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6/x+y =7/x-y + 3 and 1/2(x+y)=1/3(x-y) |
| Answer» According to question given system of equations are{tex}\\frac { 6 } { x + y } = \\frac { 7 } { x - y } + 3{/tex}Let {tex}\\frac{1}{{x + y}} = u{/tex}\xa0and {tex}\\frac{1}{{x - y}} = v{/tex}. Then, the given system of equations becomes6u = 7v + 3{tex}\\Rightarrow{/tex}\xa06u - 7v = 3 ...........(i)Given equation,{tex}\\frac{u}{2} = \\frac{v}{3}{/tex}{tex}\\Rightarrow{/tex}\xa03u = 2v{tex}\\Rightarrow{/tex}\xa03u - 2v = 0 .......(ii)Multiplying equation (ii) by 2, we get6u - 4v = 0 ....(iii)Subtracting equation (i) from equation (iii), we get-7v + 4v = 3{tex}\\Rightarrow{/tex}\xa0-3v = 3{tex}\\Rightarrow{/tex}\xa0v = -1Put the value of v = -1 in equation (ii), we get3u - 2 {tex}{/tex}× (-1) = 0{tex}\\Rightarrow{/tex}\xa03u + 2 = 0{tex}\\Rightarrow{/tex}\xa03u = -2{tex}\\Rightarrow u = \\frac{{ - 2}}{3}{/tex}Now,{tex}u = \\frac{{ - 2}}{3}{/tex}{tex}\\Rightarrow \\frac{1}{{x + y}} = \\frac{{ - 2}}{3}{/tex}{tex}\\Rightarrow x + y = \\frac{{ - 3}}{2}{/tex}\xa0.......(iv)and, v = -1{tex}\\Rightarrow \\frac{1}{{x - y}} = -1{/tex}{tex}\\Rightarrow{/tex}\xa0x - y = -1 ........(v)Adding equation (iv) and equation (v), we get{tex}2x = \\frac{{ - 3}}{2} - 1{/tex}{tex}\\Rightarrow 2x = \\frac{{ - 3 - 2}}{2}{/tex}{tex}\\Rightarrow 2x = \\frac{{ - 5}}{2}{/tex}{tex}\\Rightarrow x = \\frac{{ - 5}}{4}{/tex}Put the value of\xa0{tex}x = \\frac{{ - 5}}{4}{/tex}\xa0in equation (v), we get{tex}\\frac{{ - 5}}{4} - y = - 1{/tex}{tex}\\Rightarrow \\frac{{ - 5}}{4} + 1 = y{/tex}{tex}\\Rightarrow \\frac{{ - 5 + 4}}{4} = y{/tex}{tex}\\Rightarrow \\frac{{ - 1}}{4} = y{/tex}{tex}\\Rightarrow y = \\frac{{ - 1}}{4}{/tex}Hence, the solution of the system of equation is {tex}x = \\frac{{ - 5}}{4},y = \\frac{{ - 1}}{4}{/tex}. | |