**Exercise 3.5**

**In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution, find it from 1 to 4:**

**(1) \(x-3y-3=0\)**

**Â Â Â \(3x-9y-2=0\)**

**Soln:**

The given system may be written as

\(x-3y-3=0\)

\(3x-9y-2=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=1,b_{1}=-3,c_{1}=-3\)

\(a_{2}=3,b_{2}=-9,c_{2}=-2\)

We have,

\(\frac{a_{1}}{a_{2}}=\frac{1}{3}\)

\(\frac{b_{1}}{b_{2}}=\frac{-3}{-9}=\frac{1}{3}\)

And , \(\frac{c_{1}}{c_{2}}=\frac{-3}{-2}=\frac{3}{2}\)

\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\)

Therefore, the given equation has no solution.

**(2) \(2x+y-5=0\)**

**Â Â Â \(4x+2y-10=0\)**

Soln:

The given system may be written as

\(2x+y-5=0\)

\(4x+2y-10=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=1,c_{1}=-5\)

\(a_{2}=4,b_{2}=2,c_{2}=-10\)

We have,

\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}\)

\(\frac{b_{1}}{b_{2}}=\frac{1}{2}\)

And , \(\frac{c_{1}}{c_{2}}=\frac{-5}{-10}=\frac{1}{2}\)

So, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}\)

Therefore, the given equation has infinitely many solution.

**(3) \(3x-5y=20\)**

**Â Â Â \(6x-10y=40\)**

Soln:

The given system may be written as

\(3x-5y=20\)

\(6x-10y=40\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=3,b_{1}=-5,c_{1}=-20\)

\(a_{2}=6,b_{2}=-10,c_{2}=-40\)

We have,

\(\frac{a_{1}}{a_{2}}=\frac{3}{6}=\frac{1}{2}\)

\(\frac{b_{1}}{b_{2}}=\frac{-5}{-10}=\frac{1}{2}\)

And , \(\frac{c_{1}}{c_{2}}=\frac{-20}{-40}=\frac{1}{2}\)

So, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}\)

Therefore, the given equation has infinitely many solution.

**(4) \(x-2y-8=0\)**

**Â Â Â \(5x-10y-10=0\)**

Soln:

The given system may be written as

\(x-2y-8=0\)

\(5x-10y-10=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=1,b_{1}=-2,c_{1}=-8\)

\(a_{2}=5,b_{2}=-10,c_{2}=-10\)

We have,

\(\frac{a_{1}}{a_{2}}=\frac{1}{5}\)

\(\frac{b_{1}}{b_{2}}=\frac{-2}{-10}=\frac{1}{5}\)

And , \(\frac{c_{1}}{c_{2}}=\frac{-8}{-10}=\frac{4}{5}\)

\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\)

Therefore, the given equation has no solution.

**Find the value of k for each of the following system of equations which have a unique solution (5-8)**

**(5) \(kx+2y-5=0\)**

**Â Â Â Â Â \(3x+y-1=0\)**

Soln:

The given system may be written as

\(kx+2y-5=0\)

\(3x+y-1=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=k,b_{1}=2,c_{1}=-5\)

\(a_{2}=3,b_{2}=1,c_{2}=-1\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

\(\frac{k}{3} Â \neq \frac{2}{1}\)

\(\Rightarrow k\neq 6\)

Therefore, the given system will have unique solution for all real values of k other than 6.

**(6) \(4x+ky+8=0\)**

**Â Â Â Â \(2x+2y+2=0\)**

Soln:

The given system may be written as

\(4x+ky+8=0\)

\(2x+2y+2=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=4,b_{1}=k,c_{1}=8\)

\(a_{2}=2,b_{2}=2,c_{2}=2\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

\(\frac{4}{2} Â \neq \frac{k}{2}\)

\(\Rightarrow k\neq 4\)

Therefore, the given system will have unique solution for all real values of k other than 4.

**(7) \(4x-5y=k\)**

**Â Â Â Â \(2x-3y=12\)**

Soln:

The given system may be written as

\(4x-5y-k=0\)

\(2x-3y-12=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=4,b_{1}=-5,c_{1}=-k\)

\(a_{2}=2,b_{2}=-3,c_{2}=-12\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

\(\frac{4}{2} Â \neq \frac{-5}{-3}\)

\(\Rightarrow k\) can have any real values.

Therefore, the given system will have unique solution for all real values of k.

