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  • (i)Do the following pair of linear equations have no solution Justify your answer. 2x + 4y = 3 ; 12y + 6x = 6

Do the following pair of linear equations have no solution? Justify your answer.

(i) 2x + 4y = 3 ; 12y + 6x = 6
(ii) x = 2y; y = 2x
(iii) 3x + y – 3 = 0 ;2x+\frac{2}{3}y= 2

Answers (1)

(i)Solution:
Equations are 2x + 4y = 3
12y + 6x = 6
In equation 2x + 4y – 3 = 0
a1 = 2, b1 = 4, c1 = –3
In equation 12y + 6x – 6 = 0
a2 = 6, b2 = 12, c2 = –6
\frac{a_{1}}{b_{2}}= \frac{2}{6}= \frac{1}{3},\frac{b_{1}}{b_{2}}= \frac{4}{12}= \frac{1}{3},\frac{c_{1}}{c_{2}}= \frac{-3}{-6}= \frac{1}{2}
For no solution \frac{a_{1}}{b_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}} also, Here we have \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}

Hence the given equation have no solution.
(ii)Solution:
Equations are x – 2y = 0
–2x + y = 0
In equation
x – 2y = 0
a1 = 1, b1 = –2, c1 = 0
In equation
–2x + y = 0
a2 = –2, b2 = 1, c2 = 0
\frac{a_{1}}{b_{2}}= \frac{1}{-2},\frac{b_{1}}{b_{2}}= \frac{-2}{1}= \frac{c_{1}}{c_{2}}= \frac{0}{0}
For no solution  but\frac{a_{1}}{b_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}} here

\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}
Hence the given pair of equations a unique solution.

(iii)Solution:
Equations are  3x + y = 3
2x+\frac{2}{3}y= 2
In equation
3x + y – 3 = 0
a1 = 3, b1 = 1, c1 = –3
In equation
2x+\frac{2}{3}y-2= 0
a2 = 2, b2= \frac{2}{3}, c2 = –2
\frac{a_{1}}{b_{2}}= \frac{3}{2},\frac{b_{1}}{b_{2}}= \frac{3}{2}= \frac{c_{1}}{c_{2}}= \frac{-3}{-2}= \frac{3}{2}

For no solution \frac{a_{1}}{b_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}} but here
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
Hence the given pair of equations have infinity many solutions.

 

 

Posted by

infoexpert27

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