A) \[5x-6y=13;\,\,3x-2y=6\]
B) \[~x\text{ }+\text{ }\frac{1}{2}y=1;2x+2y=1\]
C) \[3+3x=y;-x-y=6\]
D) \[3x-4y=1,8y-6x=4\]
Correct Answer: D
Solution :
(d): In (d), eqns. can be written as \[\frac{x}{\left( \frac{1}{3} \right)}+\frac{y}{\left( \frac{-1}{4} \right)}=1\Rightarrow \left( \frac{1}{3},0 \right)\] and \[\left( 0,\frac{-1}{4} \right)\] are points) Other eqns. is \[\frac{x}{\left( \frac{-2}{3} \right)}+\frac{y}{\left( \frac{1}{2} \right)}=1\] so that points are \[\left( \frac{-2}{3},0 \right)\] and \[\left( 0,\frac{1}{2} \right)\]: You may also cross check, for your own satisfaction.
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