Question

(i) ax + by=c
bx + ay = 1 + c​

Answer 1

Answer:

Step-by-step explanation:

ax + by = c       . . . . . . . . .(1)

bx + ay = 1 + c  . . . . . . . . .(2)

Add (1) and (2)

(a+b) x + (a +b) y = 1 + 2c

∴ x + y = $\frac{(1 + 2c)}{(a + b)}$ . . . . . . . . . .(3)

Substract (2) from (1)

(a -b) x - (a -b) y = -1

∴ x - y = $\frac{ - 1}{(a - b)}$      . . . . . . . . .(4)

Add (3) and (4)

2x = $\frac{( 1 + 2c )}{(a + b)}$ - $\frac{1}{(a - b)}$

    = $\frac{a - b + 2ac - 2bc - a - b}{(a+b) (a -b )}$

    = $\frac{ 2(ac - b - bc)}{(a^{2} - b^{2}) }$

 x = $\frac{(ac - b - bc)}{(a^{2} - b^{2} )}$

From (1)

 y = $\frac{( 1 + 2c )}{a + b)}$  - $\frac{( ac - b - bc )}{(a^{2} - b^{2} )}$

    = $\frac{(1 + 2c) (a - b) - (ac - b- bc)}{(a^{2} - b^{2} )}$

    = $\frac{a - b + 2ac - 2bc - ac + b + bc}{(a^{2} - b^{2} )}$

    = $\frac{( a + ac - bc)}{(a^{2} - b^{2} )}$

Answer 2

Step-by-step explanation:

Given :

ax + by = c----(i)

bx + ay = 1 + c ---(ii)

To solve :

For x and y.

Solution :

multiply equation (i) by a and equation (ii) by b and subtracting (ii) from (i) we get,

a^2 x + aby - b^2 x - bay = ac - b - bc

=> x ( a^ - b^2 ) = c ( a - b) - b

=> x = [ c (a - b) - b ]/ (a^2 - b^2)]

Similarly, multiply equation (i) by b and equation (ii) by a and subtracting (ii) from (i) we get,

abx + b^2 y - abx - a^2 y = bc - a - ac

=> y (b^2 - a^2) = c(b - a) - a

=> y = [ c(b - a) - a]/(b^2 - a^2)