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#### Question 1:

Complete the following activity to solve the simultaneous equations.
5x + 3y = 9 -----(I)
2x + 3y = 12 ----- (II)

Disclaimer: There is error in the Q. In (II) there should have been 2x $-$ 3y = 12
5x + 3y = 9 -----(I)
2x $-$ 3y = 12 ----- (II)
7x = 21
x = 3
Putting the value of x = 3 in (I) we get
$5\left(3\right)+3y=9\phantom{\rule{0ex}{0ex}}⇒15+3y=9\phantom{\rule{0ex}{0ex}}⇒3y=9-15=-6\phantom{\rule{0ex}{0ex}}⇒y=-2$
Thus, (x, y) = (3, $-$6).

#### Question 2:

Solve the following simultaneous equations.
(1) 3a + 5b = 26; a + 5b = 22
(2) x + 7y = 10; 3x – 2y = 7
(3) 2x – 3y = 9; 2x + y = 13
(4) 5m – 3n = 19; m – 6n = –7
(5) 5x + 2y = –3; x + 5y = 4
(6)
(7) 99x + 101y = 499; 101x + 99y = 501
(8) 49x – 57y = 172; 57x – 49y = 252

(1) 3a + 5b = 26;                           .....(I)
a + 5b = 22                                    .....(II)
Subtracting (II) from (I)
2a = 4
$⇒$ a = 2
Putting the value of a = 2 in (II)
5b = 22 $-$ 2 = 20
$⇒$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
$⇒$y = 1
Putting the value of y in (IV) we get
3x – 2 = 7
$⇒$3x = 7 + 2 = 9
$⇒$3x = 9
$⇒$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
$⇒-4y=-4\phantom{\rule{0ex}{0ex}}⇒y=1$
Putting this value in (I) we get
$2x-3\left(1\right)=9\phantom{\rule{0ex}{0ex}}⇒2x=9+3=12\phantom{\rule{0ex}{0ex}}⇒x=\frac{12}{2}=6$
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
$10m-m-6n-\left(-6n\right)=38-\left(-7\right)\phantom{\rule{0ex}{0ex}}⇒9m=45\phantom{\rule{0ex}{0ex}}⇒m=\frac{45}{9}=5$
Putting the value of m = 5 in (II) we get
$5-6n=-7\phantom{\rule{0ex}{0ex}}⇒-6n=-7-5\phantom{\rule{0ex}{0ex}}⇒-6n=-12\phantom{\rule{0ex}{0ex}}⇒n=\frac{-12}{-6}=2$
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
$5x-5x+2y-25y=-3-20\phantom{\rule{0ex}{0ex}}⇒-23y=-23\phantom{\rule{0ex}{0ex}}⇒y=\frac{-23}{-23}=1$
Putting the value of y = 1 in (II) we get
$x+5\left(1\right)=4\phantom{\rule{0ex}{0ex}}⇒x+5=4\phantom{\rule{0ex}{0ex}}⇒x=4-5=-1$
Thus, (x, y) = (−1, 1)

(6)

Multiply (I) with 3 and (II) with 4

Multiply (IV) with 3
24x + 3y = 33         .....(V)
Subtracting (V) from (III)
$x-24x+3y-3y=10-33\phantom{\rule{0ex}{0ex}}⇒-23x=-23\phantom{\rule{0ex}{0ex}}⇒x=1$
Putting the value of x = 1 in (III)
$1+3y=10\phantom{\rule{0ex}{0ex}}⇒3y=10-1=9\phantom{\rule{0ex}{0ex}}⇒y=\frac{9}{3}=3$
Thus, (x, y) = (1, 3)

(7) 99x + 101y = 499                       .....(I)
101x + 99y = 501                             .....(II)

Subtracting (II) from (I)

$x+y=5\phantom{\rule{0ex}{0ex}}-x+y=-1\phantom{\rule{0ex}{0ex}}⇒2y=4\phantom{\rule{0ex}{0ex}}⇒y=2$
Putting the value of y = 2 in (III) we get
$x+2=5\phantom{\rule{0ex}{0ex}}⇒x=5-2=3$
Thus, (x, y) = (3, 2)

(8) 49x – 57y = 172                         .....(I)
57x – 49y = 252                              .....(II)

Subtracting (II) from (I) we have

$x-y=4\phantom{\rule{0ex}{0ex}}x+y=10\phantom{\rule{0ex}{0ex}}⇒2x=14\phantom{\rule{0ex}{0ex}}⇒x=7$
Putting the value of x = 7 in (IV) we get
$7+y=10\phantom{\rule{0ex}{0ex}}⇒y=10-7\phantom{\rule{0ex}{0ex}}⇒y=3$
Thus, (x, y) = (7, 3).

