# RD Sharma Solutions Class 10 Pair Of Linear Equations In Two Variables 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. $x-3y-3=0$

$3x-9y-2=0$

Solution:

Given:

$x-3y-3=0$

$3x-9y-2=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=1,b_{1}=-3,c_{1}=-3$

$a_{2}=3,b_{2}=-9,c_{2}=-2$

According to the question, we get,

$\frac{a_{1}}{a_{2}}=\frac{1}{3}$

$\frac{b_{1}}{b_{2}}=\frac{-3}{-9}=\frac{1}{3}$

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

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

∴, the given system of equation has no solution.

2. $2x+y-5=0$

$4x+2y-10=0$

Solution:

Given:

$2x+y-5=0$

$4x+2y-10=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=1,c_{1}=-5$

$a_{2}=4,b_{2}=2,c_{2}=-10$

According to the question, we get,

$\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}$

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

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

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

∴, the given system of equation has infinitely many solution.

3. $3x-5y=20$

$6x-10y=40$

Solution:

Given:

$3x-5y=20$

$6x-10y=40$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=3,b_{1}=-5,c_{1}=-20$

$a_{2}=6,b_{2}=-10,c_{2}=-40$

According to the question, we get,

$\frac{a_{1}}{a_{2}}=\frac{3}{6}=\frac{1}{2}$

$\frac{b_{1}}{b_{2}}=\frac{-5}{-10}=\frac{1}{2}$

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

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

∴, the given system of equation has infinitely many solution.

4. $x-2y-8=0$

$5x-10y-10=0$

Solution:

Given:

$x-2y-8=0$

$5x-10y-10=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=1,b_{1}=-2,c_{1}=-8$

$a_{2}=5,b_{2}=-10,c_{2}=-10$

According to the question, we get,

$\frac{a_{1}}{a_{2}}=\frac{1}{5}$

$\frac{b_{1}}{b_{2}}=\frac{-2}{-10}=\frac{1}{5}$

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

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

∴, the given system of 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. $kx+2y-5=0$

$3x+y-1=0$

Solution:

Given:

$kx+2y-5=0$

$3x+y-1=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=k,b_{1}=2,c_{1}=-5$

$a_{2}=3,b_{2}=1,c_{2}=-1$

According to the question,

For unique solution, we hav\)

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

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

$\Rightarrow k\neq 6$

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

6. $4x+ky+8=0$

$2x+2y+2=0$

Solution:

Given:

$4x+ky+8=0$

$2x+2y+2=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=4,b_{1}=k,c_{1}=8$

$a_{2}=2,b_{2}=2,c_{2}=2$

According to the question,

For unique solution, we hav\)

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

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

$\Rightarrow k\neq 4$

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

7. $4x-5y=k$

$2x-3y=12$

Solution:

Given:

$4x-5y-k=0$

$2x-3y-12=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=4,b_{1}=-5,c_{1}=-k$

$a_{2}=2,b_{2}=-3,c_{2}=-12$

According to the question,

For unique solution, we hav\)

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

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

$\Rightarrow k$ can have any real values.

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

8. $x+2y=3$

$5x+ky+7=0$

Solution:

Given:

$x+2y=3$

$5x+ky+7=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=1,b_{1}=2,c_{1}=-3$

$a_{2}=5,b_{2}=k,c_{2}=7$

According to the question,

For unique solution, we hav\)

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

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

$\Rightarrow k \neq 10$

∴, the given  system of equations 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. $2x+3y-5=0$

$6x-ky-15=0$

Solution:

Given:

$2x+3y-5=0$

$6x-ky-15=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=3,c_{1}=-5$

$a_{2}=6,b_{2}=k,c_{2}=-15$

According to the question,

For unique solution, we hav\)

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

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

$\Rightarrow k = 9$

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

10. $4x+5y=3$

$kx+15y=9$

Solution:

Given:

$4x+5y=3$

$kx+15y=9$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=4,b_{1}=5,c_{1}=3$

$a_{2}=k,b_{2}=15,c_{2}=9$

According to the question,

For unique solution, we hav\)

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

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

$\frac{4}{k} = \frac{1}{3}$

$\Rightarrow k \neq 12$

∴, the given  system of equations will have infinitely many solutions if k=12.

