Solve:
`ax - by = 1`
`bx + ay = 0`

Solve: ax + by = a - b bx - ay = a + b

The Co-ordinations of the point of intersection of lines ax + by =9 and bx + ay =5 is ( 3,-1) find the values of a and b.

Solve the following linear equations: ax + by = c bx + ay = 1+c

{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}

Find the value of x and y by solving given linear equations ax + by = a^2 ; bx + ay = b^2

If a !=b , then the system of equation ax + by + bz = 0 bx + ay + bz = 0 and bx + by + az = 0 will have a non-trivial solution, if

For a gt b gt c gt 0 if the distance between (1,1) and the point of intersection of the lines ax + by +c=0 and bx + ay+c=0 is less than 2sqrt2 then

The base BC of a triangle ABCis bisected at the point (a, b) and equation to the sides AB and AC are respectively ax + by=1 and bx + ay = 1. Equation of the median through A is

Solve: ax+by=a-b,quad bx-ay=a+b