**(8) \(x+2y=3\)**

**Â Â Â Â \(5x+ky+7=0\)**

Soln:

The given system may be written as

\(x+2y=3\)

\(5x+ky+7=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=1,b_{1}=2,c_{1}=-3\)

\(a_{2}=5,b_{2}=k,c_{2}=7\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

\(\frac{1}{5} Â \neq \frac{2}{k}\)

\(\Rightarrow k \neq 10\)

Therefore, the given system will have unique solution for all real values of k other than 10.

**Find the value of k for which each of the following system of equations having infinitely many solution: (9-19)**

**(9) \(2x+3y-5=0\)**

**Â Â Â Â \(6x-ky-15=0\)**

Soln:

The given system may be written as

\(2x+3y-5=0\)

\(6x-ky-15=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=3,c_{1}=-5\)

\(a_{2}=6,b_{2}=k,c_{2}=-15\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}\)

\(\frac{2}{6} Â \neq \frac{3}{k}\)

\(\Rightarrow k = 9\)

Therefore, the given system of equation will have infinitely many solutions, if k=9.

**(10) \(4x+5y=3\)**

**Â Â Â Â \(kx+15y=9\)**

Soln:

The given system may be written as

\(4x+5y=3\)

\(kx+15y=9\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=4,b_{1}=5,c_{1}=3\)

\(a_{2}=k,b_{2}=15,c_{2}=9\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}\)

\(\frac{4}{k} = \frac{5}{15} = \frac{-3}{-9}\)

\(\frac{4}{k} = \frac{1}{3}\)

\(\Rightarrow k \neq 12\)

Therefore, the given system will have infinitely many solutions if k=12.

**(11) \(kx-2y+6=0\)**

**Â Â Â Â \(4x+3y+9=0\)**

Soln:

The given system may be written as

\(kx-2y+6=0\)

\(4x+3y+9=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=k,b_{1}=-2,c_{1}=6\)

\(a_{2}=4,b_{2}=-3,c_{2}=9\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}\)

\(\frac{k}{4} = \frac{-2}{-3} = \frac{2}{3}\)

\(\Rightarrow k = \frac{8}{3} \)

Therefore, the given system of equations will have infinitely many solutions, if \(k = \frac{8}{3} \).

**(12) \(8x+5y=9\)**

**Â Â Â Â \(kx+10y=19\)**

Soln:

The given system may be written as

\(8x+5y=9\)

\(kx+10y=19\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=8,b_{1}=5,c_{1}=-9\)

\(a_{2}=k,b_{2}=10,c_{2}=-18\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= Â \frac{c_{1}}{c_{2}}\)

\(\frac{8}{k} = \frac{5}{10} = \frac{-9}{-18}= \frac{1}{2}\)

\(\Rightarrow k = 16 \)

Therefore, the given system of equations will have infinitely many solutions, if \(k = 16 \).

**(13) \(2x-3y=7\)**

**Â Â Â Â \((k+2)x-(2k+1)y=3(2k-1)\)**

Soln:

The given system may be written as

\(2x-3y=7\)

\((k+2)x-(2k+1)y=3(2k-1)\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=-3,c_{1}=-7\)

\(a_{2}=k,b_{2}=-(2k+1),c_{2}=-3(2k-1)\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= Â \frac{c_{1}}{c_{2}}\)

\(\frac{2}{k+2} = \frac{-3}{-(2k+1)} = \frac{-7}{-3(2k-1}\)

\(\frac{2}{k+2} = \frac{-3}{-(2k+1)} \; and\; \frac{-3}{-(2k+1)}= \frac{-7}{-3(2k-1}\)

\(\Rightarrow 2(2k+1)=3(k+2)\; and \; 3\times 3(2k-1)=7(2k+1)\)

\(\Rightarrow 4k+2=3k+6 \; and \; 18k-9=14k+7\)

\(\Rightarrow k = 4\; and \; 4k=16 \Rightarrow k = 4 \)

Therefore, the given system of equations will have infinitely many solutions, if \(k = 4 \).

**(14) \(2x+3y=2\)**

**Â Â Â Â \((k+2)x+(2k+1)y=2(k-1)\)**

Soln:

The given system may be written as

\(2x+3y=2\)

\((k+2)x+(2k+1)y=2(k-1)\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=3,c_{1}=-2\)

\(a_{2}=(k+2),b_{2}=(2k+1),c_{2}=-2(k-1)\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= Â \frac{c_{1}}{c_{2}}\)

\(\frac{2}{k+2} = \frac{3}{(2k+1)} = \frac{-2}{-2(k-1}\)

\(\frac{2}{k+2} = \frac{3}{(2k+1)} \; and\; \frac{3}{(2k+1)}= \frac{2}{2(k-1}\)

\(\Rightarrow 2(2k+1)=3(k+2)\; and \; 3(k-1)=(2k+1)\)

\(\Rightarrow 4k+2=3k+6 \; and \; 3k-3=2k+1\)

\(\Rightarrow k = 4\; and \; k=4 \)

Therefore, the given system of equations will have infinitely many solutions, if \(k = 4 \).

