Question

(i) ax + by = C
bx + ay =1+ c​

Answer 1

Answer:

Step-by-step explanation:

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

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

Therefore,

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Answer:

Answer:

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

Step-by-step explanation:

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

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Answer 2

Step-by-step explanation:
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

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

Therefore,




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