11. $kx-2y+6=0$

$4x+3y+9=0$

Solution:

Given:

$kx-2y+6=0$

$4x+3y+9=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=k,b_{1}=-2,c_{1}=6$

$a_{2}=4,b_{2}=-3,c_{2}=9$

According to the question,

For unique solution, we hav\)

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

$\frac{k}{4} = \frac{-2}{-3} = \frac{2}{3}$

$\Rightarrow k = \frac{8}{3}$

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

12. $8x+5y=9$

$kx+10y=19$

Solution:

Given:

$8x+5y=9$

$kx+10y=19$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=8,b_{1}=5,c_{1}=-9$

$a_{2}=k,b_{2}=10,c_{2}=-18$

According to the question,

For unique solution, we hav\)

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

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

$\Rightarrow k = 16$

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

13. $2x-3y=7$

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

Solution:

Given:

$2x-3y=7$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=-3,c_{1}=-7$

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

According to the question,

For unique solution, we hav\)

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

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

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

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

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

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

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

14. $2x+3y=2$

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

Solution:

Given:

$2x+3y=2$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=3,c_{1}=-2$

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

According to the question,

For unique solution, we hav\)

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

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

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

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

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

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

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

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

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

Solution:

Given:

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

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=1,b_{1}=(k+1),c_{1}=-4$

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

According to the question,

For unique solution, we hav\)

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

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

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

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

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

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

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

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

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

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

thus, k=2 satisfies both the condition.

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

16. $kx+3y=2k+1$

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

Solution:

Given:

$kx+3y=2k+1$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

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

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

According to the question,

For unique solution, we hav\)

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

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

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

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

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

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

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

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

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

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

Solution:

Given:

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

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=(k-2),c_{1}=-k$

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

According to the question,

For unique solution, we hav\)

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

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

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

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

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

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

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

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

18. $2x+3y=7$

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

Solution:

Given:

$2x+3y=7$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=3,c_{1}=-7$

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

According to the question,

For unique solution, we hav\)

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

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

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

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

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

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

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

∴, if $k = 5 [/latex the given system of equations will have infinitely many solutions. 19. $$2x+3y=k$ $(k-1)x+(k+2)y= 3k$ Solution: Given: $2x+3y=k$ $(k-1)x+(k+2)y= 3k$ The equations are of the form $a_{1}x+b_{1}y-c_{1}=0$ $a_{2}x+b_{2}y-c_{2}=0$ Here, $a_{1}=2,b_{1}=3,c_{1}=-k$ $a_{2}=k-1,b_{2}=k+2,c_{2}=-3k$ According to the question, For unique solution, we hav$$

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

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

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

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

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

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

∴, if $k = 7$, then the given system of equations will have infinitely many solutions.

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

20. $kx-5y=2$

$6x+2y= 7$

Solution:

Given:

$kx-5y=2$

$6x+2y= 7$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=k,b_{1}=-5,c_{1}=-2$

$a_{2}=6,b_{2}=2,c_{2}=-7$

According to the question,

For no solution, we hav\)

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

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

$\Rightarrow 2k = -30$

$\Rightarrow k = -15$

∴, if $k = -15$, then the given system of equations will have no solutions.

21. $x+2y=0$

$2x+ky= 5$

Solution:

Given:

$x2y=0$

$2x+ky= 5$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=1,b_{1}=2,c_{1}=0$

$a_{2}=2,b_{2}=k,c_{2}=-5$

According to the question,

For no solution, we hav\)

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

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

$\Rightarrow k = 4$

∴, if $k = 4$, then the given system of equations will have no solutions.

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

$kx+3y-5=0$

Solution:

Given:

$3x-4y+7=0$

$kx+3y-5=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=3,b_{1}=-4,c_{1}=7$

$a_{2}=k,b_{2}=3,c_{2}=-5$

According to the question,

For no solution, we hav\)

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

$\frac{3}{k} = \frac{-4}{3}$

$\Rightarrow k = \frac{-9}{4}$

∴, if $k = \frac{-9}{4}$, then the given system of equations will have no solutions.

23. $2x-ky+3=0$

$3x+2y-1=0$

Solution:

Given:

$2x-ky+3=0$

$3x+2y-1=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=-k,c_{1}=3$

$a_{2}=3,b_{2}=2,c_{2}=-1$

According to the question,

For no solution, we hav\)

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

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

$\Rightarrow k = \frac{-4}{3}$

∴, if $k = \frac{-4}{3}$, then the given system of equations will have no solutions.

