InterviewSolution
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InterviewSolution
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ax+by =a2 , bx +ay=b2 . solve by cross multiplication method |
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Answer» We can write the given system of equations asax + by - a2 = 0........(1)bx + ay - b2 = 0........(2)From equation (1) & (2), we havea1 = a, b1 = b, c1 = -a2a2 = b, b2 = a, and c2 = -b2Therefore, by cross-multiplication, we get{tex}\\Rightarrow \\frac{x}{{b \\times ( - {b^2}) - ( - {a^2}) \\times a}}{/tex}{tex} = \\frac{{ - y}}{{a \\times ( - {b^2}) - ( - {a^2}) \\times b}}{/tex}{tex}= \\frac{1}{{a \\times a - b \\times b}}{/tex}{tex}\\Rightarrow \\frac{x}{{ - {b^3} + {a^3}}} = \\frac{{ - y}}{{ - a{b^2} + {a^2}b}} = \\frac{1}{{{a^2} - {b^2}}}{/tex}Now,{tex}\\frac{x}{{ - {b^3} + {a^3}}} = \\frac{1}{{{a^2} - {b^2}}}{/tex}{tex}\\Rightarrow x = \\frac{{{a^3} - {b^3}}}{{{a^2} - {b^2}}}{/tex}{tex} = \\frac{{(a - b)\\left( {{a^2} + ab + {b^2}} \\right)}}{{(a - b)(a + b)}}{/tex}{tex} = \\frac{{{a^2} + ab + {b^2}}}{{a + b}}{/tex}also{tex}\\frac{{ - y}}{{ - a{b^2} + {a^2}b}} = \\frac{1}{{{a^2} - {b^2}}}{/tex}{tex}\\Rightarrow - y = \\frac{{{a^2}b - a{b^2}}}{{{a^2} - {b^2}}}{/tex}{tex}\\Rightarrow y = \\frac{{a{b^2} - {a^2}b}}{{{a^2} - {b^2}}}{/tex}{tex} = \\frac{{ab(b - a)}}{{(a - b)(a + b)}}{/tex}{tex} = \\frac{{ - ab(a - b)}}{{(a - b)(a + b)}}{/tex}{tex} = \\frac{{ - ab}}{{a + b}}{/tex}Therefore, {tex}x = \\frac{{{a^2} + ab + {b^2}}}{{a + b}},\\;y = \\frac{{ - ab}}{{a + b}}{/tex}\xa0is the solution of the given system of the equations. bilkul sahi answer hai |
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