Question

friends please help me to solve the sums

Answer 1

ANSWER:
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Hey friend,

Refer the attachment

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Hope this helps you.

Answer 2

(1)

   4x + 3y = 4    ------------ (1)

    6x + 5y = 8   ----------- (2)

Now,

On solving (1) * 3 & (2) * 2, we get

12x + 9y = 12     ----------------- (3) 

12x + 10y = 16    ----------------- (4)

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

          -y = -4 

          y = 4.


Substitute y = 4 in (3), we get

12x + 9y = 12

12x + 9(4) = 12

12x + 36 = 12

12x = 12 - 36

12x = -24

x = -2.


Therefore the value of x = -2 and y = 4.



(2) 

3x + y = 1     ----------- (1)

2x - 11y = 3   ---------- (2)

Now,

On solving (1) * 2 & (2) * 3, we get

6x + 2y = 2    ---------------- (3)

6x - 33y = 9    ---------------- (4)

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

        35y = -7

       $y = \frac{-1}{5}$


Substitute y = -1/5 in (3), we get

6x + 2y = 2

6x + 2(-1/5) = 2

6x - 2/5 = 2

6x = 2 + 2/5

6x = 12/5

x = 2/5.


Therefore the value of x = 2/5 and y = -1/5.


Hope this helps!