Question

Please solve question no. 11th
Solve the following systems by Elimination by equating Coefficients method.
Any one can solve this fast

Answer 1

Hey!!!!

Good Afternoon

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We have

1. 2y + 5/x = 3 -----------(1)

and

2. 4y + 3/x = 23/5 ----------(2)

In (1) and (2) let 1/x = a

Thus

=> (5a + 2y = 3)--------(3)

=> 3a + 4y = 23/5 -------(4)

Multiply (3) by 2

=> 10a + 4y = 6 -----(5)

Subtract (5) from (4)

=> 3a + 4y - (10a + 4y) = 23/5 - 6

=> 3a + 4y - 10a - 4y = -7/5

=> -7a = -7/5

Cancelling -7 on both sides

=> a = 1/5

Thus x = 5 (Answer)

Using x in (1)

=> 2y + 1 = 3

=> 2y = 2

=> y = 1 (Answer)


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Hope this helps ✌️

Answer 2

11.

2y + 5/x = 3  ------ (1)

4y + 3/x = 23/5  ---- (2)

On solving (1) * 2, & (2), we get

4y + 10/x = 6

4y + 3/x = 23/5

-----------------------

       7/x = 7/5

       x = 5.


Substitute x = 5 in (1), we get

2y + 5/x = 3

2y + 5/5 = 3

2y + 1 = 3

2y = 2

y = 1.


Therefore y = 1 and x = 5


Hope this helps!