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Question

# A pair of linear equations which has a unique solution x=2, y=−3 is

A

x + y = -1 ; 2x - 3y = -5

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B

2x - y = 1 ; 3x + 2y = 0

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C

2x + 5y = -11 ; 4x + 10y = -22

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D

x - 4y -14 = 0 ; 5x - y - 13 = 0

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Solution

## We know that, if pair of linear equations a1x+b1y+c1=0, a2x+b2y+c2=0 have the unique solution then a1a2≠b1b2Now lets check the options,A. 2x−y=1; 3x+2y=0a1a2=23b1b2=−12Since, a1a2≠b1b2, which has unique solution.Now, lets substitute x=2, y=−3 in both equations.2x−y=1⇒ LHS =2(2)−3=4−3=1= RHS3x + 2y = 0⇒ LHS =3(2)+2(−3)=6−6=0 RHSHence, x=2, y=−3 is an unique solution of 2x−y=1; 3x+2y=0.B. 2x+5y=−11; 4x+10y=−22a1a2=24=12b1b2=510=12Since, a1a2=b1b2, which has no unique solution.C. x−4y−14=0;5x−y−13=0a1a2=15b1b2=−4−1=4Since, a1a2≠b1b2, which has unique solution.Now, lets substitute x=2, y=−3 in both equations.x−4y−14=0LHS =2−4(−3)−14=2+12−14=0= RHS5x−y−13=0LHS =5(2)−(−3)−13=10+3−13=13−13=0= RHSHence, x=2, y=−3 is an unique solution of x−4y−14=0;5x−y−13=0.D. x+y=−1;2x−3y=−5a1a2=12b1b2=1−3Since, a1a2≠b1b2, which has unique solution.Now, lets substitute x=2, y=−3 in both equations.x+y=−1LHS =2−3=−1= RHS2x−3y=−5LHS =2(2)−3(−3)=4+9=13≠ RHSHence, x=2, y=−3 is not an unique solution of x+y=−1;2x−3y=−5.Hence, Option A, Option C are correct.

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