**(15) \(x+(k+1)y=4\)**

**Â Â Â Â \((k+1)x+9y=(5k+2)\)**

Soln:

The given system may be written as

\(x+(k+1)y=4\)

\((k+1)x+9y=(5k+2)\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=1,b_{1}=(k+1),c_{1}=-4\)

\(a_{2}=(k+1),b_{2}=9,c_{2}=-(5k+2)\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= Â \frac{c_{1}}{c_{2}}\)

\(\frac{1}{k+1} = \frac{(k+1)}{9} = \frac{-4}{-(5k+2}\)

\(\frac{1}{k+1} = \frac{k+1}{9} \; and\; \frac{k+1}{9}= \frac{4}{5k+2}\)

\(\Rightarrow 9=(k+1)^{2}\; and \; (k+1)(5k+2)=36\)

\(\Rightarrow 9=k^{2}+2k+1 \; and \; 5k^{2}+2k+5k+2=36\)

\(\Rightarrow k^{2}+2k-8=0 \; and \; 5k^{2}+7k-34=0\)

\(\Rightarrow k^{2}+4k-2k-8=0 \; and \; 5k^{2}+17k-10k-34=0\)

\(\Rightarrow k(k+4)-2(k+4)=0\; and \; (5k+17)-2(5k+17)=0\)

\(\Rightarrow (k+4)(k-2)=0\; and \; (5k+17)(k-2)=0\)

\(\Rightarrow k=-4 \; or k=2 \; and \; k=\frac{-17}{5}\; or k=2\)

thus, k=2 satisfies both the condition.

Therefore, the given system of equations will have infinitely many solutions, if \(k = 2 \).

**(16) \(kx+3y=2k+1\)**

**Â Â Â Â \(2(k+1)x+9y=(7k+1)\)**

Soln:

The given system may be written as

\(kx+3y=2k+1\)

\(2(k+1)x+9y=(7k+1)\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=k,b_{1}=3,c_{1}=-(2k+1)\)

\(a_{2}=2(k+1),b_{2}=9,c_{2}=-(7k+1)\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= Â \frac{c_{1}}{c_{2}}\)

\(\frac{1}{2(k+2)} = \frac{3}{9} = \frac{-(2k+1)}{-(7k+1}\)

\(\frac{1}{2(k+2)} = \frac{3}{9} \; and\; \frac{3}{9} = \frac{(2k+1)}{(7k+1}\)

\(\Rightarrow 9k=3 \times 2(k+1)\; and \; 3(7k+1)=9(2k+1)\)

\(\Rightarrow 9k-6k=6 \; and \; 21k-18k=9-3\)

\(\Rightarrow 3k = 6\; and \; 3k=6 \)

\(\Rightarrow k = 2\; and \; k=2 \)

Therefore, the given system of equations will have infinitely many solutions, if \(k = 2 \).

**(17) \(2x+(k-2)y=k\)**

**Â Â Â Â \(6x+(2k-1)y= (2k+5)\)**

Soln:

The given system may be written as

\(2x+(k-2)y=k\)

\(6x+(2k-1)y= (2k+5)\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=(k-2),c_{1}=-k\)

\(a_{2}=6,b_{2}=(2k-1),c_{2}=-(2k+5)\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= Â \frac{c_{1}}{c_{2}}\)

\(\frac{2}{6} = \frac{k-2}{2k-1} = \frac{-k}{-2(2k+5)}\)

\(\frac{2}{6} = \frac{k-2}{2k-1} \; and\; \frac{k-2}{2k-1} = \frac{k}{(2k+5}\)

\(\frac{1}{3}=\frac{k-2}{2k-1}\; and \; 2k^{2}+5k-4k-10=2k^{2}-k\)

\(\Rightarrow 2k-3k=-6+1 \; and \; k+k=10\)

\(\Rightarrow -k = -5\; and \; 2k=10 \)

\(\Rightarrow k = 5\; and \; k=5 \)

Therefore, the given system of equations will have infinitely many solutions, if \(k = 5 \).

