Question

solve x and y: bx + ay -2ab=0, ax -by -A square + b square =0

Answer 1

GIVEN
bx + ay - 2ab = 0 ........... (1)
ax - by - a^2 + b^2 = 0 .... (2)

MULTIPLY
Equation (1) by 'a' & (2) by 'b'

ACHIEVE
CASE 1.
(bx + ay - 2ab = 0) a
abx + a^2y - 2a^2b = 0

CASE 2.
( ax - by - a^2 + b^2 = 0) b
abx - b^2y - a^2b + b^3 = 0
abx - b^2y - b (a^2 - b^2) = 0

SUBTRACT (CASE 2) from (CASE 1)

[abx + a^2y - 2a^2b = 0] - [abx - b^2y - b (a^2 - b^2) = 0]
= (abx - abx) + (a^2y + b^2y) - [2a^2b - b(a^2 - b^2)] = 0
y (a^2 + b^2) - b(2a^2 - a^2 + b^2) = 0
y (a^2 + b^2) - b(a^2 + b^2) = 0
y (a^2 + b^2) = b(a^2 + b^2)
y = [b(a^2 + b^2)] / (a^2 + b^2)
y = b ........................ (3)

SUBSITUTE (3) in (1)

bx + a x b - 2ab = 0
bx - ab = 0
bx = ab
x = ab / b
x = a ......... (4)

Therefore, the values from (3) & (4) are
y = b & x = a

Answer 2

GIVEN

bx + ay - 2ab = 0 ........... (1)

ax - by - a^2 + b^2 = 0 .... (2)

MULTIPLY

Equation (1) by 'a' & (2) by 'b'

ACHIEVE

CASE 1.

(bx + ay - 2ab = 0) a

abx + a^2y - 2a^2b = 0

CASE 2.

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

abx - b^2y - a^2b + b^3 = 0

abx - b^2y - b (a^2 - b^2) = 0

SUBTRACT (CASE 2) from (CASE 1)

[abx + a^2y - 2a^2b = 0] - [abx - b^2y - b (a^2 - b^2) = 0]

= (abx - abx) + (a^2y + b^2y) - [2a^2b - b(a^2 - b^2)] = 0

y (a^2 + b^2) - b(2a^2 - a^2 + b^2) = 0

y (a^2 + b^2) - b(a^2 + b^2) = 0

y (a^2 + b^2) = b(a^2 + b^2)

y = [b(a^2 + b^2)] / (a^2 + b^2)

y = b ........................ (3)

SUBSITUTE (3) in (1)

bx + a x b - 2ab = 0

bx - ab = 0

bx = ab

x = ab / b

x = a ......... (4)

Therefore, the values from (3) & (4) are

y = b & x = a

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