Mathematics Part i Solutions Solutions for Class 10 Maths Chapter 1 - Linear Equations In Two Variables

Answer:

Disclaimer: There is error in the Q. In (II) there should have been 2x $-$ 3y = 12
5x + 3y = 9 -----(I)
2x $-$ 3y = 12 ----- (II)
Add (I) and (II)
7x = 21
x = 3
Putting the value of x = 3 in (I) we get
$$\begin{gathered} 5\left(3\right)+3y=9 \\ \Rightarrow 15+3y=9 \\ \Rightarrow 3y=9-15=-6 \\ \Rightarrow y=-2 \end{gathered}$$
Thus, (x, y) = (3, $-$6). 

Answer:

(1) 3a + 5b = 26;                           .....(I)
a + 5b = 22                                    .....(II)
Subtracting (II) from (I)
2a = 4
$\Rightarrow$ a = 2
Putting the value of a = 2 in (II)
5b = 22 $-$ 2 = 20
$\Rightarrow$b = $\frac{20}{5}=4$ 
Thus, a = 2 and b = 4. 

(2) x + 7y = 10;                             .....(I)
3x – 2y = 7                                    .....(II)  
Multiplying (I) with 3
3x + 21y = 30;                             .....(III)
3x – 2y = 7                                    .....(IV)  
Subtracting (IV) from (III) we get
23y = 23
$\Rightarrow$y = 1
Putting the value of y in (IV) we get
3x – 2 = 7
$\Rightarrow$3x = 7 + 2 = 9
$\Rightarrow$3x = 9
$\Rightarrow$x = 3
Thus, (x, y) = (3, 1)

(3) 2x – 3y = 9                         .....(I)
2x + y = 13                              .....(II)   
Subtracting (II) from (I) we get
– 3y − y = 9 − 13
$$\begin{gathered} \Rightarrow -4y=-4 \\ \Rightarrow y=1 \end{gathered}$$
Putting this value in (I) we get
$$\begin{gathered} 2x-3\left(1\right)=9 \\ \Rightarrow 2x=9+3=12 \\ \Rightarrow x=\frac{12}{2}=6 \end{gathered}$$
Thus, (x, y) = (6, 1)

(4) 5m – 3n = 19               .....(I)
m – 6n = –7                      .....(II)
Multiplying (I) with 2 we get
10m – 6n = 38               .....(III)
m – 6n = –7                      .....(IV)
Subtracting (IV) from (III) we get
$$\begin{gathered} 10m-m-6n-\left(-6n\right)=38-\left(-7\right) \\ \Rightarrow 9m=45 \\ \Rightarrow m=\frac{45}{9}=5 \end{gathered}$$
Putting the value of m = 5 in (II) we get
 $$\begin{gathered} 5-6n=-7 \\ \Rightarrow -6n=-7-5 \\ \Rightarrow -6n=-12 \\ \Rightarrow n=\frac{-12}{-6}=2 \end{gathered}$$
Thus, (m, n) = (5, 2).   

(5) 5x + 2y = –3                      .....(I)
x + 5y = 4                                .....(II)   
Multiply (II) with 5 we get
5x + 25y = 20                          .....(III)
Subtracting (III) from (I) we get
$$\begin{gathered} 5x-5x+2y-25y=-3-20 \\ \Rightarrow -23y=-23 \\ \Rightarrow y=\frac{-23}{-23}=1 \end{gathered}$$
Putting the value of y = 1 in (II) we get
$$\begin{gathered} x+5\left(1\right)=4 \\ \Rightarrow x+5=4 \\ \Rightarrow x=4-5=-1 \end{gathered}$$
Thus, (x, y) = (−1, 1)

