`ax+by=1`
`bx+ay=2`

ax+by=5 bx+ay=3

ax+by=2; bx+ay=3

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

If ax+by=6,bx-ay=2 and x^(2)+y^(2)=4, then the value of (a^(2)+b^(2)) would be:

If ax+by=6, bx-ay=2and x^(2)+y^(2)=4 then what is (a^(2)+b^(2)) ?

Solve the following system of equations in x and yax+by=1bx+ay=((a+b)^(2))/(a^(2)+b^(2))-1 or,bx+ay=(2ab)/(a^(2)+b^(2))

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

If ax+by=1 and bx+ay=(2ab)/(a^(2)+b^(2)) then (x^(2)+y^(2))(a^(2)+b^(2)) is equal to

ax+by=a+b bx+ay=b^2

ax+by=a-b;bx-ay=a+b