**(18) \(2x+3y=7\)**

**Â Â Â Â \((k+1)x+(2k-1)y= (4k+1)\)**

Soln:

The given system may be written as

\(2x+3y=7\)

\((k+1)x+(2k-1)y= (4k+1)\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=3,c_{1}=-7\)

\(a_{2}=k+1,b_{2}=2k-1,c_{2}=-(4k+1)\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= Â \frac{c_{1}}{c_{2}}\)

\(\frac{2}{k+1} = \frac{3}{2k-1} = \frac{-7}{-(4k+1)}\)

\(\frac{2}{k+1} = \frac{3}{2k-1} \; and\; Â \frac{3}{2k-1} = \frac{7}{(4k+1)}\)

\(2(2k-1)}=3(k+1) \; and \; 3(4k+1)=7(2k-1)\)

\(\Rightarrow 4k-2=3k+3 \; and \; 12k+3=14k-7\)

\(\Rightarrow k = 5\; and \; 2k=10 \)

\(\Rightarrow k = 5\; and \; k=5 \)

Therefore, the given system of equations will have infinitely many solutions, if \(k = 5 \).

**(19) \(2x+3y=k\)**

**Â Â Â Â \((k-1)x+(k+2)y= 3k\)**

Soln:

The given system may be written as

\(2x+3y=k\)

\((k-1)x+(k+2)y= 3k\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=3,c_{1}=-k\)

\(a_{2}=k-1,b_{2}=k+2,c_{2}=-3k\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= Â \frac{c_{1}}{c_{2}}\)

\(\frac{2}{k-1} = \frac{3}{k+2} = \frac{-k}{-3k}\)

\(\frac{2}{k-1} = \frac{3}{k+2} \; and\; Â \frac{3}{k+1} = \frac{-k}{-3k}\)

\(2(k+2)}=3(k-1) \; and \; 3 \times 3=k+2\)

\(\Rightarrow 2k+4=3k-3 \; and \; 9=k+2\)

\(\Rightarrow k = 7\; and \; k=7 \)

Therefore, the given system of equations will have infinitely many solutions, if \(k = 7 \).

**Find the value of k for which the following system of equation has no solution : (20-25)**

**(20) \(kx-5y=2\)**

**Â Â Â Â \(6x+2y= 7\)**

Soln:

The given system may be written as

\(kx-5y=2\)

\(6x+2y= 7\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=k,b_{1}=-5,c_{1}=-2\)

\(a_{2}=6,b_{2}=2,c_{2}=-7\)

For no solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

\(\frac{k}{6} = \frac{-5}{2} \neq \frac{2}{7}\)

\(\Rightarrow 2k = -30 \)

\(\Rightarrow k = -15 \)

Therefore, the given system of equations will have no solutions, if \(k = -15 \).

**(21) \(x+2y=0\)**

**Â Â Â Â \(2x+ky= 5\)**

Soln:

The given system may be written as

\(x2y=0\)

\(2x+ky= 5\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=1,b_{1}=2,c_{1}=0\)

\(a_{2}=2,b_{2}=k,c_{2}=-5\)

For no solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

\(\frac{1}{2} = \frac{2}{k} \neq \frac{2}{7}\)

\(\Rightarrow k = 4 \)

Therefore, the given system of equations will have no solutions, if \(k = 4 \).

**(22) \(3x-4y+7=0\)**

**Â Â Â Â \(kx+3y-5=0\)**

Soln:

The given system may be written as

\(3x-4y+7=0\)

\(kx+3y-5=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=3,b_{1}=-4,c_{1}=7\)

\(a_{2}=k,b_{2}=3,c_{2}=-5\)

For no solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

\(\frac{3}{k} = \frac{-4}{3} \)

\(\Rightarrow k = \frac{-9}{4}\)

Therefore, the given system of equations will have no solutions, if \(k = \frac{-9}{4} \).

**(23) \(2x-ky+3=0\)**

**Â Â Â Â \(3x+2y-1=0\)**

Soln:

The given system may be written as

\(2x-ky+3=0\)

\(3x+2y-1=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=-k,c_{1}=3\)

\(a_{2}=3,b_{2}=2,c_{2}=-1\)

For no solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

\(\frac{2}{3} = \frac{-k}{2} \)

\(\Rightarrow k = \frac{-4}{3}\)

Therefore, the given system of equations will have no solutions, if \(k = \frac{-4}{3} \).

