InterviewSolution
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InterviewSolution
| 1. |
(a-b)x+(a+b)y=2a square-2b square(a+b)(x+y)=4ab |
| Answer» The given system of equations are:(a - b)x + (a + b)y = 2a2 - 2b2So, (a - b)x + (a + b)y - 2a2 - 2b2 = 0\xa0(a - b)x + (a + b)y - 2(a2 - b2) = 0 ........(i)And (a + b)(x + y) = 4abSo, (a +b)x + (a + b)y - 4ab = 0 ..........(ii)The given system of equation is in the form ofa1x + b1y - c1 = 0and a2x + b2y - c2 = 0Compare (i) and (ii) , we geta1 = a - b, b1 = a + b, c1 = -2(a2 + b2)a2 = a + b, b2 = a + b, c2 = -4abBy cross-multiplication method{tex}\\frac{x}{{2(a + b)({a^2} - {b^2} + 2ab)}}{/tex}\xa0{tex} = \\frac{{ - y}}{{2(a - b)({a^2} + {b^2})}}{/tex}\xa0{tex} = \\frac{1}{{ - 2b(a + b)}}{/tex}Now, {tex}\\frac{x}{{2(a + b)({a^2} - {b^2} + 2ab)}} = \\frac{1}{{ - 2b(a + b)}}{/tex}\xa0{tex}{/tex}{tex}⇒ x = \\frac{{2ab - {a^2} + {b^2}}}{b}{/tex}And, {tex}\\frac{{ - y}}{{2(a - b)({a^2} + {b^2})}} = \\frac{1}{{ - 2b(a + b)}} {/tex}\xa0{tex}⇒ y = \\frac{{(a - b)({a^2} - {b^2})}}{{b(a + b)}}{/tex}The solution of the system of equations are\xa0{tex}\\frac{{2ab - {a^2} + {b^2}}}{b}{/tex}\xa0and {tex}\\frac{{(a - b)({a^2} - {b^2})}}{{b(a + b)}}{/tex}\xa0respectively. | |