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2(ax-by)+a+4b=0 & 2(bx+ay)+b-4a=0 solve by cross multiplication |
| Answer» The given system of equation may be written as2(ax - by) + a + 4b = 0So, 2ax - 2by + a + 4b ..............(i)2(bx + ay) + b - 4a = 0so, 2bx+2ay+b-4a=0................(ii)compare (i) and (ii) with standard form, we geta1 = 2a, b1 = -2b, c1 = a + 4ba2 = 2b, b2 = 2a, c2 = b - 4aBy cross multiplication method{tex} \\frac{x}{{ - 2{b^2} + 8ab - 2{a^2} - 8ab}}{/tex}\xa0{tex}= \\frac{{ - y}}{{2ab - 8{a^2} - 2ab - 8{b^2}}}{/tex}\xa0{tex} = \\frac{1}{{4{a^2} + 4{b^2}}}{/tex}{tex} \\frac{x}{{ - 2{b^2} - 2{a^2}}} = \\frac{{ - y}}{{ - 8{a^2} - 8{b^2}}} = \\frac{1}{{4{a^2} + 4{b^2}}}{/tex}Now, {tex}\\frac{x}{{ - 2{b^2} - 2{a^2}}} = \\frac{1}{{4{a^2} + 4{b^2}}} {/tex}{tex}⇒ x = \\frac{{ - 1}}{2}{/tex}And, {tex}\\frac{{ - y}}{{ - 8{a^2} - 8{b^2}}} = \\frac{1}{{4{a^2} + 4{b^2}}} {/tex}{tex}⇒ y = 2{/tex}Therefore, the solution of the given pair of equations are {tex}\\frac{{ - 1}}{2}{/tex}\xa0and 2 respectively. | |