**(24) \(2x+ky-11=0\)**

**Â Â Â Â \(5x-7y-5=0\)**

Soln:

The given system may be written as

\(2x+ky-11=0\)

\(5x-7y-5=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=k,c_{1}=-11\)

\(a_{2}=5,b_{2}=-7,c_{2}=-5\)

For no solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

\(\frac{2}{5} = \frac{-k}{-7} \)

\(\Rightarrow k = \frac{-14}{5}\)

Therefore, the given system of equations will have no solutions, if \(k = \frac{-14}{5} \).

**(25) \(kx+3y=3\)**

**Â Â Â Â \(12x+ky=6\)**

Soln:

The given system may be written as

\(kx+3y=3\)

\(12x+ky=6\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=k,b_{1}=3,c_{1}=-3\)

\(a_{2}=12,b_{2}=k,c_{2}=-6\)

For no solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

\(\frac{k}{12} = \frac{3}{k} \neq \frac{3}{6} \) Â Â Â â€¦â€¦(i)

\(\Rightarrow k^{2} = 36\)

\(\Rightarrow k = +6 \; or \; -6\)

From (i)

\( \frac{k}{12} \neq \frac{3}{6}\)

\(\Rightarrow k \neq 6\)

Therefore, the given system of equations will have no solutions, if \(k = -6 \).

**(26) For what value of a, the following system of equation will be inconsistent?**

**\(4x+6y-11=0\)**

**\(2x+ay-7=0\)**

Soln:

The given system may be written as

\(4x+6y-11=0\)

\(2x+ay-7=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=4,b_{1}=6,c_{1}=-11\)

\(a_{2}=2,b_{2}=a,c_{2}=-7\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \)

\(\frac{4}{2} = \frac{6}{a} \)

\(\Rightarrow a = 3\)

Therefore, the given system of equations will be inconsistent, if \(a = 3 \).

**(27) For what value of a, the following system of equation have no solution?**

**\(ax+3y=a-3\)**

**\(12x+ay=a\)**

Soln:

The given system may be written as

\(ax+3y=a-3\)

\(12x+ay=a\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=a,b_{1}=3,c_{1}=-(a-3)\)

\(a_{2}=12,b_{2}=a,c_{2}=-a\)

For unique solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

\(\frac{a}{12} = \frac{3}{a} \neq \frac{-(a-3)}{-a} \)

\(\frac{3}{a} \neq frac{-(a-3)}{-a} \)

\(\Rightarrow a-3 \neq 3\)

\(\Rightarrow a \neq 6\)

And,

\(\frac{a}{12} = \frac{3}{a} \)

\(\Rightarrow a^{2}=36 \)

\(\Rightarrow a=+6\; or \; -6 \)

âˆµ \( a \neq 6\)

\(\Rightarrow a= -6 \)

Therefore, the given system of equations will have no solution, if \(a = -6 \).

**(28) Find the value of a,for which the following system of equation have **

**(i) Unique solution **

**(ii) No solution**

**\(kx+2y=5\)**

**\(3x+y=1\)**

Soln:

The given system may be written as

\(kx+2y-5=0\)

\(3x+y-1=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=k,b_{1}=2,c_{1}=-5\)

\(a_{2}=3,b_{2}=1,c_{2}=-1\)

(i) For unique solution, we have

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

\(\frac{k}{3} \neq \frac{2}{1} \)

\( k \neq 6 \)

Therefore, the given system of equations will have unique solution, if \(k \neq 6 \).

(ii) For no solution, we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \)

\(\frac{k}{3} = \frac{2}{1} \neq \frac{-5}{-1} \)

\(\frac{k}{3} = \frac{2}{1} \)

\(\Rightarrow k=6 \)

Therefore, the given system of equations will have no solution, if \(a = 6 \).

**(29) For what value of c, the following system of equation have infinitely many solution (where \( c \neq 0 \) )?**

**\(6x+3y=c-3\)**

**\(12x+cy=c\)**

Soln:

The given system may be written as

\(6x+3y-(c-3)=0\)

\(12x+cy-c=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=6,b_{1}=3,c_{1}=-(c-3)\)

\(a_{2}=12,b_{2}=c,c_{2}=-c\)

For infinitely many solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}\)

\(\frac{6}{12} = \frac{3}{c} = \frac{-(c-3)}{-c} \)

\(\frac{1}{2} = \frac{3}{c}\; and \; \frac{3}{c} Â = \frac{-(c-3)}{-c} \)

\(\Rightarrow c=6 \; and \; c-3=3 \)

\(\Rightarrow c=6 \; and \; c=6 \)

Therefore, the given system of equations will have infinitely many solution, if \( c = 6 \).

