$We\ are\ given,\\3x-2y=-6\\5x-2y=7\\Considering\ this\ system\ of\ equations,\\Let\ 3x-2y=-6\ be\ E_1\\Let\ 5x-2y=7\ be\ E_2,\\Hence,\\Multiplying\ E2\ with\ -1, we\ get:\\3x-2y=-6\\-1(5x-2y)=7*-1\\Hence,\\3x-2y=-6\\-5x+2y=-7\\Now,\\Lets\ add\ E_1\ and\ E_2\ together\\Hence,\\(3x-2y)+(-5x+2y)=(-6)+(-7)\\Hence,\\3x-2y-5x+2y=-13\\Arranging\ And\ Adding\ Like\ terms,\ we\ have:\\3x-5x-2y+2y=-13\\Hence,\\-2x+0y=-13\\-2x=-13\\x=\frac{-13}{-2}=\frac{13}{2}$

$Now,\ we\ can\ substitute\ x=\frac{13}{2}\ in\ E_1\ or\ E_2,\\But,\\Here,\ we'll\ substitute\ it\ in\ E1,\\Hence,\\Substituting\ x=\frac{13}{2}\ in\ E1,\ we\ have:\\3*\frac{13}{2}-2y=-6\\\frac{39}{2}-2y=-6\\-2y=-6- \frac{39}{2}\\-2y=\frac{-12-39}{2}\\-2y=\frac{-51}{2}\\y=\frac{-51}{2*-2}=\frac{51}{4}\\Hence,\\x=\frac{13}{2}, y=\frac{51}{4}$

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2 years ago