Page 44 - Algebra

b1 = 11 = 4 b2 11
4 c1 −6
c2=−3 =4 2
So,
a1 =b1 =c1
a2 b2 c2
Therefore, the equation has infinitely many solutions. It means that the graph would be
coincident. And, only graph (b) is coincident in nature.
10. Calculate the ratio of the coefficients for all the equations.
(i) 10x –16y + 2 = 0 and 6x – y + = 0
10x –16y + 2 = 0 and 30x – 48y + 7 = 0
a1 =10, b1 = –16, c1 = 2
a2 = 30, b2 = – 48, c2 =7
a1 = 10 = 1 a2 30 3
b1 −16 1
==
b2 −48 3 c1 = 2 = 2
c2 7 7
So,
a1 =b1 ≠c1
a2 b2 c2
It means that the system of equation has no solutions.
(ii) 3x + 7y = 2 and 6x + 14y = 4
a1 =3, b1 = 7, c1 = –2
a2 = 6, b2 = 14, c2 = – 4
a1 = 3 = 1
a2 6 2
b1 = 7 = 1
b2 14 2
c1 −2 1
==
c2 −4 2
48 7
55
So,
a1 =b1 =c1
a2 b2 c2
It means that the system of equation has infinitely many solutions
(iii) 3x + 7y = 2 and 6x + 14y = 5
a1 =3, b1 = 7, c1 = –2
a2 = 6, b2 = 14, c2 = – 5
a1 = 3 = 1
a2 6 2
b1 = 7 = 1
b2 14 2
c1 −2 2
==
c2 −5 5
So,
a1 =b1 ≠c1 a2 b2 c2
Page 43 of 177
Algebra I & II