bx + ay = a + b

ax ( 1/a-b - 1/a+b ) + by ( 1/b-a - 1/b+a ) = 2 ; a is not equal to b, a is not equal to -b

Question

bx + ay = a + b
ax ( 1/a-b - 1/a+b ) + by ( 1/b-a - 1/b+a ) = 2 ; a is not equal to
b, a is not equal to -b

Ajanta Trivedi · Accepted Answer

the given equations are:

$b x + a y = a + b ................. (1) a x (\frac{1}{a - b} - \frac{1}{a + b}) + b y (\frac{1}{b - a} - \frac{1}{b + a}) = 2 w h e r e a \neq b a n d a \neq - b a x (\frac{a + b - (a - b)}{a^{2} - b^{2}}) + b y (\frac{b + a - (b - a)}{b^{2} - a^{2}}) = 2 a x (\frac{2 b}{a^{2} - b^{2}}) + b y (\frac{2 a}{b^{2} - a^{2}}) = 2 2 a b x - 2 a b y = 2 (a^{2} - b^{2})$ multiplying the eq by $a^{2} - b^{2}$ .

$b x - b y = \frac{(a^{2} - b^{2})}{a}$ ............(2)[dividing the eq by 2a]

subtracting the eq(2) by eq(1):

$a y + b y = (a + b) - (\frac{a^{2} - b^{2}}{a}) (a + b) y = (a + b) [1 - \frac{a - b}{a}] y = \frac{a - a + b}{a} \Rightarrow y = \frac{b}{a}$

substitute y = b/a in eq(1):

$b x + a * \frac{b}{a} = a + b b x + b = a + b b x = a x = \frac{a}{b}$

hope this helps you.

cheers!!

bx + ay = a + b ax ( 1/a-b - 1/a+b ) + by ( 1/b-a - 1/b+a ) = 2 ; a is not equal to b, a is not equal to -b

bx + ay = a + b

ax ( 1/a-b - 1/a+b ) + by ( 1/b-a - 1/b+a ) = 2 ; a is not equal to b, a is not equal to -b