RD Sharma Solutions Class 10 Pair Of Linear Equations In Two Variables Exercise 3.5

RD Sharma Solutions Class 10 Chapter 3 Exercise 3.5

RD Sharma Class 10 Solutions Chapter 3 Ex 3.5 PDF Free Download

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) $x-3y-3=0$

$3x-9y-2=0$

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) $2x+y-5=0$

$4x+2y-10=0$

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) $3x-5y=20$

$6x-10y=40$

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) $x-2y-8=0$

$5x-10y-10=0$

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) $kx+2y-5=0$

$3x+y-1=0$

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

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

$\frac{k}{3} \neq \frac{2}{1}$

$\Rightarrow k\neq 6$

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

(6) $4x+ky+8=0$

$2x+2y+2=0$

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

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

$\frac{4}{2} \neq \frac{k}{2}$

$\Rightarrow k\neq 4$

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

(7) $4x-5y=k$

$2x-3y=12$

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

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

$\frac{4}{2} \neq \frac{-5}{-3}$

$\Rightarrow k$ can have any real values.

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

(8) $x+2y=3$

$5x+ky+7=0$

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

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

$\frac{1}{5} \neq \frac{2}{k}$

$\Rightarrow k \neq 10$

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) $2x+3y-5=0$

$6x-ky-15=0$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

$\frac{2}{6} \neq \frac{3}{k}$

$\Rightarrow k = 9$

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

(10) $4x+5y=3$

$kx+15y=9$

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) $kx-2y+6=0$

$4x+3y+9=0$

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) $8x+5y=9$

$kx+10y=19$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

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

$\Rightarrow k = 16$

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

(13) $2x-3y=7$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow k = 4\; and \; 4k=16 \Rightarrow k = 4$

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

(14) $2x+3y=2$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow k = 4\; and \; k=4$

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

(15) $x+(k+1)y=4$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

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

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

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

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

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

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

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

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

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

thus, k=2 satisfies both the condition.

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

(16) $kx+3y=2k+1$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow 3k = 6\; and \; 3k=6$

$\Rightarrow k = 2\; and \; k=2$

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

(17) $2x+(k-2)y=k$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow -k = -5\; and \; 2k=10$

$\Rightarrow k = 5\; and \; k=5$

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

(18) $2x+3y=7$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow k = 5\; and \; 2k=10$

$\Rightarrow k = 5\; and \; k=5$

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

(19) $2x+3y=k$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

$\frac{2}{k-1} = \frac{3}{k+2} = \frac{-k}{-3k}$

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

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

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

$\Rightarrow k = 7\; and \; k=7$

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) $kx-5y=2$

$6x+2y= 7$

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) $x+2y=0$

$2x+ky= 5$

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) $3x-4y+7=0$

$kx+3y-5=0$

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) $2x-ky+3=0$

$3x+2y-1=0$

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) $2x+ky-11=0$

$5x-7y-5=0$

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) $kx+3y=3$

$12x+ky=6$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\frac{k}{12} = \frac{3}{k} \neq \frac{3}{6}$    ……(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,

$\frac{a}{12} = \frac{3}{a}$

$\Rightarrow a^{2}=36$

$\Rightarrow a=+6\; or \; -6$

$a \neq 6$

$\Rightarrow a= -6$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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

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

$\Rightarrow c=6 \; and \; c-3=3$

$\Rightarrow c=6 \; and \; c=6$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\frac{2}{3} = \frac{k}{-5} neq \frac{-1}{-7}$

$\frac{2}{3} = \frac{k}{-5} \; and \; \frac{k}{-5} neq \frac{1}{7}$

$\Rightarrow k= \frac{-10}{3} \; and \; k neq \frac{-5}{7}$

$\Rightarrow k=\frac{-10}{3}$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{2}{3} = \frac{k}{-5} = \frac{-1}{-7}$

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.

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{1}{2} = \frac{2}{k} = \frac{7}{14}$

$\frac{1}{2} = \frac{2}{k} = \frac{1}{2}$

$\Rightarrow k=4$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow 4a = 17 \; and \; 5b = 11$

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

(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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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

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

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

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

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow p=5q \; and \; p-2q=1$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{2a-1}{3} = \frac{-3}{b-2} = \frac{-5}{-3}$

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

$2a-1 =5 \; and \; -9=5(b-2)$

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

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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

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

$\Rightarrow 6=2b+1 \; and \; 2a+5=3$

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{a-1}{6} = \frac{3}{1-2b} = \frac{2}{6}$

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

$\Rightarrow a-1=2 \; and \; 1-2b=9$

$\Rightarrow a=3 \; and \; b= -4$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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

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

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

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

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow a=5 \; and \; a=5$

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

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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

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

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

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

$\Rightarrow a=7 \; and \; a=7$

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

$a= 7$ and $b=1$..

Practise This Question

We have three numbers 2, 4, 7. Find the LCM of these numbers.