#### Question 1:

Complete the following table to draw graph of the equations–
(I) x + y = 3 (II) x y = 4

 x 3 y 5 3 (x, y) (3, 0) (0, 3)

 x –1 0 y 0 –4 (x, y) (0, –4)

 x 3 y 5 3 (x, y) (3, 0) (0, 3)

 x –1 0 y 0 –4 (x, y) (0, –4) #### Question 2:

Solve the following simultaneous equations graphically.
(1) x + y = 6 ; x y = 4
(2) x + y = 5 ; x y = 3
(3) x + y = 0 ; 2x y = 9
(4) 3x y = 2 ; 2x y = 3
(5) 3x – 4y = –7 ; 5x – 2y = 0
(6) 2x – 3y = 4 ; 3y x = 4

(1)  = 6;

x 0 6 2
y 6 0 4

– = 4

 x 4 5 0 y 0 1 −4 Point of intersection of the two lines is (5, 1).

(2) = 5

 x 0 5 2 y 5 0 3

– = 3
 x 3 0 5 y 0 −3 2 Point of intersection of the two lines is (4, 1)
(3) = 0

 x 3 1 2 y −3 −1 −2

2– = 9
 x 3 0 1 y −3 −9 −7 ​Point of intersection of the two lines is (3, −3).

(4) 3– = 2

 x 0 1 2 y −2 1 4

2– = 3

 x 0 1 2 y −3 −1 1 ​Point of intersection of the two lines is (−1, −5).

(5) 3– 4= –7

 x 1 0 −2.3 y 2.5 1.75 0

5– 2= 0

 x 0 2 4 y 0 5 10 ​Point of intersection of the two lines is (1, 2.5).

(6) 2– 3= 4

 x 2 3.5 1 y 0 1 −0.6

3– = 4

 x −4 2 −1 y 0 2 1 ​Point of intersection of the two lines is (8, 4).

#### Question 1:

Fill in the blanks with correct number

Thus, we have

#### Question 2:

Find the values of following determinants.

(1)

(2)

(3) $\left|\begin{array}{cc}\frac{7}{3}& \frac{5}{3}\\ \frac{3}{2}& \frac{1}{2}\end{array}\right|$

(1)

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

(2)

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

#### Question 3:

Solve the following simultaneous equations using Cramer’s rule.
(1) 3x – 4y = 10 ; 4x + 3y = 5
(2) 4x + 3y – 4 = 0 ; 6x = 8 – 5y
(3) x + 2y = –1 ; 2x – 3y = 12
(4) 6x – 4y = –12 ; 8x – 3y = –2
(5) 4m + 6n = 54 ; 3m + 2n = 28
(6)

(1) 3– 4= 10
4+ 3= 5
$D=\left|\begin{array}{cc}3& -4\\ 4& 3\end{array}\right|=3×3-\left(-4\right)×4=9+16=25\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}10& -4\\ 5& 3\end{array}\right|=10×3-\left(-4\right)×5=30+20=50\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}3& 10\\ 4& 5\end{array}\right|=3×5-10×4=15-40=-25\phantom{\rule{0ex}{0ex}}$
$x=\frac{{D}_{x}}{D}=\frac{50}{25}=2\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{-25}{25}=-1\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(2,-1\right)$

(2) 4+ 3– 4 = 0 ; 6= 8 – 5y
$D=\left|\begin{array}{cc}4& 3\\ 6& 5\end{array}\right|=4×5-6×3=20-18=2\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}4& 3\\ 8& 5\end{array}\right|=4×5-3×8=20-24=-4\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}4& 4\\ 6& 8\end{array}\right|=4×8-6×4=32-24=8$

