Let →α=(x+4y)→a+(2x+y+1)→b and →β=(y−2x+2)→a+(2x−3y−1)→b where →a and →b are non-zero, non-collinear. If 3→α=2→β then
A
x=1,y=2
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B
x=2,y=1
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C
x=−1,y=2
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D
x=2,y=−1
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Solution
The correct option is Dx=2,y=−1 α=(x+4y)a+(2x+y+1)bandβ=(y−2x+2)a+(2x−3y−1)b 3α=2β ⇒3(x+4y)=2(y−2x+2) and 3(2x+y+1)=2(2x−3y−1) ⇒7x+10y−4=0 and 2x+9y+5=0 solving above two equations gives x=2 and y=−1 Hence, option D.