(6)
 $$\begin{gathered} \frac{1}{3}x+y=\frac{10}{3} .....\left(\mathrm{I}\right) \\ 2x+\frac{1}{4}y=\frac{11}{4} .....\left(\mathrm{II}\right) \end{gathered}$$
Multiply (I) with 3 and (II) with 4
$$\begin{gathered} x+3y=10 .....\left(\mathrm{III}\right) \\ 8x+y=11 .....\left(\mathrm{IV}\right) \end{gathered}$$
Multiply (IV) with 3
24x + 3y = 33         .....(V)
Subtracting (V) from (III)
$$\begin{gathered} x-24x+3y-3y=10-33 \\ \Rightarrow -23x=-23 \\ \Rightarrow x=1 \end{gathered}$$
Putting the value of x = 1 in (III)
$$\begin{gathered} 1+3y=10 \\ \Rightarrow 3y=10-1=9 \\ \Rightarrow y=\frac{9}{3}=3 \end{gathered}$$
 Thus, (x, y) = (1, 3)

(7) 99x + 101y = 499                       .....(I)
101x + 99y = 501                             .....(II)
Adding (I) and (II) 
$$\begin{gathered} 200x+200y=1000 \\ \Rightarrow x+y=5 .....\left(\mathrm{III}\right) \end{gathered}$$                            
Subtracting (II) from (I) 
$$\begin{gathered} 99x-101x+101y-99y=499-501 \\ \Rightarrow -2x+2y=-2 \\ \Rightarrow -x+y=-1 .....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV)
$$\begin{gathered} x+y=5 \\ -x+y=-1 \\ \Rightarrow 2y=4 \\ \Rightarrow y=2 \end{gathered}$$
Putting the value of y = 2 in (III) we get
$$\begin{gathered} x+2=5 \\ \Rightarrow x=5-2=3 \end{gathered}$$
Thus, (x, y) = (3, 2)

(8) 49x – 57y = 172                         .....(I)
57x – 49y = 252                              .....(II)
Adding (I) and (II) 
$$\begin{gathered} 49x+57x-57y-49y=172+252 \\ \Rightarrow 106x-106y=424 \\ \Rightarrow x-y=4 .....\left(\mathrm{III}\right) \end{gathered}$$
Subtracting (II) from (I) we have
$$\begin{gathered} 49x-57y-57y-\left(-49y\right)=252-172 \\ \Rightarrow -8x-8y=-80 \\ \Rightarrow -x-y=-10 \\ \Rightarrow x+y=10 .....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV)

$$\begin{gathered} x-y=4 \\ x+y=10 \\ \Rightarrow 2x=14 \\ \Rightarrow x=7 \end{gathered}$$
Putting the value of x = 7 in (IV) we get
$$\begin{gathered} 7+y=10 \\ \Rightarrow y=10-7 \\ \Rightarrow y=3 \end{gathered}$$
Thus, (x, y) = (7, 3).     
 



 

Answer:

Answer:

(1)  x + y = 6; 
 

Answer:

$\left|\begin{array}{cc}3 & 2\\ 4 & 5\end{array}\right|=3\left(5\right)-2\left(4\right)=15-8=7$
Thus, we have
$\left|\begin{array}{cc}3 & 2\\ 4 & 5\end{array}\right|=3\times \boxed{ 5 } – \boxed{ 2 }\times 4=\boxed{ 15 }–8=\boxed{ 7 }$

Answer:

(1) $\left|\begin{array}{cc}-1 & 7\\ 2 & 4\end{array}\right|$

= $-1\left(4\right)-7\left(2\right)=-4-14=-18$

(2) $\left|\begin{array}{cc} 5 & 3\\ -7 & 0\end{array}\right|=5\times 0-3\times \left(-7\right)=0+21=21$ 

(3) $\left|\begin{array}{cc}\frac{7}{3}& \frac{5}{3}\\ \frac{3}{2}& \frac{1}{2}\end{array}\right|=\frac{7}{3}\times \frac{1}{2}-\frac{5}{3}\times \frac{3}{2}=\frac{7}{6}-\frac{5}{2}=\frac{7-15}{6}=\frac{-8}{6}=\frac{-4}{3}$
 

Answer:

(1) 3x – 4y = 10
4x + 3y = 5
$$\begin{gathered} D=\left|\begin{array}{cc}3& -4\\ 4& 3\end{array}\right|=3\times 3-\left(-4\right)\times 4=9+16=25 \\ {D}_x=\left|\begin{array}{cc}10& -4\\ 5& 3\end{array}\right|=10\times 3-\left(-4\right)\times 5=30+20=50 \\ {D}_y=\left|\begin{array}{cc}3& 10\\ 4& 5\end{array}\right|=3\times 5-10\times 4=15-40=-25 \end{gathered}$$
$$\begin{gathered} x=\frac{D_x}{D}=\frac{50}{25}=2 \\ y=\frac{D_y}{D}=\frac{-25}{25}=-1 \\ \left(x,y\right)=\left(2,-1\right) \end{gathered}$$

(2) 4x + 3y – 4 = 0 ; 6x = 8 – 5y
$$\begin{gathered} D=\left|\begin{array}{cc}4& 3\\ 6& 5\end{array}\right|=4\times 5-6\times 3=20-18=2 \\ {D}_x=\left|\begin{array}{cc}4& 3\\ 8& 5\end{array}\right|=4\times 5-3\times 8=20-24=-4 \\ {D}_y=\left|\begin{array}{cc}4& 4\\ 6& 8\end{array}\right|=4\times 8-6\times 4=32-24=8 \end{gathered}$$
$$\begin{gathered} x=\frac{D_x}{D}=\frac{-4}{2}=-2 \\ y=\frac{D_y}{D}=\frac{8}{2}=4 \\ \left(x,y\right)=\left(-2,4\right) \end{gathered}$$

(3) x + 2y = –1 ; 2x – 3y = 12
$$\begin{gathered} D=\left|\begin{array}{cc}1& 2\\ 2& -3\end{array}\right|=1\times \left(-3\right)-2\times 2=-3-4=-7 \\ {D}_x=\left|\begin{array}{cc}-1& 2\\ 12& -3\end{array}\right|=\left(-1\right)\times \left(-3\right)-2\times 12=3-24=-21 \\ {D}_y=\left|\begin{array}{cc}1& -1\\ 2& 12\end{array}\right|=1\times 12-\left(-1\right)\times 2=12+2=14 \end{gathered}$$
$$\begin{gathered} x=\frac{D_x}{D}=\frac{-21}{-7}=3 \\ y=\frac{D_y}{D}=\frac{14}{-7}=-2 \\ \left(x,y\right)=\left(3,-2\right) \end{gathered}$$

(4) 6x – 4y = –12 ; 8x – 3y = –2

$$\begin{gathered} D=\left|\begin{array}{cc}6& -4\\ 8& -3\end{array}\right|=6\times \left(-3\right)-\left(-4\right)\times 8=-18+32=14 \\ {D}_x=\left|\begin{array}{cc}-12& -4\\ -2& -3\end{array}\right|=\left(-12\right)\times \left(-3\right)-\left(-4\right)\times \left(-2\right)=36-8=28 \\ {D}_y=\left|\begin{array}{cc}6& -12\\ 8& -2\end{array}\right|=6\times \left(-2\right)-\left(-12\right)\times 8=-12+96=84 \end{gathered}$$
$$\begin{gathered} x=\frac{D_x}{D}=\frac{28}{14}=2 \\ y=\frac{D_y}{D}=\frac{84}{14}=6 \\ \left(x,y\right)=\left(2,6\right) \end{gathered}$$

(5) 4m + 6n = 54 ; 3m + 2n = 28
$$\begin{gathered} D=\left|\begin{array}{cc}4& 6\\ 3& 2\end{array}\right|=4\times 2-6\times 3=8-18=-10 \\ {D}_x=\left|\begin{array}{cc}54& 6\\ 28& 2\end{array}\right|=54\times 2-6\times 28=108-168=-60 \\ {D}_y=\left|\begin{array}{cc}4& 54\\ 3& 28\end{array}\right|=4\times 28-54\times 3=112-162=-50 \end{gathered}$$
$$\begin{gathered} x=\frac{D_x}{D}=\frac{-60}{-10}=6 \\ y=\frac{D_y}{D}=\frac{-50}{-10}=5 \\ \left(x,y\right)=\left(6,5\right) \end{gathered}$$

