InterviewSolution
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InterviewSolution
| 1. |
bx+ay=a+bax (1/a -b -1/a+b) +by (1/b-a -1/b+a)=2 |
| Answer» The system of equation is given by :bx + cy = a + b ......(i){tex}ax\\left( {\\frac{1}{{a - b}} - \\frac{1}{{a + b}}} \\right) + cy\\left( {\\frac{1}{{b - a}} + \\frac{1}{{b + a}}} \\right){/tex}{tex}= \\frac{{2a}}{{a + b}}{/tex}\xa0...(ii)From equation (i)bx + cy - (a + b) = 0 ............ (iii)From equation (ii){tex} ax\\left( {\\frac{1}{{a - b}} - \\frac{1}{{a + b}}} \\right) + cy\\left( {\\frac{1}{{b - a}} + \\frac{1}{{b + a}}} \\right){/tex}\xa0{tex} - \\frac{{2a}}{{a + b}} = 0{/tex}{tex}⇒ x\\left( {\\frac{{2ab}}{{(a - b)(a + b)}}} \\right) + y\\left( {\\frac{{2ac}}{{(b - a)(b + a)}}} \\right){/tex}\xa0{tex}- \\frac{{2a}}{{a + b}} = 0{/tex}{tex} ⇒ \\frac{1}{{a + b}}\\left( {\\frac{{2abx}}{{a - b}} - \\frac{{2acy}}{{a - b}} - 2a} \\right) = 0{/tex}\xa0{tex}⇒ \\frac{{2abx}}{{a - b}} - \\frac{{2acy}}{{a - b}} - 2a = 0{/tex} 2abx - 2acy - 2a(a - b) = 0 ....(iv)From equation (iii) and (iv), we geta1 = b, b1 = c and c1 = - (a + b)and a2 = 2ab , b2 = -2ac and c3 = -2a(a - b)by cross-multiplication, we get{tex}\\frac{x}{{ - 4{a^2}c}} = \\frac{{ - y}}{{4a{b^2}}} = \\frac{{ - 1}}{{4abc}}{/tex}Now, {tex}\\frac{x}{{ - 4{a^2}c}} = \\frac{{ - 1}}{{4abc}} {/tex}{tex}⇒ x = \\frac{a}{b}{/tex}And, {tex}\\frac{{ - y}}{{4a{b^2}}} = \\frac{{ - 1}}{{4abc}} {/tex}{tex}⇒ y = \\frac{b}{c}{/tex}The solution of the system of equation are {tex}\\frac{a}{b}{/tex}\xa0and {tex}\\frac{b}{c}{/tex}. | |