Question

If bx+ay = a^2 + b^2 and ax- by = 0, then the value of (x - y) will be what?

Answer 1

Step-by-step explanation:

ay+bx=a²+b²----(1)

ax-by=0---(2)

from(2)

ax/b=y

a(ax/b)+bx=a²+b²

a²x/b+bx=a²+b²

a²x/b+bx-a²-b²=0

(a²/b+b)x-a²-b²=0

(a²/b+b)x=a²+b²

x=a²+b²/(a²/b+b) ---(A)

for y, put (A) in (2)

a[(a²+b²)/(a²/b+b)]-by=0

a/b[(a²+b²)/(a²/b+b)=y

x-y =a²+b²/(a²/b+b)-(a/b[a²+b²)/(a²/b+b)

x-y=a²+b²[1-a/b)/(a²/b+b)

x-y=a²+b²[b-a/b)/(a²+b²/b)

x-y=a²+b²(b-a)/(a²+b²)

x-y=b-a

this is your answer

Answer 2

The value of (x-y) is b-a.

Step-by-step explanation:

The given equations are

$bx + ay = a^2 + b^2$       .... (1)

$ax - by = 0$             .... (2)

We need to find the value of (x - y).

Add (equation(1) × b) and (equation(2) × a), to eliminate y.

$b(bx + ay)+a(ax-by)= b(a^2 + b^2)+a(0)$

$b^2x + aby+a^2x-aby= b(a^2+ b^2)$

$x(b^2+a^2)= b(a^2+ b^2)$

Cancel out common factors.

$x=b$

Substituting value of x=b in (2),

$a(b) - by = 0$

$ab=by$

Divide both sides by b.

$a=y$

We need to find the value of x-y.

$x-y=b-a$

Therefore, the value of (x-y) is b-a.

#Learn more

If bx+ay=a^2+b^2 and ax-by=0,then the value of (x-y) is?

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