Question

solve..... I didn't get a proper answer

Answer 1

x/a + y/b = 2

${\Rightarrow }$ay + xb = 2ab----------( 1 )

ax - by = a² - b²---------( 2 )

from-----( 1 ) &-------( 2 )

multiply ( 1 ) by "a" ( 2 ) by "b"

abx + a²y = 2a²b

abx - b²y = ba² - b³

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

y(a² + b²) = 2a²b - ba² + b³

${\Rightarrow }$y(a² + b²) = a²b + b³

${\Rightarrow }$y(a² + b²) = b(a² + b²)

${\Rightarrow }$y = b [ put in -----( 1 )]

we get,

${\Rightarrow }$ay + xb = 2ab

${\Rightarrow }$a(b) + xb = 2ab

${\Rightarrow }$xb = 2ab - ab

${\Rightarrow }$x = ab/b

${\Rightarrow }$x = a

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Hello

Solution:

$\frac{x}{a} + \frac{y}{b} = 2\\ \\ \frac{bx+ay}{ab} =2\\ \\ bx+ay = 2ab -----eq1\\ ax-by = a^{2} -b^{2} \\ \\ --------eq2$

Elimination method:

multiply eq 1 by a and eq2 by b and subtract both eqs

$abx+a^{2}y =2a^{2} b\\ \\ abx-b^{2}y = a^{2}b - b^{3} \\ \\ abx+a^{2}y -abx+b^{2}y =2a^{2} b - a^{2}b + b^{3}\\ \\ (a^{2}+b^{2}) y = a^{2}b + b^{3} \\ \\ (a^{2}+b^{2}) y = (a^{2} + b^{2}) b \\ \\ y = \frac{(a^{2} + b^{2}) b}{(a^{2} + b^{2}) } \\ \\ y = b -------eq3$

Now put the vlaue of y = b from eq 3 into eq1 or in eq2

$bx+ay = 2ab \\ \\ bx + ab = 2ab\\ \\ bx = 2ab-ab\\ \\ bx = ab\\ \\ x =\frac{ab}{b} \\ \\ x = a -----eq4$

By this way we get y =b and x = a are solution of pair of linear equation in two variable.

Hope it helps you.