Question

Hi frds !!

Solve by elimination method

◆ax +by =c
bx + ay = 1 +c

$\frac{x}{a} - \frac{y}{b } = 0 \\ \\ ax + by = {a}^{2} + {b}^{2}$

Answer 1

$\boxed { \bf Heya \: Folk } \\ \\ \boxed {\bf see \: attached \: pictures } \:$

Hope this helps you :)

Answer 2

Hey!

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1)

ax + by = c ------- (i)
bx + ay = 1 + c -------- (ii)

Now,

Multiply the equation (i) by 'b' and (ii) by 'a'

b (ax + by = c)
a (bx + ay = 1 + c)

abx + b^2y = bc ----- (iii)
abx + a^2y = a + ac ------ (iv)

Subtract (iii) and (iv)
We get,

(b^2 - a^2) y = bc - ac - a

y = bc - ac - a / b^2 - a^2

Now,

Multiply equation (i) by 'a' and equation (ii) by 'b'

a (ax + by = c)
b (bx + ay = 1 + c)

a^2x + aby = ac ----- (v)
b^2x + aby = b + bc ---- (vi)

Now, subtract (v) and (vi)
We get,

a^2x - b^2x = ac - bc - b
(a^2 - b^2) x = ac - bc - b

x = ac - bc - b / a^2 - b^2


VALUES OF 'X' AND 'Y' are -:

x = ac - bc - b / a^2 - b^2
y = bc - ac - a / b^2 - a^2


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2)

x / a - y/b = 0

We can also write it as -

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

Now,

Multiply equation (i) by 'a' and equation (ii) by 'b'


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

abx - a^2 y = 0 ----- (iii)
abx + b^2 y = a^2b + b^3 ----- (iv)

Now Subract them,
We get,

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

y = b

Putting the value in (i)

bx - ab = 0

x = ab/b
x = a

VALUES OF 'X' AND 'Y' are -:

x = a
y = b




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Hope it helps...!!!

[ I would have solved on paper, but unfortunately not having any paper pencil at present ]