$x=\frac{{D}_{x}}{D}=\frac{-4}{2}=-2\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{8}{2}=4\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(-2,4\right)$

(3) + 2= –1 ; 2– 3= 12
$D=\left|\begin{array}{cc}1& 2\\ 2& -3\end{array}\right|=1×\left(-3\right)-2×2=-3-4=-7\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}-1& 2\\ 12& -3\end{array}\right|=\left(-1\right)×\left(-3\right)-2×12=3-24=-21\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}1& -1\\ 2& 12\end{array}\right|=1×12-\left(-1\right)×2=12+2=14$
$x=\frac{{D}_{x}}{D}=\frac{-21}{-7}=3\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{14}{-7}=-2\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(3,-2\right)$

(4) 6– 4= –12 ; 8– 3= –2

$D=\left|\begin{array}{cc}6& -4\\ 8& -3\end{array}\right|=6×\left(-3\right)-\left(-4\right)×8=-18+32=14\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}-12& -4\\ -2& -3\end{array}\right|=\left(-12\right)×\left(-3\right)-\left(-4\right)×\left(-2\right)=36-8=28\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}6& -12\\ 8& -2\end{array}\right|=6×\left(-2\right)-\left(-12\right)×8=-12+96=84$
$x=\frac{{D}_{x}}{D}=\frac{28}{14}=2\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{84}{14}=6\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(2,6\right)$

(5) 4+ 6= 54 ; 3+ 2= 28
$D=\left|\begin{array}{cc}4& 6\\ 3& 2\end{array}\right|=4×2-6×3=8-18=-10\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}54& 6\\ 28& 2\end{array}\right|=54×2-6×28=108-168=-60\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}4& 54\\ 3& 28\end{array}\right|=4×28-54×3=112-162=-50$
$x=\frac{{D}_{x}}{D}=\frac{-60}{-10}=6\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{-50}{-10}=5\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(6,5\right)$

(6)
$D=\left|\begin{array}{cc}2& 3\\ 1& \frac{-1}{2}\end{array}\right|=2×\left(-\frac{1}{2}\right)-3×1=-1-3=-4\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}2& 3\\ \frac{1}{2}& \frac{-1}{2}\end{array}\right|=2×\left(-\frac{1}{2}\right)-3×\frac{1}{2}=-1-\frac{3}{2}=\frac{-5}{2}\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}2& 2\\ 1& \frac{1}{2}\end{array}\right|=2×\frac{1}{2}-2×1=1-2=-1$
$x=\frac{{D}_{x}}{D}=\frac{\frac{-5}{2}}{-4}=\frac{5}{8}\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{-1}{-4}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(\frac{5}{8},\frac{1}{4}\right)$

#### Question 1:

Solve the following simultaneous equations.

So, the equation becomes

Multiply (I) with 4 we get

(II) − (III)

Thus,
$\frac{1}{x}=u=9\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{9}\phantom{\rule{0ex}{0ex}}\frac{1}{y}=v=1\phantom{\rule{0ex}{0ex}}⇒y=1\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(\frac{1}{9},1\right)$

Let
So, the equation becomes

Multiplying (I) with 5 and (II) with 2 we get

Adding (III) and (IV) we get
$u=\frac{16}{80}=\frac{1}{5}$
Putting this value in (I)
$10×\frac{1}{5}+2v=4\phantom{\rule{0ex}{0ex}}⇒2+2v=4\phantom{\rule{0ex}{0ex}}⇒v=1$

Adding (III) and (IV) we get
$2u=4\phantom{\rule{0ex}{0ex}}⇒u=2$
Putting the value of in III
$2+v=3\phantom{\rule{0ex}{0ex}}⇒v=1$
$\frac{1}{x-2}=u=2\phantom{\rule{0ex}{0ex}}⇒x-2=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{2}$
$\frac{1}{y+3}=1\phantom{\rule{0ex}{0ex}}⇒y+3=1\phantom{\rule{0ex}{0ex}}⇒y=-2$
$\left(x,y\right)=\left(\frac{5}{2},-2\right)$

Let

So, the equations become

$8u=4\phantom{\rule{0ex}{0ex}}⇒u=\frac{1}{2}$
Putting the value of in (I)
$\frac{1}{2}+2v=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒v=\frac{1}{4}$

(III) + (IV) we get
$6x=6\phantom{\rule{0ex}{0ex}}⇒x=1\phantom{\rule{0ex}{0ex}}y=-1$

#### Question 1:

Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.

