Ax+by=a-b;bx-ay=a+b by elimination

1.	Ax+by=a-b;bx-ay=a+b by elimination
Answer» Given\xa0pair of linear equation is\xa0{tex}ax+by-a+b=0{/tex} .....(i)and {tex}bx-ay-a-b=0{/tex} ........... (ii)Multiplying\xa0{tex}ax+by-a+b=0{/tex}\xa0by a and\xa0{tex}bx-ay-a-b=0{/tex}\xa0by b, and adding them, we geta2x + aby - a2 + ab = 0 and b2x - aby - ab - b2 = 0(a2x + aby - a2 + ab ) + (b2x - aby - ab - b2 ) = 0a2x + aby - a2 + ab + b2x - aby - ab - b2 = 0a2x + b2x - a2- b2 = 0{tex} \\Rightarrow ({a^2} + {b^2})x = ({a^2} + {b^2}){/tex}{tex} \\Rightarrow x = \\frac{{({a^2} + {b^2})}}{{({a^2} + {b^2})}} = 1{/tex}On putting x =1 in first equation, we get{tex}ax+by-a+b=0{/tex}{tex}a + by = a - b {/tex}{tex}\\Rightarrow y = - \\frac{b}{b} = - 1{/tex}Hence, x=1 and y=-1, which is the required unique solution.

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