24. $2x+ky-11=0$

$5x-7y-5=0$

Solution:

Given:

$2x+ky-11=0$

$5x-7y-5=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=k,c_{1}=-11$

$a_{2}=5,b_{2}=-7,c_{2}=-5$

According to the question,

For no solution, we hav\)

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

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

$\Rightarrow k = \frac{-14}{5}$

∴, if $k = \frac{-14}{5}$, then the given system of equations will have no solutions.

25. $kx+3y=3$

$12x+ky=6$

Solution:

Given:

$kx+3y=3$

$12x+ky=6$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=k,b_{1}=3,c_{1}=-3$

$a_{2}=12,b_{2}=k,c_{2}=-6$

According to the question,

For no solution, we hav\)

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

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

$\Rightarrow k^{2} = 36$

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

From (i)

$\frac{k}{12} \neq \frac{3}{6}$

$\Rightarrow k \neq 6$

∴, if $k = -6$, then the given system of equations will have no solutions.

26. For what value of a, the following system of equation will be inconsistent?

$4x+6y-11=0$

$2x+ay-7=0$

Solution:

Given:

$4x+6y-11=0$

$2x+ay-7=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=4,b_{1}=6,c_{1}=-11$

$a_{2}=2,b_{2}=a,c_{2}=-7$

According to the question,

For unique solution, we hav\)

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

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

$\frac{4}{2} = \frac{6}{a}$

$\Rightarrow a = 3$

∴, if $a = 3$, then the given system of equations will be inconsistent.

27. For what value of a, the following system of equation have no solution?

$ax+3y=a-3$

$12x+ay=a$

Solution:

Given:

$ax+3y=a-3$

$12x+ay=a$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=a,b_{1}=3,c_{1}=-(a-3)$

$a_{2}=12,b_{2}=a,c_{2}=-a$

According to the question,

For unique solution, we hav\)

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

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

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

$\Rightarrow a-3 \neq 3$

$\Rightarrow a \neq 6$

And,

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

$\Rightarrow a^{2}=36$

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

∵ $a \neq 6$

$\Rightarrow a= -6$

∴, if $a = -6$, then the given system of equations will have no solution.

28. Find the value of a, for which the following system of equation have

(i) Unique solution

(ii) No solution

$kx+2y=5$

$3x+y=1$

Solution:

Given:

$kx+2y-5=0$

$3x+y-1=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=k,b_{1}=2,c_{1}=-5$

$a_{2}=3,b_{2}=1,c_{2}=-1$

(i) For unique solution,

According to the question, we hav\)

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

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

$k \neq 6$

∴, if $k \neq 6$, then the given system of equations will have unique solution.

(ii) For no solution,

According to the question, we hav\)

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

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

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

$\Rightarrow k=6$

∴, if $a = 6$, then the given system of equations will have no solution.

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

$6x+3y=c-3$

$12x+cy=c$

Solution:

Given:

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

$12x+cy-c=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=6,b_{1}=3,c_{1}=-(c-3)$

$a_{2}=12,b_{2}=c,c_{2}=-c$

According to the question,

For infinitely many solution, we hav\)

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

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

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

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

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

∴, if $c = 6$, then the given system of equations will have infinitely many solution.

30. Find the value of k, for which the following system of equation have

(i) Unique solution

(ii) No solution

(iii) Infinitely many solution

$2x+ky=1$

$3x-5y=7$

Solution:

Given:

$2x+ky=1$

$3x-5y=7$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=k,c_{1}=-1$

$a_{2}=3,b_{2}=-5,c_{2}=-7$

(i) For unique solution,

According to the question, we have,

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

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

$k \neq \frac{-10}{3}$

∴, if $k \neq \frac{-10}{3}$, then the given system of equations will have unique solution.

(ii) For no solution,

According to the question, we hav\)

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

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

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

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

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

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

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

According to the question, we hav\)

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

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

Since, $\frac{a_{1}}{a_{2}} \neq \frac{c_{1}}{c_{2}}$ ,

∴, 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?

$x+2y+7=0$

$2x+ky+14=0$

Solution:

Given:

$x+2y+7=0$

$2x+ky+14=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=1,b_{1}=2,c_{1}=7$

$a_{2}=2,b_{2}=k,c_{2}=14$

According to the question,

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

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

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

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

$\Rightarrow k=4$

∴, if $k = 4$, then the given system of equations will have infinitely many solution.

32. Obtain the condition for the following system of linear equations to have a unique solution

$ax+by=c$

$lx+my=n$

Solution:

Given:

$ax+by-c=0$

$lx+my-n=0$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=a,b_{1}=b,c_{1}=-c$

$a_{2}=l,b_{2}=m,c_{2}=-n$

According to the question,

For unique solution, we hav\)

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

$\Rightarrow \frac{a}{l} \neq \frac{b}{m}$

$\Rightarrow am \neq bl$

∴, if $am \neq bl$, then the given system of equations will have unique solution.