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
$⇒y=\frac{28}{4}=7$
Putting the value of y = 7 in (I) we get
$7-x=3\phantom{\rule{0ex}{0ex}}⇒-x=3-7\phantom{\rule{0ex}{0ex}}⇒-x=-4\phantom{\rule{0ex}{0ex}}⇒x=4$
Thus, the two numbers are 4 and 7.

#### Question 2:

Complete the following. The length of the given rectangle is $2x+y+8$ and $4x-y$

Breadth of the rectangle is 2and + 4.

Subtracting (II) from (I)

Length = $4x-y=4\left(12\right)-8=40$
Breadth = $2×8=16$
Perimeter = $2\left(\mathrm{length}+\mathrm{breadth}\right)=2\left(40+16\right)=112$ units
Area =

#### Question 3:

The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.

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

Subtracting (I) from (II)

Adding (III) and (IV) we get

Thus, the age of the father is 40 years and age of his son is 15 years.

#### Question 4:

The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.

Let the fraction be $\frac{x}{y}$
Denominator of a fraction is 4 more than twice its numerator.
So,

Also, denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6.
So,

Subtracting (I) from (II)
$12x-2x-y-\left(-y\right)=66-\left(-4\right)\phantom{\rule{0ex}{0ex}}⇒10x=70\phantom{\rule{0ex}{0ex}}⇒x=\frac{70}{10}=7\phantom{\rule{0ex}{0ex}}⇒x=7$
Putting the value of = 7 in (I)
$2\left(7\right)-y=-4\phantom{\rule{0ex}{0ex}}⇒14-y=-4\phantom{\rule{0ex}{0ex}}⇒y=14+4=18$
Thus, the fraction obtained is $\frac{7}{18}$.

#### Question 5:

Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons. But 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. Find the weight of each type of box.

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,

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.

Subtracting (I) from (II) we get

Thus, weight of box A = 30 kg and that of box B = 55 kg.

#### Question 6:

Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.

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

It takes 5 hours to complete the journey so,

Multiplying (II) with 3

Subtracting (III) from (I) we get
$3x-3x+35y-3y=95-15\phantom{\rule{0ex}{0ex}}⇒32y=80\phantom{\rule{0ex}{0ex}}⇒y=2.5$
Putting the value of = 2.5 in (II) we get
$x+2.5=5\phantom{\rule{0ex}{0ex}}⇒x=2.5$
Distance travelled by Vishal by bus = .

#### Question 1:

Choose correct alternative for each of the following questions
(1) To draw graph of 4x +5y = 19, Find y when x = 1.
 A)  4 (B)  3 (C)  2 (D)  –3

(2) For simultaneous equations in variables x and y, Dx = 49, Dy = –63, D = 7 then what is x ?
 A)  7 (B) –7 (C)  $\frac{1}{7}$ (D)  $\frac{-1}{7}$

(3) Find the value of $\left|\begin{array}{cc}5& 3\\ -7& -4\end{array}\right|$
 A)  –1 (B)  –41 (C)  41 (D)  1

(4) To solve x + y = 3 ; 3x – 2y – 4 = 0 by determinant method find D.
 A)  5 (B)  1 (C)  –5 (D)  –1

(5) ax + by = c and mx + ny = d and anbm then these simultaneous equations have -
 (A) Only one common solution. (A) No solution. (C) Infinite number of solutions. (D) Only two solution.

(1) 4x +5y = 19
When
x = 1, then y will be
$4\left(1\right)+5y=19\phantom{\rule{0ex}{0ex}}⇒4+5y=19\phantom{\rule{0ex}{0ex}}⇒5y=19-4=15\phantom{\rule{0ex}{0ex}}⇒5y=15\phantom{\rule{0ex}{0ex}}⇒y=\frac{15}{5}=3$
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)
$\left|\begin{array}{cc}5& 3\\ -7& -4\end{array}\right|=5×\left(-4\right)-3×\left(-7\right)=-20+21=1$
Hence, the correct answer is option (D).