(6)  $2x+3y=2 ; x-\frac{y}{2}=\frac{1}{2}$
$$\begin{gathered} D=\left|\begin{array}{cc}2& 3\\ 1& \frac{-1}{2}\end{array}\right|=2\times \left(-\frac{1}{2}\right)-3\times 1=-1-3=-4 \\ {D}_x=\left|\begin{array}{cc}2& 3\\ \frac{1}{2}& \frac{-1}{2}\end{array}\right|=2\times \left(-\frac{1}{2}\right)-3\times \frac{1}{2}=-1-\frac{3}{2}=\frac{-5}{2} \\ {D}_y=\left|\begin{array}{cc}2& 2\\ 1& \frac{1}{2}\end{array}\right|=2\times \frac{1}{2}-2\times 1=1-2=-1 \end{gathered}$$
$$\begin{gathered} x=\frac{D_x}{D}=\frac{{\displaystyle \frac{-5}{2}}}{-4}=\frac{5}{8} \\ y=\frac{D_y}{D}=\frac{-1}{-4}=\frac{1}{4} \\ \left(x,y\right)=\left(\frac{5}{8},\frac{1}{4}\right) \end{gathered}$$
 

Answer:

$\left(1\right) \frac{2}{x}-\frac{3}{y}=15; \frac{8}{x}+\frac{5}{y}=77$
$\mathrm{Let} \frac{1}{x}=u \mathrm{and} \frac{1}{y}=v$
So, the equation becomes
$$\begin{gathered} 2u-3v=15 .....\left(\mathrm{I}\right) \\ 8u+5v=77 .....\left(\mathrm{II}\right) \end{gathered}$$
Multiply (I) with 4 we get
$8u-12v=60 .....\left(\mathrm{III}\right)$
(II) − (III)
$$\begin{gathered} 8u-8u+5v-\left(-12v\right)=77-60 \\ \Rightarrow 17v=17 \\ \Rightarrow v=1 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} v \mathrm{in} \left(\mathrm{I}\right) \\ 2\mathrm{u}-3\left(1\right)=15 \\ \Rightarrow 2\mathrm{u}=15+3=18 \\ \Rightarrow \mathrm{u}=9 \end{gathered}$$
Thus, 
$$\begin{gathered} \frac{1}{x}=u=9 \\ \Rightarrow x=\frac{1}{9} \\ \frac{1}{y}=v=1 \\ \Rightarrow y=1 \\ \left(x,y\right)=\left(\frac{1}{9},1\right) \end{gathered}$$

$\left(2\right) \frac{10}{x+y}+\frac{2}{x-y}=4; \frac{15}{x+y}-\frac{5}{x-y}=-2$
Let $\frac{1}{x+y}=u \mathrm{and} \frac{1}{x-y}=v$
So, the equation becomes 
$$\begin{gathered} 10u+2v=4 .....\left(\mathrm{I}\right) \\ 15u-5v=-2 .....\left(\mathrm{II}\right) \end{gathered}$$
Multiplying (I) with 5 and (II) with 2 we get
$$\begin{gathered} 50u+10v=20 .....\left(\mathrm{III}\right) \\ 30u-10v=-4 .....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV) we get
$u=\frac{16}{80}=\frac{1}{5}$
Putting this value in (I)
$$\begin{gathered} 10\times \frac{1}{5}+2v=4 \\ \Rightarrow 2+2v=4 \\ \Rightarrow v=1 \end{gathered}$$

$$\begin{gathered} \frac{1}{x+y}=\frac{1}{5} \mathrm{and} \frac{1}{x-y}=1 \\ \Rightarrow x+y=5 \mathrm{and} x-y=1 \\ \mathrm{Solving} \mathrm{these} \mathrm{equations} \mathrm{we} \mathrm{get} \\ x=3 \mathrm{and} y=2 \end{gathered}$$