33. Determine the values of a and b so that the following system of linear equations have infinitely many solutions:

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

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

Solution:

Given:

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

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=(2a-1),b_{1}=3,c_{1}=-5$

$a_{2}=3,b_{2}=b-1,c_{2}=-2$

According to the question,

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:

$2x-3y=7$

$(a+b)x-(a+b-3)y=4a+b$

Solution:

Given:

$2x-3y-7=0$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=-3,c_{1}=-7$

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

According to the question,

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 hav\)

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

$\Rightarrow -5b-30-4b=-21$

$\Rightarrow 9b=-9$

$\Rightarrow b=-1$

As a=-6-b

$\Rightarrow a=-6+1=-5$

∴, if a=-5 and b=-1, then the given system of equation will have infinitely many solution.

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

$2x-3y=9$

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

Solution:

Given:

$2x-3y-9=0$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=3,c_{1}=-9$

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

According to the question,

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 hav\)

3q=1

$\Rightarrow q=\frac{1}{3}$

Substituting the value of p in $p=5q$ we hav\)

$p= \frac{5}{3}$

∴, if $p= \frac{5}{3}$ and $q= \frac{1}{3}$ then the given system of equation will have infinitely many solution.

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

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

$3x +(b-2)y=3$

Solution:

Given:

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

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2a-1,b_{1}=3,c_{1}=-5$

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

According to the question,

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

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

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

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

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

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

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

∴, if$a= 3$ and $b= \frac{1}{5}$, then the given system of equation will have infinitely many solution.

(ii) $2x-(2a+5)y=5$

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

Solution:

Given:

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

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=-(2a+5),c_{1}=-5$

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

According to the question,

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

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

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

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

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

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

∴, if $a= -1$ and $b= \frac{5}{2}$, then the given system of equation will have infinitely many solution.

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

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

Solution:

Given:

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

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=a-1,b_{1}=3,c_{1}=-2$

$a_{2}=6,b_{2}=1-2b ,c_{2}= -6$

According to the question,

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

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

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

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

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

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

∴, if $a= 3$ and $b= -4$, then the given system of equation will have infinitely many solution.

(iv) $3x+4y=12$

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

Solution:

Given:

$3x+4y-12=0$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=3,b_{1}=4,c_{1}=-12$

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

According to the question,

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 hav\)

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

$\Rightarrow -10b=-12b+2$

$\Rightarrow b=1$

Thus a=5

∴, if $a= 5$ and $b=1$, then the given system of equation will have infinitely many solution.

(v)

$2x+3y=7$

$(a-b)x+(a+b)y=3a+b-2$

Solution:

Given:

$2x+3y=7$

$(a-b)x+(a+b)y=3a+b-2$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

(vi) $2x+3y-7=0$

$(a-1)x+(a+1)y=3a-1$

Solution:

Given:

$2x+3y-7=0$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=3,c_{1}=-7$

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

According to the question,

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

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

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

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

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

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

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

∴, if $a= 5$ and $b=1$, then the given system of equation will have infinitely many solution.

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

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

Solution:

Given:

$2x+3y-7=0$

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

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here, $a_{1}=2,b_{1}=3,c_{1}=-7$

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

According to the question,

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

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

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

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

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

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

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

∴, if $a= 7$ and $b=1$, then the given system of equation will have infinitely many solution.

(viii)

$x+2y=1$

$(a-b)x+(a+b)y=a+b-2$

Given:

$x+2y=1$ …(i)

$(a-b)x+(a+b)y=a+b-2$ …(ii)

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

3b+b=4

⇒  4b=4

⇒  b=1

Substituting the value of b in Eq(iii)

a=3 x 1

⇒  a=3

(ix)

$2x+3y=7$

$2ax+ ay=28-by$

Given:

$2x+3y=7$

$2ax+ ay=28-by$

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

37. For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have

(i) no solution ?

(ii) infinitely many solutions ?

(iii) a unique solutions ?

Solution:

Given:

λx + y = λ2

x + λy = 1

The equations are of the form

$a_{1}x+b_{1}y-c_{1}=0$

$a_{2}x+b_{2}y-c_{2}=0$

Here,

a1=λ;  b1=1; c12

a2=1;  b2=λ; c2=1