Math, asked by Afifasagir, 1 year ago

Solve the following pair of eqaution: ax+by=1; bx+ay=2ab/a^2+b^2

Answers

Answered by ashishkhare687
5
ax+by=1................(i)
bx+ay=2ab/a² + b² .......................(ii)

ax+by=1    .................. *a
a²x + aby = a ....................(iii)

bx+ay=2ab/(a² + b²).......................* b
b ² x +aby = 2ab² /(a² + b²)..........(iv)

subtracting eq (iv) from (iii)
a²x + aby = a - [b ² x +aby = 2ab² /(a² + b²)]
a²x - b²x + aby - aby  = a -2ab² /(a² + b²)
(a² - b²)x = [a(a² + b²) - 2ab²]/(a² + b²)
(a² - b²)x = [a³ + ab² - 2ab²]/(a² + b²)
(a² - b²)x = [a³ - ab²]/(a² + b²)
x = [a(a² - b²)] / (a² + b²)(a² - b²)
x = a/ (a² + b²)

putting x in (i)
ax+by=1
a [a/ (a² + b²)] +by = 1
a²/(a² + b²) + by = 1
by = 1 - a²/(a² + b²)
by = [a² + b² - a²]/a² + b²
by = b²/a² + b²
y = b²/(a² + b²)(b)
y = b/a² + b²

x  = a/ (a² + b²)
y = b/a² + b²
Similar questions