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1/7x+ 1/6y=31/2x-1/3y=5 |
| Answer» {tex}\\frac{1}{7x}{/tex}\xa0+\xa0{tex}\\frac{1}{6y}{/tex}\xa0= 3 .......(i)and\xa0{tex}\\frac{1}{2x}{/tex}\xa0-\xa0{tex}\\frac{1}{3y}{/tex}\xa0= 5 ...........(ii)Multiplying equation (ii) by\xa0{tex}\\frac{1}{2}{/tex}, we get\xa0{tex}\\frac{1}{4x}{/tex}\xa0-\xa0{tex}\\frac{1}{6y}{/tex}\xa0=\xa0{tex}\\frac{5}{2}{/tex} ..........(iii)Adding eq. (i) and (iii), we get{tex}\\frac{1}{4x}{/tex}\xa0+\xa0{tex}\\frac{1}{7x}{/tex}\xa0=\xa0{tex}\\frac{5}{2}{/tex}\xa0+ 3{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{7 + 4}{28x}{/tex}\xa0=\xa0{tex}\\frac{11}{2}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{11}{28x}{/tex}\xa0=\xa0{tex}\\frac{11}{2}{/tex}{tex}\\Rightarrow{/tex}\xa028x = 2\xa0{tex}\\Rightarrow{/tex}\xa0x =\xa0{tex}\\frac{1}{14}{/tex}Putting the value of x in eq.(i), we get{tex}\\frac{1}{7(\\frac1{14})}{/tex}\xa0+\xa0{tex}\\frac{1}{6y}{/tex}\xa0= 3y =\xa0{tex}\\frac{1}{6}{/tex}Hence x = {tex}\\frac{1}{14}{/tex} and y = {tex}\\frac{1}{6}{/tex} is the solution of given system of equations. | |