Question

solve:ax-by=cd; bx-ay=ef​

Answer 1

Answer:

Given system of equations:

ax+by=c ---(1)

bx+ay=1+c ----(2)

multiply equation (1) by a, and equation (2) by b , we get

a²x+aby=ac---(3)

b²x+aby=b+bc---(4)

Subtract (4) from (3) , we get

(a²-b²)x= ac-b-bc

\implies x =\frac{ac-b-bc}{a^{2}-b^{2}}⟹x=

a

2

−b

2

ac−b−bc

Now,

multiply equation (1) by b, and equation (2) by a , we get

abx+b²y=bc---(5)

abx+a²y=a+ac---(6)

Subtract (5) from (6), we get

(a²-b²)y = a+ac-bc

\implies y = \frac{a+ac-bc}{a^{2}-b^{2}}⟹y=

a

2

−b

2

a+ac−bc

Therefore,

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

a

2

−b

2

ac−b−bc

y = \frac{a+ac-bc}{a^{2}-b^{2}}y=

a

2

−b

2

a+ac−bc

Answer 2

Answer:Wrong question

Step-by-step explanation:

This is wrong