**(30) Find the value of k,for which the following system of equation have **

**(i) Unique solution **

**(ii) No solution**

**(iii) Infinitely many solution**

**\(2x+ky=1\)**

**\(3x-5y=7\)**

Soln:

The given system may be written as

\(2x+ky=1\)

\(3x-5y=7\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=k,c_{1}=-1\)

\(a_{2}=3,b_{2}=-5,c_{2}=-7\)

(i) For unique solution, we have

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

\(\frac{2}{3} \neq \frac{-k}{-5} \)

\( k \neq \frac{-10}{3} \)

Therefore, the given system of equations will have unique solution, if \(k \neq \frac{-10}{3} \).

(ii) For no solution, we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \)

\(\frac{2}{3} = \frac{k}{-5} neq \frac{-1}{-7} Â \)

\(\frac{2}{3} = \frac{k}{-5} \; and \; \frac{k}{-5} neq \frac{1}{7} Â \)

\(\Rightarrow k= Â \frac{-10}{3} \; and \; k neq \frac{-5}{7} \)

\(\Rightarrow k=\frac{-10}{3} \)

Therefore, the given system of equations will have no solution, if \(k = Â \frac{-10}{3} \).

(iii) For the given system to have infinitely many solution,we have

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{2}{3} = \frac{k}{-5} = \frac{-1}{-7} Â \)

Clearly \(\frac{a_{1}}{a_{2}} \neq \frac{c_{1}}{c_{2}} \) ,

So there is no value of k for which the given system of equation has infinitely many solution.

**(31) For what value of k, the following system of equation will represent the coincident lines?**

**\(x+2y+7=0\)**

**\(2x+ky+14=0\)**

Soln:

The given system may be written as

\(x+2y+7=0\)

\(2x+ky+14=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=1,b_{1}=2,c_{1}=7\)

\(a_{2}=2,b_{2}=k,c_{2}=14\)

The given system of equation will represent the coincident lines if they have infinitely many solution.

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{1}{2} = \frac{2}{k} = \frac{7}{14} Â \)

\(\frac{1}{2} = \frac{2}{k} = \frac{1}{2} Â \)

\(\Rightarrow k=4 \)

Therefore, the given system of equations will have infinitely many solution, if \( k = 4 \).

**(32) (30) Find the value of k,for which the following system of equation have unique solution.**

**\(ax+by=c\)**

**\(lx+my=n\)**

Soln:

The given system may be written as

\(ax+by-c=0\)

\(lx+my-n=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=a,b_{1}=b,c_{1}=-c\)

\(a_{2}=l,b_{2}=m,c_{2}=-n\)

For unique solution, we have

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

\(\Rightarrow \frac{a}{l} \neq \frac{b}{m} \)

\(\Rightarrow am \neq bl \)

Therefore, the given system of equations will have unique solution, if \( am \neq bl \).

**(33) Find the value of a and b such that the following system of linear equation have infinitely many solution:**

**\((2a-1)x+3y-5=0\)**

**\(3x+(b-1)y-2=0\)**

Soln:

The given system of equation may be written as,

\((2a-1)x+3y-5=0\)

\(3x+(b-1)y-2=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=(2a-1),b_{1}=3,c_{1}=-5\)

\(a_{2}=3,b_{2}=b-1,c_{2}=-2\)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{(2a-1)}{3} = \frac{3}{b-1} = \frac{-5}{-2} Â \)

\(\frac{(2a-1)}{3} = \frac{5}{2} \; and \; \frac{3}{b-1} = \frac{5}{2} \)

\(\Rightarrow 2(2a-1) = 15 \; and \; 6 = 5(b-1) \)

\(\Rightarrow 4a-2 = 15 \; and \; 6 = 5b-5 \)

\(\Rightarrow 4a = 17 \; and \; 5b = 11 \)

\(\Rightarrow a =\frac{17}{4} \; and \; b = \frac{11}{5} \)

**(34) Find the value of a and b such that the following system of linear equation have infinitely many solution:**

**\(2x-3y=7\)**

**\((a+b)x-(a+b-3)y=4a+b\)**

Soln:

The given system of equation may be written as,

\(2x-3y-7=0\)

\((a+b)x-(a+b-3)y-(4a+b)=0\)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=-3,c_{1}=-7\)

