InterviewSolution
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InterviewSolution
| 1. |
Find the value of x and y by cross multiplication methodbx/a + ay/b = a2+b2x+y-2ab |
| Answer» The given pair of equations are:{tex}\\frac{b}{a}x + \\frac{a}{b}y = {a^2} + {b^2} {/tex}So,\xa0{tex}\\frac{b}{a}x + \\frac{a}{b}y -[ {a^2} + {b^2} ] = 0{/tex} ...................(i)And x + y = 2abx + y - 2ab = 0 ....................(ii)Here,{tex}{a_1} = \\frac{b}{a},{b_1} = \\frac{a}{b}{/tex}, c1 = -(a2 + b2)a2 = 1, b2 = 1, c2 = -(2ab)By cross-multiplication method{tex}\\begin{array}{l}\\;\\frac x{{\\displaystyle\\frac ab}\\times-(2ab)\\;-1\\lbrack-(a^2\\;+\\;b^2)\\rbrack}=\\;\\frac y{-(a^2\\;+\\;b^2)\\;-{\\displaystyle\\frac ba}\\lbrack\\;-(2ab)\\rbrack}=\\;\\frac1{{\\displaystyle\\frac ba}-{\\displaystyle\\frac ab}}\\\\\\frac x{\\displaystyle\\frac{-2a^2b}b\\;\\;+(a^2\\;+\\;b^2)}=\\;\\frac y{-(a^2\\;+\\;b^2)\\;+{\\displaystyle\\frac{2ab^2}a}}=\\;\\frac1{\\displaystyle\\frac{b^2\\;-\\;a^2}{ab}\\;}\\\\\\end{array}{/tex}{tex} \\frac{x}{{{\\frac ba - \\frac ab} }} = \\frac{{ - y}}{{ - {b^2} + {a^2}}} = \\frac{1}{{\\frac{{{b^2} - {a^2}}}{{ab}}}} {/tex}{tex} \\frac{x}{{{b^2} - {a^2}}} = \\frac{{ - y}}{{ - {b^2} + {a^2}}} = \\frac{1}{{\\frac{{{b^2} - {a^2}}}{{ab}}}} {/tex}{tex} \\frac{x}{{{b^2} - {a^2}}} = \\frac{1}{{\\frac{{{b^2} - {a^2}}}{{ab}}}} {/tex}{tex}⇒ x = ab{/tex}And, {tex}\\frac{{ - y}}{{ - {b^2} + {a^2}}} = \\frac{1}{{\\frac{{{b^2} - {a^2}}}{{ab}}}} {/tex}{tex}⇒ y = ab{/tex}The solutions of the given pair of equations is x= ab and y = ab . | |