Step-by-step explanation:

The given system of equations may be written as

ax+by−c=0

bx+ay−(1+c)=0

By cross-multiplication, we have

b×−(1+c)−a×−c

x

=

a×−(1+c)−b×−c

−y

=

a×a−b×b

1

⇒

−b(1+c)+ac

x

=

−a(1+c)+bc

−y

=

a

2

−b

2

1

⇒

ac−bc−b

x

=

ac−bc+a

y

=

a

2

−b

2

1

⇒

c(a−b)−b

x

=

c(a−b)+a

y

=

(a−b)(a+b)

1

⇒x=

(a−b)(a+b)

c(a−b)−b

and y=

(a−b)(a+b)

c(a−b)+a

⇒x=

a+b

c

−

(a−b)(a+b)

b

and y=

a+b

c

+

(a−b)(a+b)

a

Hence, the SOLUTION of the given system of equations is

x=

a+b

c

−

a

2

−b

2

b

,y=

a+b

c

+

a

2

−b

2

a

.

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R.D. Sharma Solutions for CLASS 10 Maths CHAPTER 3

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Solve the given pair of linear equations

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+b

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