(4) x + y = 3 ; 3x – 2y – 4 = 0
$D=\left|\begin{array}{cc}1& 1\\ 3& -2\end{array}\right|=1×\left(-2\right)-1×3=-2-3=-5$
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.

#### Question 2:

Complete the following table to draw the graph of 2x –  6y = 3

 x –5 y 0 (x, y)

2x –  6y = 3

 x –5 $\frac{3}{2}$ y $\frac{-13}{6}$ 0 (x, y) $\left(-5,-\frac{13}{6}\right)$ $\left(\frac{3}{2},0\right)$ #### Question 3:

Solve the following simultaneous equations graphically.
(1) 2x + 3y = 12 ; x y = 1
(2) x – 3y = 1 ; 3x – 2y + 4 = 0
(3) 5x – 6y + 30 = 0 ; 5x + 4y – 20 = 0
(4) 3xy – 2 = 0 ; 2x + y = 8
(5) 3x + y = 10 ;  x y = 2

(1) 2x + 3y = 12

 x 0 6 3 y 4 0 2

– y = 1

 x 0 1 3 y −1 0 2 The solution of the given equations is the point of intersection of the two line i.e(3, 2).

(2) x – 3y = 1

 x 1 4 7 y 0 1 2

3x – 2y + 4 = 0

 x 0 2 4 y 2 5 8 The solution of the given equations is the point of intersection of the two line i.e ($-2,-1$).

(3) 5x – 6y + 30 = 0

 x 0 –6 y 5 0

5x + 4y – 20 = 0

 x 0 4 y 5 0 The solution of the given equations is the point of intersection of the two line i.e (0, 5).

(4) 3x – y – 2 = 0

 x 0 1 y –2 1

2x + y = 8

 x 0 4 y 8 0 The solution of the given equations is the point of intersection of the two line i.e (2, 4).

(5) 3x + y = 10

 x 0 1 y 10 7

– y = 2

 x 0 2 y –2 0 The solution of the given equations is the point of intersection of the two line i.e (3, 1).

#### Question 4:

Find the values of each of the following determinants.

 (1)  $\left|\begin{array}{cc}4& 3\\ 2& 7\end{array}\right|$ (2)  $\left|\begin{array}{cc}5& -2\\ -3& 1\end{array}\right|$ (3)  $\left|\begin{array}{cc}3& -1\\ 1& 4\end{array}\right|$

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

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

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

#### Question 5:

Solve the following equations by Cramer’s method.
(1) 6x – 3y = –10 ; 3x + 5y – 8 = 0
(2) 4m – 2n = –4 ; 4m + 3n = 16
(3) 3x – 2y = $\frac{5}{2}$ ; $\frac{1}{3}x+3y=-\frac{4}{3}$
(4) 7x + 3y = 15 ; 12y – 5x = 39
(5) $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$

(1) 6x – 3= –10 ; 3x + 5y – 8 = 0
$D=\left|\begin{array}{cc}6& -3\\ 3& 5\end{array}\right|=6×5-\left(-3\right)×3=30+9=39\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}-10& -3\\ 8& 5\end{array}\right|=-10×5-\left(-3\right)×8=-50+24=-26\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}6& -10\\ 3& 8\end{array}\right|=6×8-\left(-10\right)×3=48+30=78\phantom{\rule{0ex}{0ex}}x=\frac{{D}_{x}}{D}=\frac{-26}{39}=\frac{-2}{3}\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{78}{39}=2\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(\frac{-2}{3},2\right)$

(2) 4m – 2= –4 ; 4m + 3n = 16
$D=\left|\begin{array}{cc}4& -2\\ 4& 3\end{array}\right|=4×3-\left(-2\right)×4=12+8=20\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}-4& -2\\ 16& 3\end{array}\right|=-4×3-\left(-2\right)×16=-12+32=20\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}4& -4\\ 4& 16\end{array}\right|=4×16-\left(-4\right)×4=64+16=80\phantom{\rule{0ex}{0ex}}x=\frac{{D}_{x}}{D}=\frac{20}{20}=1\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{80}{20}=4\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(1,4\right)$
(3) 3x – 2$\frac{5}{2}$ ; $\frac{1}{3}x+3y=-\frac{4}{3}$
$D=\left|\begin{array}{cc}3& -2\\ \frac{1}{3}& 3\end{array}\right|=9+\frac{2}{3}=\frac{29}{3}\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}\frac{5}{2}& -2\\ \frac{-4}{3}& 3\end{array}\right|=\frac{15}{2}-\frac{8}{3}=\frac{29}{6}\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}3& \frac{5}{2}\\ \frac{1}{3}& \frac{-4}{3}\end{array}\right|=-4-\frac{5}{6}=\frac{-29}{6}\phantom{\rule{0ex}{0ex}}x=\frac{{D}_{x}}{D}=\frac{\frac{29}{6}}{\frac{29}{3}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{\frac{-29}{6}}{\frac{29}{3}}=\frac{-1}{2}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(\frac{1}{2},\frac{-1}{2}\right)$

