Solve the following system of equations using cross-multiplication mathod:-

1.  ax+by-(a-b)=0  ;  bx-ay-(a+b)=0

 The given equation is 

ax + by - (a - b) = 0

 bx -ay - (a + b) = 0

Then  x/ ( -b) (a + b) - (a) (a - b) = -y/ (-a) (a+ b) + b (a - b) = 1/( -a square - b square)

or x/ -ab -b square - a square + ab = -y/ - a square - ab +ab -b square = 1/ (-a square - b square)

or x/ -a square - b square = - y/ - a square - b square = 1/ -a square - b square

Hence x = 1 and y = -1

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ax + by - (a - b) = 0

bx -ay - (a + b) = 0

[ a1 = a ]  [ b= b ]  [ c= -(a - b) ]

[ a2 = b ]  [ b2 = -a ]  [ c2 = -(a + b) ]

x / (b1c2 - b2c1) = y / (c1a2 - c2a1) = 1 / (a1b2 - a2b1)

x / (-ab - b2 - a2 + ab) = y / (-ab + b2 + a2 + ab) = 1 / -a2 - b2

x / -a2 - b2 = y / a2 + b2 = 1 / -a2 - b2

x / -a2 - b2 = 1 / -a2 - b2 

So,  x = 1

y / a2 + b2 = 1 / -1(a2 + b2 )

multiply both side by -1

-y / a2 + b2 = 1 / a2 + b2 

or  -y = 1

So,  y = -1

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