Question

solve it fast..................................​

Answer 1

Answer:

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Step-by-step explanation:

3x+2y=11 ---------------->(1)

2x+3y=4 ----------------->(2)

Adding eq(1) and eq(2) we get,

   3x+2y=11

+  2x+3y=4

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

   5x+5y=15

Dividing both sides of the above equation we get,

 x+y=3 ----------------------->(3)

By subtracting eq(2) from eq(1),

   3x+2y=11          [ since we have to subtract the sign of eq (2) will

-   2x+3y=4             change....i.e the eq will be --> -2x-3y= -4]

--------------------      

  x - y = 7 ----------------->(4)

Now, by adding eq(3) and (4) ,

   x+y=3

+  x - y = 7

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

  2x = 10

  ∴x = 5

substituting x = 5 in eq (3) ,

x + y = 3

∴5 + y = 3

∴y = 3-5

∴ y = -2

(x,y) = (5, -2) is the solution of given simultaneous equations.

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Answer 2

Answer:

3x+2y=11 ---------------->(1)

2x+3y=4 ----------------->(2)

Adding eq(1) and eq(2) we get,

3x+2y=11

+ 2x+3y=4

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

5x+5y=15

Dividing both sides of the above equation we get,

x+y=3 ----------------------->(3)

By subtracting eq(2) from eq(1),

3x+2y=11 [ since we have to subtract the sign of eq (2) will

- 2x+3y=4 change....i.e the eq will be --> -2x-3y= -4]

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

x - y = 7 ----------------->(4)

Now, by adding eq(3) and (4) ,

x+y=3

+ x - y = 7

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

2x = 10

∴x = 5

substituting x = 5 in eq (3) ,

x + y = 3

∴5 + y = 3

∴y = 3-5

∴ y = -2

(x,y) = (5, -2) is the solution of given simultaneous equations