(4) 7x + 3y = 15 ; 12y – 5x = 39
$D=\left|\begin{array}{cc}7& 3\\ -5& 12\end{array}\right|=7×12-\left(-5\right)×3=84+15=99\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}15& 3\\ 39& 12\end{array}\right|=15×12-39×3=180-117=63\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}7& 15\\ -5& 39\end{array}\right|=7×39-\left(-5\right)×15=273+75=348\phantom{\rule{0ex}{0ex}}x=\frac{{D}_{x}}{D}=\frac{63}{99}=\frac{7}{11}\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{348}{99}=\frac{116}{33}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(\frac{7}{11},\frac{116}{33}\right)$

(5) $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$

From (I) and (II)
$D=\left|\begin{array}{cc}1& -1\\ 5& -11\end{array}\right|=-11×1-\left(-1\right)×5=-11+5=-6\phantom{\rule{0ex}{0ex}}{D}_{x}=\left|\begin{array}{cc}-4& -1\\ -56& -11\end{array}\right|=-11×\left(-4\right)-\left(-1\right)×\left(-56\right)=44-56=-12\phantom{\rule{0ex}{0ex}}{D}_{y}=\left|\begin{array}{cc}1& -4\\ 5& -56\end{array}\right|=-56×1-\left(-4\right)×5=-56+20=-36\phantom{\rule{0ex}{0ex}}x=\frac{{D}_{x}}{D}=\frac{-12}{-6}=2\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{y}}{D}=\frac{-36}{-6}=6\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(2,6\right)$

#### Question 6:

Solve the following simultaneous equations.
(1)
(2)
(3)
(4)
(5)

(1)
Let

Multiply (II) with 2

$\left(\mathrm{I}\right)-\left(\mathrm{III}\right)$
$6u=1\phantom{\rule{0ex}{0ex}}⇒u=\frac{1}{6}$
Putting the value of in II.
$3×\frac{1}{6}+2v=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}+2v=0\phantom{\rule{0ex}{0ex}}⇒v=\frac{-1}{4}$
$\frac{1}{x}=u\phantom{\rule{0ex}{0ex}}⇒x=6\phantom{\rule{0ex}{0ex}}\frac{1}{y}=v\phantom{\rule{0ex}{0ex}}⇒y=-4\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(6,-4\right)$

(2)
Let

(I) + (II)

(II) − (I)

(III) + (IV)

(3)
Multiply by xy

Putting the value of y in (IV)
$2-x=1\phantom{\rule{0ex}{0ex}}⇒x=1\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(1,2\right)$

(4)

Let $\frac{1}{x}=u,\frac{1}{y}=v$

Multiply (I) with 7 and (II) with 2

Putting the value of in (I)
$7\left(1\right)-2u=5\phantom{\rule{0ex}{0ex}}⇒u=1\phantom{\rule{0ex}{0ex}}\frac{1}{x}=u=1\phantom{\rule{0ex}{0ex}}⇒x=1\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(1,1\right)$

(5)

(I) + (II)
$20u=2\phantom{\rule{0ex}{0ex}}⇒u=\frac{1}{10}$
Putting the value of in (II)
$10×\frac{1}{10}-4v=-3\phantom{\rule{0ex}{0ex}}⇒1+3=4v\phantom{\rule{0ex}{0ex}}⇒v=1$

