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First clear the system of fractions, then solve by substitution x/2-y/3=5/6 x/5-y/4=1/10
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x/2-y/3=5/6 (x)/(2)-(y)/(3)=(5)/(6) (x)/(2)=(y)/(3)+(5)/(6) x=(2y+5)/(3) ----------------------- x/5-y/4=1/10 (2y+5)/(3)/5-y/4=1/10 ((2y+5)/(3))/(5)-(y)/(4)=(1)/(10) (2y+5)/(15)-(y)/(4)=(1)/(10) (2y+5)/(15)*(4)/(4)-(y)/(4)*(15)/(15)=(1)/(10) ((2y+5)(4))/(60)-(y(15))/(60)=(1)/(10) ((2y+5)(4)-(y(15)))/(60)=(1)/(10) (-7y+20)/(60)=(1)/(10) -7y+20=6 -7y=-14 y=2 --------------------------------------- x=(2y+5)/(3) x=(2(2)+5)/(3) x=(4+5)/(3) x=9/3 x=3
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