Solve the following simultaneous equations by elimination method:-
(a)2x + 3y = 12 ; x – y = 1 .
(b) 3x + y = 10
(c) x – 3y = 1; 3x – 2y + 4 = 0
(d) 5x – 6y + 30 = 0 ; 5x + 4y – 20 = 0 (e) 3x – y – 2 = 0 ; 2x + y = 8
Answers
Answer:
x – 3y – 3 = 0
3x – 9y – 2 =0
a1/a2 = 1/3
b1/b2 = – 3/-9 = 1/3 and
c1/c2 = – 3/-2 = 3/2
a1/a2 = b1/b2 ≠ c1/c2
Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.
2x + y = 5
3x + 2y = 8
a1/a2 = 2/3
b1/b2 = 1/2 and
c1/c2 = – 5/–8 = 5/8
a1/a2 ≠ b1/b2
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication method,
x/b1c2– b2c1 = y/c1a2 – c2a1 = 1/a1b2 – a2b1
x/-8–(–10) = y/–15 + 16 = 1/4 – 3
x/2 = y/1 = 1
x/2 = 1, y/1 = 1
∴ x = 2, y = 1.
3x – 5y = 20
6x – 10y = 40
a1/a2 = 3/6 = 1/2
b1/b2 = – 5/–10 = 1/2 and
c1/c2 = – 20/–40 = 1/2
a1/a2 = b1/b2 = c1/c2
Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.
x – 3y – 7 = 0
3x – 3y – 15= 0
a1/a2 = 1/3
b1/b2 = – 3/–3 = 1 and
c1/c2 = – 7/–15 = 7/15
a1/a2 ≠ b1/b2
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication,
x/45 – (21) = y/–21 – (–15) = 1/–3 – (–9)
x/24 = y/–6 = 1/6
x/24 = 1/6 and y/–6 = 1/6
x = 4 and y = – 1
∴ x = 4, y = – 1.
Hope it helped u..