Multiply (III) with 2 and (IV) with 3

(V) − (VI)
$17y=17\phantom{\rule{0ex}{0ex}}⇒y=1$
Putting the value of in (VI)
$6x-9=3\phantom{\rule{0ex}{0ex}}⇒6x=12\phantom{\rule{0ex}{0ex}}⇒x=2\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(2,1\right)$

#### Question 7:

Solve the following word problems.
(1) A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.
(2) Kantabai bought $1\frac{1}{2}$ kg tea and 5 kg sugar from a shop. She paid Rs 50 as return fare for rickshaw. Total expense was Rs 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid Rs 880 for that. Find the rate of sugar and tea per kg.
(3) To find number of notes that Anushka had, complete the following activity. (4) Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
(5) In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is Rs 720. Find daily wages of skilled and unskilled workers.
(6) Places A and B are 30 km apart and they are on a st raight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.

(1) Let the number at the unit's place be and the digit at the ten's place be y.
The number is thus 10y + x
After interchanging the digits the number becomes 10x + y.
Given that two digit number and the number with digits interchanged add up to 143.
So, 10y + x + 10x + y = 143

Also, in the given number the digit in unit’s place is 3 more than the digit in the ten’s place.
So,
Adding (I) and (II) we get
$2x=16\phantom{\rule{0ex}{0ex}}⇒x=8$
Putting the value of in (I) we get
$8+y=13\phantom{\rule{0ex}{0ex}}⇒y=13-8=5$
Thus, the number is 58.

(2) Let the rate of tea be Rs per kg and that of sugar be Rs per kg.
When Kantabai bought the items by going to the shop,

When Kantabai bought the items online then

Multiplying (I) with 2 and (II) with 3 we get

(IV) $-$ (III)
$y=40$
Putting the value of y = 40 in (II)
$2x+7\left(40\right)=880\phantom{\rule{0ex}{0ex}}⇒2x=880-280=600\phantom{\rule{0ex}{0ex}}⇒x=300$
Thus, tea is at 300 Rs per kg and sugar is 40 Rs per kg.

(3) Disclaimer: There is error in the question given. Instead of Rs 10 notes there should be Rs 100 notes.
Let the number of notes of Rs 100 be x and that of Rs 50 be y

When number of notes is interchanged so,

Multiply (I) with 2

Subtracting (III) from (II) we get
$3x=60\phantom{\rule{0ex}{0ex}}⇒x=20$$3x=60\phantom{\rule{0ex}{0ex}}⇒x=20$
Putting the value of x in (I) we get
$y=10$
Thus, there are 20 Rs 100 notes and 10 Rs 50 notes.

(4) Let the present age of Manish be years and that of Savita be years.
Sum of their present ages = 31

Their age 3 years ago was
Manish's age = $x-3$
Savita's age = $y-3$
Manish’s age 3 years ago was 4 times the age of Savita.

(I) $-$ (II) we get
$5y=40\phantom{\rule{0ex}{0ex}}⇒y=8$
Putting the value of in (I)  we get
$x+8=31\phantom{\rule{0ex}{0ex}}⇒x=23$
Thus, age of Manish is 23 years and age of Savita is 8 years.

(5) Ratio of salary of skilled to unskilled workers = 5 : 3
Let one day salary of skilled person be x and that of unskilled person be y.
Their total one day salary Rs 720

Also,

Multiplying (I) with 3 we get

(III) $-$ (II)
$8y=2160\phantom{\rule{0ex}{0ex}}⇒y=270\phantom{\rule{0ex}{0ex}}$
Putting value of in (I) we get
$x=450$
One day salary of skilled person Rs 450 and that of unskilled person Rs 270.

(6) Let the speed of Hamid be km/h and that of Joseph be km/h.
When both travel in same direction so, the distance covered by them together will be 30 km.
We know $\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$
They meet each other after 20 min = $\frac{20}{60}=\frac{1}{3}$ hours

When Joseph started from point B but moved in opposite direction.
Distance travelled by Hamid $-$ Distance travelled by Joseph = 30

Adding (I) and (II) we get
$2x=100\phantom{\rule{0ex}{0ex}}⇒x=50$
Putting the value of in (II) we get
$50-y=10\phantom{\rule{0ex}{0ex}}⇒y=40$
Thus, speed of Hamid is 50 km/h and that of Joseph is 40 km/h.

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