(i) ax + by = C
bx + ay =1+ c
Answers
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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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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