Question

Which pair of the simultaneous equations, have unique solution ?
A.x+y+2=0,x+y+3=0
B.2x-y+3=0,-6x+3y+5=0
C.2x-y+3=0,-4x+2y-6=0
D.x+y+1=0,x-y+1=0

Answer 1

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Your answer is in the pic.
Answer is D.

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Answer 2

Given:

A.  x+y+2=0,    x+y+3=0

B.  2x-y+3=0,  -6x+3y+5=0

C.  2x-y+3=0,  -4x+2y-6=0

D.  x+y+1=0,     x-y+1=0

To find: the pair of equation having unique solution.

Step-by-step explanation:

Step 1 of 1

For a pair of equation having unique solution,

$\frac{a_{1} }{b_{1} } \neq \frac{a_{2} }{b_{2} }$

A.  x+y+2=0,    x+y+3=0

$\frac{a_{1} }{b_{1} } =\frac{1}{1} and \frac{a_{2} }{b_{2} }=\frac{1}{1}$

B.  2x-y+3=0,  -6x+3y+5=0

$\frac{a_{1} }{b_{1} } =\frac{2}{-1} and \frac{a_{2} }{b_{2} }=\frac{-6}{3}=\frac{2}{-1}$

C.  2x-y+3=0,  -4x+2y-6=0

$\frac{a_{1} }{b_{1} } =\frac{2}{-1} and \frac{a_{2} }{b_{2} }=\frac{-4}{2}=\frac{-2}{1}$

D.  x+y+1=0,     x-y+1=0

$\frac{a_{1} }{b_{1} } =\frac{1}{1} and \frac{a_{2} }{b_{2} }=\frac{-1}{1}\\\\\frac{1}{1} \neq \frac{-1}{1}$

Therefore, x+y+1=0,     x-y+1=0 will have unique solution.