=> 1/y = 2

=> y = 1/2

Therefore,x = 1/3 and y = 1/2

Answer:-

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

Check :-

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

LHS = 2/y + 5/x

=> 2/(1/2)+5/(1/3)

=> (2×2)+(5×3)

=> 4+15

= 19

LHS= RHS

and

5/y - 3/x

=> 5/(1/2) - 3/(1/3)

=> (5×2) -(3×3)

=> 10-9

=> 1

=> LHS = RHS

Verified the given relations in the given problem.

Used Method:-

Substitution method