$\left(3\right) \frac{27}{x-2}+\frac{31}{y+3}=85; \frac{31}{x-2}+\frac{27}{y+3}=89$
$\mathrm{Let} \frac{1}{x-2}=u \mathrm{and} \frac{1}{y+3}=v$
$$\begin{gathered} 27u+31v=85 .....\left(\mathrm{I}\right) \\ 31u+27v=89 .....\left(\mathrm{II}\right) \\ \mathrm{Adding} \left(\mathrm{I}\right) \mathrm{and} \left(\mathrm{II}\right) \\ 58u+58v=174 \\ u+v=3 .....\left(\mathrm{III}\right) \\ \mathrm{Subtracting} \left(\mathrm{II}\right) \mathrm{from} \left(\mathrm{I}\right) \\ 4u-4v=4 \\ \Rightarrow u-v=1 .....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV) we get
$$\begin{gathered} 2u=4 \\ \Rightarrow u=2 \end{gathered}$$
Putting the value of u in III
$$\begin{gathered} 2+v=3 \\ \Rightarrow v=1 \end{gathered}$$
$$\begin{gathered} \frac{1}{x-2}=u=2 \\ \Rightarrow x-2=\frac{1}{2} \\ \Rightarrow x=\frac{5}{2} \end{gathered}$$
$$\begin{gathered} \frac{1}{y+3}=1 \\ \Rightarrow y+3=1 \\ \Rightarrow y=-2 \end{gathered}$$
$\left(x,y\right)=\left(\frac{5}{2},-2\right)$

$\left(4\right) \frac{1}{3x+y}+\frac{2}{3x-y}=\frac{3}{4}; \frac{1}{2\left(3x+y\right)}-\frac{1}{2\left(3x-y\right)}=-\frac{1}{8}$
Let $\frac{1}{3x+y}=u \mathrm{and} \frac{1}{3x-y}=v$
$u+2v=\frac{3}{4} \mathrm{and} \frac{1}{2}u-\frac{1}{2}v=\frac{-1}{8}$
So, the equations become
$$\begin{gathered} 4u+4v=3 .....\left(\mathrm{I}\right) \\ 4u-4v=1 .....\left(\mathrm{II}\right) \end{gathered}$$
Adding (I) and (II)
$$\begin{gathered} 8u=4 \\ \Rightarrow u=\frac{1}{2} \end{gathered}$$
Putting the value of u in (I)
$$\begin{gathered} \frac{1}{2}+2v=\frac{3}{4} \\ \Rightarrow v=\frac{1}{4} \end{gathered}$$
$$\begin{gathered} \frac{1}{3x+y}=u \mathrm{and} \frac{1}{3x-y}=v \\ \Rightarrow \frac{1}{3x+y}=\frac{1}{2} \\ 3x+y=2 .....\left(\mathrm{III}\right) \\ \mathrm{Also}, \frac{1}{3x-y}=\frac{1}{4} \\ \Rightarrow 3x-y=4 .....\left(\mathrm{IV}\right) \end{gathered}$$
(III) + (IV) we get
$$\begin{gathered} 6x=6 \\ \Rightarrow x=1 \\ y=-1 \end{gathered}$$



















 

Answer:

Let the smaller number be x and the larger number be y. 
Given that the two numbers differ by 3 so, 
$y-x=3$                                             .....(I)
Also, sum of twice the smaller number and thrice the greater number is 19
So, $2x+3y=19$                                  ......(II)
The two equations obtained are
$y-x=3$
$2x+3y=19$
Multiplying (I) by 3 we get
$3y-3x=9$                                          .....(III)
Adding (III) and (II) we have
4y = 28
$\Rightarrow y=\frac{28}{4}=7$
Putting the value of y = 7 in (I) we get
$$\begin{gathered} 7-x=3 \\ \Rightarrow -x=3-7 \\ \Rightarrow -x=-4 \\ \Rightarrow x=4 \end{gathered}$$
Thus, the two numbers are 4 and 7. 
 

