Solve the following systems of equations:




Multiplying eq1 by 5 and eq2 by 2 and subtracting


13/(3x – 2y) = 13


3x – 2y = 1------- (3)


Multiplying eq2 by 3 and subtracting eq1 from it


13/(3x +2y) = 13/5


3x + 2y = 5 ------ (4)


(3) + (4)


6x = 6


x = 1


Thus, 3 + 2 = 5


y = 1


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