Question

Solve for x and y :

6(ax+by)= 3a + 2b
6(bx - ay)= 3b - 2a

plzz solve it fast.....

Answer 1

hello , friend.....

here, is Ur answer ...
Sol.

The given equation are:
6(ax+by)=3a+2b
6(bx-ay)=3b-2a

Equation (1) and (2) can be re -written as.
$ax + by \: = \frac{3a + 2b}{6} \\ bx - ay \: = \frac{3b - 2a}{6}$

multiple (1a) by a and (2a) by b and adding them , we have

$( {a}^{2} x + aby) + ( {b}^{2} - aby) = \frac{3 {a}^{2} + 2ab }{6} + \frac{3 {a}^{2} - 2 ab }{6} \\ \\ = > ( {a}^{2} + {b}^{2} )x = \frac{3 {a}^{2} + 3 {b}^{2} }{6} \\ \\ = > ( {a}^{2} + {b}^{2} )x = \frac{3}{6} ( {a}^{2} + {b}^{2} ) \\ \\ = > x = \frac{1}{2} \frac{( {a}^{2} + {b}^{2} )}{( {a}^{2} + {b}^{2} )} \\ \\ = > x = \frac{1}{2}$
subtracting x = 1/2 in (1a) , we get

$\frac{1}{2} a + by = \frac{3a + 2b}{6} \\ \\ = > by \: = \frac{3a + 2b}{6} - \frac{1}{2} a \\ \\ = > by = \frac{3a + 2b - 3a}{6} \\ \\ = > by \: = \frac{2b}{6} \\ \\ = > y = \frac{1}{3} \times \frac{b}{b} \\ \\ = > y = \frac{1}{3}$
Hence, the solution of the given system of equations is x= 1/2 and y = 1/3

Hope it's helps you .
☺☺

Answer 2

6(ax + by) = 3a + 2b 

6ax + 6by = 3a + 2b  --------- (1)

6(bx - ay) = 3b - 2a

6bx - 6ay = 3b - 2a   ------------- (2)


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


6ax^2 + 6aby = 3a^2 + 2ab

6b^2x - 6aby = 3b^2 - 2ab

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

6ax^2 + 6b^2x = 3a^2 + 3b^2

6x(a^2 + b^2) = 3(a^2 + b^2)

x = 3/6

x = 1/2.


Substitute x = 1/2 in above equations, we get

6ax + 6by = 3a + 2b

6a(1/2) + 6by = 3a + 2b

3a + 6by = 3a + 2b

6by = 2b

y = 2b/6b

y = 1/3.



Therefore the value of x = 1/2 and y = 1/3.


Hope this helps!