Answer:

The length of the given rectangle is $2x+y+8$ and $4x-y$
$$\begin{gathered} 2x+y+8=4x-y \\ \Rightarrow y+y+8=4x-2x \\ \Rightarrow 8+2y=2x \\ \Rightarrow 2x-2y=8 \\ \mathrm{Dividing} \mathrm{by} 2 \\ x-y=4 .....\left(\mathrm{I}\right) \end{gathered}$$
Breadth of the rectangle is 2y and x + 4. 
$$\begin{gathered} 2y=x+4 \\ \Rightarrow x-2y=-4 .....\left(\mathrm{II}\right) \end{gathered}$$
Subtracting (II) from (I) 
$$\begin{gathered} x-x-y-\left(-2y\right)=4-\left(-4\right) \\ \Rightarrow -y+2y=8 \\ \Rightarrow y=8 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} y=8 \mathrm{in} \left(\mathrm{I}\right) \mathrm{we} \mathrm{get} \\ x-8=4 \\ \Rightarrow x=4+8=12 \end{gathered}$$
Length = $4x-y=4\left(12\right)-8=40$
Breadth = $2\times 8=16$
Perimeter = $2\left(\mathrm{length}+\mathrm{breadth}\right)=2\left(40+16\right)=112$ units
Area = $\mathrm{length}\times \mathrm{breadth}=40\times 16=640 {\mathrm{unit}}^2$

Answer:

Let the father's age be x years and son's age be y years. 
Sum of father’s age and twice the age of his son is 70 so, 
$x+2y=70$                               ......(I)
Double the age of the father added to the age of his son the sum is 95
$2x+y=95$                               .....(II)
Adding (I) and (II) we get
$$\begin{gathered} 3x+3y=165 \\ \mathrm{Dividing} \mathrm{by} 3 \\ x+y=55 .....\left(\mathrm{III}\right) \end{gathered}$$
Subtracting (I) from (II)
$$\begin{gathered} 2x-x+y-2y=95-70 \\ \Rightarrow x-y=25 .....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV) we get
$$\begin{gathered} 2x=80 \\ \Rightarrow x=40 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} x=40 \mathrm{in} \left(\mathrm{III}\right) \\ 40+y=55 \\ \Rightarrow y=55-40 \\ \Rightarrow y=15 \end{gathered}$$
Thus, the age of the father is 40 years and age of his son is 15 years. 

Answer:

Let the fraction be $\frac{x}{y}$. 
Denominator of a fraction is 4 more than twice its numerator. 
So,
 $$\begin{gathered} y=4+2x \\ \Rightarrow 2x-y=-4 .....\left(\mathrm{I}\right) \end{gathered}$$                                           
Also, denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6.
So, 
$$\begin{gathered} y-6=12\left(x-6\right) \\ \Rightarrow y-6=12x-72 \\ \Rightarrow 12x-y=72-6=66 \\ \Rightarrow 12x-y=66 .....\left(\mathrm{II}\right) \end{gathered}$$
Subtracting (I) from (II)
$$\begin{gathered} 12x-2x-y-\left(-y\right)=66-\left(-4\right) \\ \Rightarrow 10x=70 \\ \Rightarrow x=\frac{70}{10}=7 \\ \Rightarrow x=7 \end{gathered}$$
Putting the value of x = 7 in (I)
$$\begin{gathered} 2\left(7\right)-y=-4 \\ \Rightarrow 14-y=-4 \\ \Rightarrow y=14+4=18 \end{gathered}$$
Thus, the fraction obtained is $\frac{7}{18}$.  
 

Answer:

Let the weight of box A be x and that of box B be y. 
When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons i.e 10000 kg.
So, 
$$\begin{gathered} 150x+100y=10000 \\ \Rightarrow 15x+10y=1000 \\ \Rightarrow 3x+2y=200 .....\left(\mathrm{I}\right) \end{gathered}$$
When 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded.
$$\begin{gathered} 260x+40y=10000 \\ \Rightarrow 26x+4y=1000 \\ \Rightarrow 13x+2y=500 .....\left(\mathrm{II}\right) \end{gathered}$$   
Subtracting (I) from (II) we get
$$\begin{gathered} 13x-3x+2y-2y=500-200 \\ \Rightarrow 10x=300 \\ \Rightarrow x=30 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} x=30 \mathrm{in} \left(\mathrm{I}\right) \\ 3\left(30\right)+2y=200 \\ \Rightarrow 90+2y=200 \\ \Rightarrow 2y=200-90=110 \\ \Rightarrow y=\frac{110}{2}=55 \end{gathered}$$
Thus, weight of box A = 30 kg and that of box B = 55 kg. 

