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  • (i)Find the value(s) of p in for the following pair of equations: 3x – y – 5 = 0 and 6x – 2y – p = 0, if the lines represented by these equations are parallel.

Find the value(s) of p in for the following pair of equations:

(i) 3x – y – 5 = 0 and 6x – 2y – p = 0, if the lines represented by these equations are parallel.

(ii) – x + py = 1 and px – y = 1,if the pair of equations has no solution.

(iii) – 3x + 5y = 7 and 2px – 3y = 1,
if the lines represented by these equations are intersecting at a unique point.

(iv) 2x + 3y – 5 = 0 and px– 6y – 8 = 0,if the pair of equations has a unique solution.

(v) 2x + 3y = 7 and 2px + py = 28 – qy,if the pair of equations have infinitely many solutions.

 



 

 

Answers (1)

(i)Solution:
In equation 3x – y – 5 = 0
a1 = 3, b1 = –1, c1 = –5
In equation 6x – 2y – p = 0
a2 = 6, b2 = –2, c2 = –p

\frac{a_{1}}{a_{2}}= \frac{3}{6}= \frac{1}{2};\frac{b_{1}}{b_{2}}= \frac{-1}{-2}= \frac{1}{2};\frac{c_{1}}{c_{2}}= \frac{-5}{-p}= \frac{5}{p}

If the lines are parallel \frac{a_{1}}{a_{1}}= \frac{b_{1}}{b_{1}}\neq \frac{c_{1}}{c_{1}}
\frac{1}{2}= \frac{1}{2}\neq \frac{5}{p}

\frac{1}{2}\neq \frac{5}{p}\Rightarrow p\neq 10
Any real value of p except 10.
(ii)Solution:
Here the equations are
–x + py = 1, px – y = 1
In equation
–x + py = 1
a1 = –1, b1 = p, c1 = –1
In equation
px – y = 1
a2 = p, b2 = –1, c2 = –1

\frac{a_{1}}{a_{2}}= \frac{-1}{p};\frac{b_{1}}{b_{2}}= \frac{p}{-1}= -p,\frac{c_{1}}{c_{2}}= \frac{-1}{1}= 1

For no solution \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}

\frac{-1}{p}= -p\neq 1

\frac{+1}{p}= +p                           
p^{2} = 1
p = ± 1
p = + 1 (p\neq -1 because if p = –1 then \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}})

(iii)Solution:
Here the equation are –3x + 5y = 7, 2px – 3y = 1
In equation –3x + 5y = 7
a1 = –3, b1 = 5, c1 = –7
In equation 2px – 3y = 1
a2 = 2p, b2 = –3, c2 = –1

\frac{a_{1}}{a_{2}}= \frac{-3}{2p};\frac{b_{1}}{b_{2}}= \frac{5}{-3},\frac{c_{1}}{c_{2}}= \frac{-7}{-1}= 7
For unique solution \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}

\frac{-3}{2p}\neq \frac{-5}{3}
-9\neq -10p
p\neq \frac{9}{10}
p\neq 0\cdot 9
Hence p can have any real value for unique solution except 0.9

(iv)Solution:
Here the equation are 2x + 3y – 5 = 0
px – 6y –8 = 0
In equation 2x + 3y – 5 = 0
a1 = 2, b1 = 3, c1 = –5
In equation px – 6y – 8 = 0
a2 = p, b2 = –6, c2 = –8

\frac{a_{1}}{a_{2}}= \frac{2}{p};\frac{b_{1}}{b_{2}}= \frac{3}{-6}= \frac{-1}{2};\frac{c_{1}}{c_{2}}= \frac{-5}{-8}= \frac{5}{8}

For unique solution \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}

\frac{2}{p}\neq \frac{-1}{2}
4\neq -p
p\neq -4
Hence the equations have unique solution for every real value of p except –4.

(v)Solution:

Here the given equations are 2x + 3y = 7
a1 = 2, b1 = 3, c1 = –7
In equation 2px + py = 28 – qy
2px + py + qy = 28
2px + y(p + q) = 28
a2 = 2p, b2 = p + q, c2 = –28
\frac{a_{1}}{a_{2}}= \frac{2}{2p}= \frac{1}{p};\frac{b_{1}}{b_{2}}= \frac{3}{p+q};\frac{c_{1}}{c_{2}}= \frac{-7}{-28}= \frac{1}{4}
For infinite many solution \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
\frac{1}{p}= \frac{3}{p+q}= \frac{1}{4}
\frac{1}{p}= \frac{1}{4}

4 = p
p = 4   

\frac{3}{p+q}= \frac{1}{4}
p + q = 12
Put p =4
q=12-4
q=8
For infinitely many solutions, p = 4, q = 8
 

 

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