Question -
Answer -
(i) 2x┬а+ 4y┬а=3
12y┬а+ 6x┬а=6
(ii)┬аx┬а=2y
y┬а= 2x
(iii) 3x┬а+┬аy┬атАУ3 = 0
2x┬а+ 2/3y┬а=2
Solution:
The Condition for nosolution =┬аa1/a2┬а= b1/b2┬атЙаc1/c2┬а(parallel lines)
(i) Yes.
Given pair ofequations are,
2x+4y тАУ 3 = 0 and 6x +12y тАУ 6 = 0
Comparing theequations with ax+ by +c = 0;
We get,
a1┬а=2, b1┬а= 4, c1┬а= тАУ 3;
a2┬а=6, b2┬а= 12, c2┬а= тАУ 6;
a1┬а/a2┬а=2/6 = 1/3
b1┬а/b2┬а=4/12 = 1/3
c1┬а/c2┬а=тАУ 3/ тАУ 6 = ┬╜
Here, a1/a2┬а=b1/b2┬атЙа┬аc1/c2, i.eparallel lines
Hence, the given pairof linear equations has no solution.
(ii) No.
Given pair ofequations,
x = 2y or x тАУ 2y = 0
y = 2x or 2x тАУ y = 0;
Comparing theequations with ax+ by +c = 0;
We get,
a1┬а=1, b1┬а= тАУ 2, c1┬а= 0;
a2┬а=2, b2┬а= тАУ 1, c2┬а= 0;
a1┬а/a2┬а=┬╜
b1┬а/b2┬а=-2/-1 = 2
Here, a1/a2┬атЙа┬аb1/b2.
Hence, the given pairof linear equations has unique solution.
(iii) No.
Given pair ofequations,
3x + y тАУ 3 = 0
2x + 2/3 y = 2
Comparing theequations with ax+ by +c = 0;
We get,
a1┬а=3, b1┬а= 1, c1┬а= тАУ 3;
a2┬а=2, b2┬а= 2/3, c2┬а= тАУ 2;
a1┬а/a2┬а=2/6 = 3/2
b1┬а/b2┬а=4/12 = 3/2
c1┬а/c2┬а=тАУ 3/-2 = 3/2
Here, a1/a2┬а=b1/b2┬а= c1/c2, i.e coincidentlines