Answer:

We know $\mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}}$
Average speed of bus = 60km/h.
Let the time taken in bus be x hours. 
Average speed of bus = 700km/h.
Let the time taken in bus be y hours. 
Total distance covered = 1900 km
$$\begin{gathered} 60x+700y=1900 \\ \Rightarrow 6x+70y=190 \\ \Rightarrow 3x+35y=95 .....\left(\mathrm{I}\right) \end{gathered}$$
It takes 5 hours to complete the journey so, 
$x+y=5 .....\left(\mathrm{II}\right)$
Multiplying (II) with 3
$3x+3y=15 .....\left(\mathrm{III}\right)$
Subtracting (III) from (I) we get
$$\begin{gathered} 3x-3x+35y-3y=95-15 \\ \Rightarrow 32y=80 \\ \Rightarrow y=2.5 \end{gathered}$$
Putting the value of y = 2.5 in (II) we get
$$\begin{gathered} x+2.5=5 \\ \Rightarrow x=2.5 \end{gathered}$$
Distance travelled by Vishal by bus = $\mathrm{speed}\times \mathrm{time}=60\times 2.5=150 \mathrm{km}$.

Answer:

(1) 4x +5y = 19
When x = 1, then y will be 
$$\begin{gathered} 4\left(1\right)+5y=19 \\ \Rightarrow 4+5y=19 \\ \Rightarrow 5y=19-4=15 \\ \Rightarrow 5y=15 \\ \Rightarrow y=\frac{15}{5}=3 \end{gathered}$$
Hence, the correct answer is option (B). 

(2) $x=\frac{D_x}{D}=\frac{49}{7}=7$
Hence, the correct answer is option (A). 

(3) $\left|\begin{array}{cc}5& 3\\ -7& -4\end{array}\right|=5\times \left(-4\right)-3\times \left(-7\right)=-20+21=1$
Hence, the correct answer is option (D).

(4) x + y = 3 ; 3x – 2y – 4 = 0
$D=\left|\begin{array}{cc}1& 1\\ 3& -2\end{array}\right|=1\times \left(-2\right)-1\times 3=-2-3=-5$
Hence, the correct answer is option (C). 

(5) ax + by = c and mx + ny = d
$D=\left|\begin{array}{cc}a& b\\ m& n\end{array}\right|=an-bm$
an ≠ bm 
So, D ≠ 0. 
So, the given equations have a unique solution or only one common solution.
Hence, the correct answer is option A. 

Answer:

2x –  6y = 3

Answer:

(1) 2x + 3y = 12
 

Answer:

(1)  $\left|\begin{array}{cc}4& 3\\ 2& 7\end{array}\right|=4\times 7-3\times 2=28-6=22$

(2) $\left|\begin{array}{cc}5& -2\\ -3& 1\end{array}\right|=5\times 1-\left(-2\right)\times \left(-3\right)=5-6=-1$

(3) $\left|\begin{array}{cc}3& -1\\ 1& 4\end{array}\right|=3\times 4-\left(-1\right)\times 1=12+1=13$

Answer:

(1) 6x – 3y = –10 ; 3x + 5y – 8 = 0
$$\begin{gathered} D=\left|\begin{array}{cc}6& -3\\ 3& 5\end{array}\right|=6\times 5-\left(-3\right)\times 3=30+9=39 \\ {D}_x=\left|\begin{array}{cc}-10& -3\\ 8& 5\end{array}\right|=-10\times 5-\left(-3\right)\times 8=-50+24=-26 \\ {D}_y=\left|\begin{array}{cc}6& -10\\ 3& 8\end{array}\right|=6\times 8-\left(-10\right)\times 3=48+30=78 \\ x=\frac{D_x}{D}=\frac{-26}{39}=\frac{-2}{3} \\ y=\frac{D_y}{D}=\frac{78}{39}=2 \\ \left(x,y\right)=\left(\frac{-2}{3},2\right) \end{gathered}$$

