[ax+by=a-b],[bx-ay=a+b]

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

Solve the following system of equations by method of cross-multiplication: ax+by=a-b,quad bx-ay=a+b

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

A, B C and D are the points of intersection with the coordinate axes of the lines ax+by=ab and bx+ay=ab , then

A, B C and D are the points of intersection with the coordinate axes of the lines ax+by=ab and bx+ay=ab, then

A, B C and D are the points of intersection with the coordinate axes of the lines ax+by=ab and bx+ay=ab, then

A,B,C,D are the points of intersection with the coordinate axes of the line ax+by=ab and bx+ay=ab. Then

if ax+by=3,bx-ay=4 and x^(2)+y^(2)=1, then the value of a^(2)+b^(2) is

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