\(a_{2}=(a+b),b_{2}=-(a+b-3) ,c_{2}=-(4a+b)\)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{2}{(a+b)} = \frac{-3}{-(a+b-3)} = \frac{-7}{-(4a+b)} Â \)

\(\frac{2}{(a+b)} = \frac{3}{(a+b-3)} \; and \; Â \frac{3}{(a+b-3)} =\frac{7}{(4a+b)} \)

\(\Rightarrow 2(a+b-3)=3(a+b) \; and \; 3(4a+b)=7 (a+b-3) \)

\(\Rightarrow 2a+2b-6=3a+3b \; and \; 12a+3b=7a+7b-21 \)

\(\Rightarrow a+b= -6 \; and \; 5a-4b= -21 \)

a+b= -6

\(\Rightarrow a=-6-b \)

Substituting the value of a in \(5a-4b= -21 \) we have

5(-b-6)-4b=-21

\(\Rightarrow -5b-30-4b=-21 \)

\(\Rightarrow 9b=-9 \)

\(\Rightarrow b=-1 \)

As a=-6-b

\(\Rightarrow a=-6+1=-5 \)

Hence the given system of equation will have infinitely many solution if

a=-5 and b=-1.

**(35) Find the value of p and q such that the following system of linear equation have infinitely many solution:**

**\(2x-3y=9\)**

**\((p+q)x+(2p-q)y=3(p+q+1) \)**

Soln:

The given system of equation may be written as,

\(2x-3y-9=0\)

\((p+q)x+(2p-q)y-3(p+q+1)=0 \)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=3,c_{1}=-9\)

\( a_{2}=(p+q),b_{2}=(2p-q) ,c_{2}= -3(p+q+1)\)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{2}{(p+q)} = \frac{3}{(2p-q)} = \frac{-9}{-3(p+q+1)} Â \)

\(\frac{2}{(p+q)} = \frac{3}{(2p-q)} \; and \; Â Â \frac{3}{(2p-q)} = \frac{3}{(p+q+1)} Â \)

\( 2(2p-q) = 3(p+q) \; and \; (p+q+1)=2p-q \)

\(\Rightarrow 4p-2q=3p+3q \; and \; -p+2q=-1 \)

\(\Rightarrow p=5q \; and \; p-2q=1 \)

Substituting the value of p in p-2q=1, we have

3q=1

\(\Rightarrow q=\frac{1}{3}\)

Substituting the value of p in \(p=5q \) we have

\( p= \frac{5}{3} \)

Hence the given system of equation will have infinitely many solution if

\( p= \frac{5}{3} \) and \( q= \frac{1}{3} \).

(**36) Find the values of a and b for which the following system of equation has infinitely many solution:**

**(i) \((2a-1)x+3y=5\)**

**\( 3x +(b-2)y=3 \)**

Soln:

The given system of equation may be written as,

\((2a-1)x+3y-5=0\)

\( 3x +(b-2)y-3=0 \)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2a-1,b_{1}=3,c_{1}=-5\)

\( a_{2}=3,b_{2}=b-2 ,c_{2}= -3(p+q+1)\)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{2a-1}{3} = \frac{-3}{b-2} = \frac{-5}{-3} Â \)

\(\frac{2a-1}{3} = \frac{5}{3} \; and \; Â \frac{-3}{b-2} = \frac{5}{3} Â \)

\( 2a-1 =5 \; and \; -9=5(b-2) Â Â \)

\(\Rightarrow a=3 \; and \; -9=5b-10 \)

\(\Rightarrow a=3 \; and \; b=\frac{1}{5} \)

Hence the given system of equation will have infinitely many solution if

\( a= 3\) and \( b= \frac{1}{5} \).

**(ii) \( 2x-(2a+5)y=5\)**

**\( (2b+1)x-9y=15 \)**

Soln:

The given system of equation may be written as,

\( 2x-(2a+5)y=5\)

\( (2b+1)x-9y=15 \)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=-(2a+5),c_{1}=-5\)

\( a_{2}=(2b+1),b_{2}=-9 ,c_{2}= -15\)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{2}{2b+1} = \frac{-(2a+5)}{-9} = \frac{-5}{-15} Â \)

\(\frac{2}{2b+1} =\frac{1}{3} \; and \; \frac{(2a+5)}{9} = \frac{1}{3} \)

\(\Rightarrow 6=2b+1 \; and \; 2a+5=3 Â Â \)

\(\Rightarrow b=\frac{5}{2} \; and \; a= -1 \)

Hence the given system of equation will have infinitely many solution if

\( a= -1\) and \( b= \frac{5}{2} \).

**(iii) \( (a-1)x+3y=2\)**

**\( 6x+(1-2b)y=6 \)**

Soln:

The given system of equation may be written as,

\( (a-1)x+3y=2\)

\( 6x+(1-2b)y=6 \)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=a-1,b_{1}=3,c_{1}=-2\)

\( a_{2}=6,b_{2}=1-2b ,c_{2}= -6\)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{a-1}{6} = \frac{3}{1-2b} = \frac{2}{6} Â \)

\(\frac{a-1}{6} =\frac{1}{3} \; and \; \frac{3}{1-2b} = \frac{1}{3} \)

\(\Rightarrow a-1=2 \; and \; 1-2b=9 Â Â \)

\(\Rightarrow a=3 \; and \; b= -4 \)

Hence the given system of equation will have infinitely many solution if

\( a= 3\) and \( b= -4 \).

**(iv) \( 3x+4y=12\)**

**\( (a+b)x+2(a-b)y=5a-1 \)**

Soln:

The given system of equation may be written as,

\( 3x+4y-12=0\)

\( (a+b)x+2(a-b)y-(5a-1)=0 \)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=3,b_{1}=4,c_{1}=-12\)

\( a_{2}=(a+b),b_{2}=2(a-b) ,c_{2}= -(5a-1)\)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{3}{a+b} = \frac{4}{2(a-b)} = \frac{12}{5a-1} Â \)

\(\frac{3}{a+b} =\frac{2}{a+b} \; and \; \frac{2}{a+b} = \frac{12}{5a-1} Â \)

\(\Rightarrow 3(a-b)=2a+2b \; and \; 2(5a-1)=12(a-b) \)

\(\Rightarrow a=5b \; and \; -2a=-12b+2 \)

Substituting a=5b in -2a=-12b+2 , we have

-2(5b)=-12b+2

\(\Rightarrow -10b=-12b+2\)

\(\Rightarrow b=1\)

Thus a=5

Hence the given system of equation will have infinitely many solution if

\( a= 5\) and \( b=1 \).

**(v) \( 2x+3y=7\)**

**\( (a-1)x+(a+1)y=3a-1 \)**

Soln:

The given system of equation may be written as,

\( 2x+3y-7=0\)

\( (a-1)x+(a+1)y-(3a-1)=0 \)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=3,c_{1}=-7\)

\( a_{2}=(a-1),b_{2}=(a+1) ,c_{2}= -(3a-1)\)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{2}{a-b} = \frac{3}{a+1)} = \frac{-7}{3a-1} Â \)

\(\frac{2}{a-b} =\frac{3}{a+1)} \; and \; \frac{3}{a+1)} = \frac{-7}{3a-1} Â \)

\(\Rightarrow 2(a+1)=3(a-1) \; and \; 3(3a-1)=7(a+1) \)

\(\Rightarrow 2a-3a=-3-2 \; and \; 9a-3=7a +7 \)

\(\Rightarrow a=5 \; and \; a=5 Â Â \)

Hence the given system of equation will have infinitely many solution if

\( a= 5\) and \( b=1 \).

**(vi) \( 2x+3y=7\)**

**\( (a-1)x+(a+2)y=3a \)**

Soln:

The given system of equation may be written as,

\( 2x+3y-7=0 \)

\( (a-1)x+(a+2)y-3a=0 \)

The given system of equation is of the form

\(a_{1}x+b_{1}y-c_{1}=0\)

\(a_{2}x+b_{2}y-c_{2}=0\)

Where, \(a_{1}=2,b_{1}=3,c_{1}=-7\)

\( a_{2}=(a-1),b_{2}=(a+2) ,c_{2}= -3a \)

The given system of equation will have infinitely many solution, if

\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

\(\frac{2}{a-b} = \frac{3}{a+2)} = \frac{-7}{-3a} Â \)

\(\frac{2}{a-b} =\frac{3}{a+2)} \; and \; \frac{3}{a+2)} = \frac{7}{3a} Â \)

\(\Rightarrow 2(a+2)=3(a-1) \; and \; 3(3a)=7(a+2) \)

\(\Rightarrow 2a+4=3a-3 \; and \; 9a=7a +14 \)

\(\Rightarrow a=7 \; and \; a=7 Â Â \)

Hence the given system of equation will have infinitely many solution if

\( a= 7\) and \( b=1 \)..