(2) 4m – 2n = –4 ; 4m + 3n = 16
$$\begin{gathered} D=\left|\begin{array}{cc}4& -2\\ 4& 3\end{array}\right|=4\times 3-\left(-2\right)\times 4=12+8=20 \\ {D}_x=\left|\begin{array}{cc}-4& -2\\ 16& 3\end{array}\right|=-4\times 3-\left(-2\right)\times 16=-12+32=20 \\ {D}_y=\left|\begin{array}{cc}4& -4\\ 4& 16\end{array}\right|=4\times 16-\left(-4\right)\times 4=64+16=80 \\ x=\frac{D_x}{D}=\frac{20}{20}=1 \\ y=\frac{D_y}{D}=\frac{80}{20}=4 \\ \left(x,y\right)=\left(1,4\right) \end{gathered}$$
(3) 3x – 2y = $\frac{5}{2}$ ; $\frac{1}{3}x+3y=-\frac{4}{3}$
$$\begin{gathered} D=\left|\begin{array}{cc}3& -2\\ \frac{1}{3}& 3\end{array}\right|=9+\frac{2}{3}=\frac{29}{3} \\ {D}_x=\left|\begin{array}{cc}\frac{5}{2}& -2\\ \frac{-4}{3}& 3\end{array}\right|=\frac{15}{2}-\frac{8}{3}=\frac{29}{6} \\ {D}_y=\left|\begin{array}{cc}3& \frac{5}{2}\\ \frac{1}{3}& \frac{-4}{3}\end{array}\right|=-4-\frac{5}{6}=\frac{-29}{6} \\ x=\frac{D_x}{D}=\frac{{\displaystyle \frac{29}{6}}}{{\displaystyle \frac{29}{3}}}=\frac{1}{2} \\ y=\frac{D_y}{D}=\frac{\frac{-29}{6}}{\frac{29}{3}}=\frac{-1}{2} \\ \left(x,y\right)=\left(\frac{1}{2},\frac{-1}{2}\right) \end{gathered}$$

(4) 7x + 3y = 15 ; 12y – 5x = 39
$$\begin{gathered} D=\left|\begin{array}{cc}7& 3\\ -5& 12\end{array}\right|=7\times 12-\left(-5\right)\times 3=84+15=99 \\ {D}_x=\left|\begin{array}{cc}15& 3\\ 39& 12\end{array}\right|=15\times 12-39\times 3=180-117=63 \\ {D}_y=\left|\begin{array}{cc}7& 15\\ -5& 39\end{array}\right|=7\times 39-\left(-5\right)\times 15=273+75=348 \\ x=\frac{D_x}{D}=\frac{{\displaystyle 63}}{{\displaystyle 99}}=\frac{7}{11} \\ y=\frac{D_y}{D}=\frac{348}{99}=\frac{116}{33} \\ \left(x,y\right)=\left(\frac{7}{11},\frac{116}{33}\right) \end{gathered}$$

(5) $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$
$$\begin{gathered} \frac{x+y-8}{2}=\frac{x+2y-14}{3} \\ \Rightarrow 3x+3y-24=2x+4y-28 \\ \Rightarrow x-y=-4 .....\left(\mathrm{I}\right) \\ \mathrm{and} \frac{x+2y-14}{3}=\frac{3x-y}{4} \\ \Rightarrow 4x+8y-56=9x-3y \\ \Rightarrow 5x-11y=-56 .....\left(\mathrm{II}\right) \end{gathered}$$

From (I) and (II)
$$\begin{gathered} D=\left|\begin{array}{cc}1& -1\\ 5& -11\end{array}\right|=-11\times 1-\left(-1\right)\times 5=-11+5=-6 \\ {D}_x=\left|\begin{array}{cc}-4& -1\\ -56& -11\end{array}\right|=-11\times \left(-4\right)-\left(-1\right)\times \left(-56\right)=44-56=-12 \\ {D}_y=\left|\begin{array}{cc}1& -4\\ 5& -56\end{array}\right|=-56\times 1-\left(-4\right)\times 5=-56+20=-36 \\ x=\frac{D_x}{D}=\frac{-12}{-6}=2 \\ y=\frac{D_y}{D}=\frac{-36}{-6}=6 \\ \left(x,y\right)=\left(2,6\right) \end{gathered}$$

Answer: