System of Equations in Real World Problems (Class- 9, Experience- 5)

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[Exercise- page 139]

E.5.1 Compare the given systems of equations with a₁x + b₁y = c₁, a₂x + b₂y = c₂ and fill in the blanks.

Solution:

Comparing the given systems of equations with a₁x + b₁y = c₁, a₁x + b₁y = c₁ and blanks are filled:

E.5.2 Sketch the graph and solve the following pairs of equations that are solvable. Write at least three solutions if they have infinite solutions: 

i) 2x + y =8 

   2x - 2y = 5 

ii) 2x + 5y = -14 

    4x - 5y = 17 

iii) x2 + y3 = 8 

     5x 4 - 3y = -3 

iv) -7x + 8y = 9 

      5x - 4y = -3

Solution:

Sketch the graph and solve the following pairs of equations-

(i)
Given,
2x + y = 8
2x - 2y = 5

Comparing with the equations a₁x + b₁y = c₁, a₂x + b₂y = c₂  we get,
 a₁  a₂ =  2  2 = 1
 b₁  b₂ =  1 -2 = -  1  2
 c₁  c₂ =  8  5

Therefore, a₁a₂ ≠ b₁b₂

∴ There is only one solution to the system of equations.

Graph Solution-
    2x + y = 8
or, y = 8 – 2x....(i)

Here, for different values of x in equation (i), we determine several values of y.
Figure-1

 

Again,
2x - 2y = 5
or, -2y = 5 - 2x
or, 2y = 2x - 5
or, y = 2x - 5   2.....(ii)

Furthermore, for different values of x in equation (ii), we determine several values of y.
Figure-2

 

On graph paper, plot the points along the x and y axes such that the values of y are opposite to the values of x. For figure-1, establish points (1,6), (2,4), and (3,2), and for figure-2, establish points (1,-1.5), (2,-0.5), and (3.5,1). By connecting the established points for figure-1, a straight line is primarily obtained, and by connecting the established points for figure-2, another straight line is obtained, intersecting each other at point (3.5,1).

 
Hence, the concluded solution- (x,y) = (3.5, 1) (Answer)


(ii)

Given,  

2x + 5y = -14
4x - 5y = 17

Comparing with the equations a₁x + b₁y = c₁, a₂x + b₂y = c₂  we get,
 a₁  a₂ =  2  4 =  1  2
 b₁  b₂ =  5 -5 = -1
 c₁  c₂ =  -14   17

Therefore, a₁a₂ ≠ b₁b₂

So, There is only one solution to the system of equations.

Graph solution-
2x + 5y = -14
or, 5y = -14 - 2x
or, y = -14 - 2x      5.....(i)

Here, for different values of x in equation (i), we determine several values of y.
Figure-1

 

Again,
4x - 5y = 17
or, -5y = 17 - 4x
or, 5y = 4x - 17
or, y = 4x - 17     5.....(ii)

 
Now, For equation (ii), we find several values of y for different values of x.
Figure-2

 

Now, on graph paper, we plot points for each square with the length of its side taken as the reciprocal of x, opposite to the value of y, obtained from Figure-1: (-7,0), (-2,-2), and (0.5,-3), and from Figure-2: (0.5,-3), (3,-1), and (8,3). We establish points on the graph paper from the points obtained for Figure-1, and obtain a straight line, and then connect the established points from Figure-2, obtaining another straight line. Here, the two straight lines intersect each other at point (0.5,-3).

 
Hence, the concluded solution- (x,y) = (0.5,-3)

 

(iii)

Given,  

 x  2 +  y  3 = 8
 5x   4 - 3y = -3 

Comparing with the equations a₁x + b₁y = c₁, a₂x + b₂y = c₂  we get,
 a₁  a₂ =  1  2 ÷  5  4 =  1  2 ✕  4  5 =  2  5
 b₁  b₂ =  1  3  ÷ (-3) =  1  3  ✕  1 -3 = -  1  9
 c₁  c₂ = -  8  3

Therefore, a₁a₂ ≠ b₁b₂

So, There is only one solution to the system of equations.

Graph solution-
 x  2 +  y  3 = 8

or, 3x + 2y     6 = 8

or, 3x + 2y = 48   [Multiplied by 6]

or, 2y = 48 - 3x

or, y = 48 - 3x    2 .....(i)

Here, for different values of x in equation (i), we determine several values of y.
Figure-1

Again,
 5x   4 - 3y = -3

or, 5x - 12y = -12

or, -12y = -12 - 5x

or, 12y = 12 + 5x

or, y = 12 + 5x    12......(ii)

Now, For equation (ii), we find several values of y for different values of x.
Figure-2

Now, on graph paper, we plot points for each square with the length of its side taken as the reciprocal of x, opposite to the value of y, obtained from Figure-1: (10,9), (8,12), and (12,6), and from Figure-2: (12,6), (6,3.5), and (0,1). We establish points on the graph paper from the points obtained for Figure-1, and obtain a straight line, and then connect the established points from Figure-2, obtaining another straight line.

Here, the two straight lines intersect each other at point (12,6).

Hence, the concluded solution (x,y)=(12,6)

(iv)

Given,    

-7x + 8y = 9
5x - 4y = -3

Comparing with the equations a₁x + b₁y = c₁, a₂x + b₂y = c₂ we get,
 a₁  a₂ = -  7  5
 b₁  b₂ = -  8  4 = -2
 c₁  c₂ = -  9  3 = -3

Therefore, a₁a₂ ≠ b₁b₂

So, There is only one solution to the system of equations.

Graph solution-
-7x + 8y = 9
or, 8y = 9 + 7x
or, y = 9 + 7x   8....(i)

Here, for different values of x in equation (i), we determine several values of y.
Figure-1

 

Again,
5x - 4y = -3
or, -4y = -3 - 5x
or, 4y = 3 + 5x
or, y = 3 + 5x   4.....(ii)

Now, For equation (ii), we find several values of y for different values of x.
Figure-2

 

Now, on graph paper, we plot points for each square with the length of its side taken as the reciprocal of x, opposite to the value of y, obtained from Figure-1 (-2,-58), (0,98) and (1,2) and from Figure-2 (-2,-1.75),(1,2) and (3,4.5) We establish points on the graph paper from the points obtained for Figure-1, and obtain a straight line, and then connect the established points from Figure-2, obtaining another straight line.

 

Here, the two straight lines intersect each other at point (1,2). Hence, the concluded solution (x,y) = (1,2)   (Answer)

E.5.3 Solve using substitution method: 

i)  7x - 3y = 31 

     9x - 5y = 41

ii)  (x + 2)(y - 3) = y(x - 1) 

      5x - 11y - 8= 0

iii)  xa + yb = 2

      ax + by = a² + b²

iv)   x14 +  y18 = 1 

      x + y   2 + 3x + 5y     2 = 2

v)  p(x + y) = q(x - y) = 2pq

Solution:

(i) 7x - 3y = 31
     9x - 5y = 41

Given,
7x - 3y = 31.......(i)
9x - 5y = 41.......(ii)

From equation no. (i), we get,
7x = 31 + 3y
or, x = 3y + 31     7…....(iii)

Here, we put the value of x in equation (ii),
    9 ✕ 3y + 31     7 – 5y = 41
or, 27y + 279       7 – 5y = 41
or, 7 ✕ (27y + 279       7 – 5y) = 7 ✕ 41   [Multiplying both sides by 7]
or, 27y + 279 - 35y = 287
or, -8y + 279 = 287
or, -8y = 287 - 279
or, -8y = 8
or, y = -1

Here, putting the value of y in equation (iii),
    x = 31 + 3.(-1)       7
or, x = 31 - 3   7
or, x = 28 7 = 4

So, solving by substitution method, (x, y) = (4, - 1)   (Ans)

(ii) (x + 2)(y - 3) = y(x - 1)
      5x - 11y - 8 = 0

Given,
(x + 2)(y - 3) = y(x - 1)......(i)
5x - 11y - 8 = 0.......(ii)

From equation no (i),
    xy + 2y - 3x - 6 = xy - y
or, xy + 2y - 3x - 6 - xy + y = 0
or, 3y = 3x + 6
or, 3y = 3(x + 2)
or, y = x + 2.......(iii)

Here, putting the value of y into equation (ii),
    5x - 11(x + 2) - 8 = 0
or, 5x - 11x - 22 - 8 = 0
or, - 6x = 22 + 8
or, - 6x = 30
or, x = - 5

Here, substituting the value of x into equation (iii),
y = - 5 + 2 = - 3

Hence, the solution by substitution is: (x, y) = (- 5, - 3)   (Ans.)

(iii) xa + yb = 2
     ax + by = a² + b²

Given,
xa + yb = 2……(i)
ax + by = a² + b²…....(ii)

From equation no. (i) , we get,
xa + yb = 2
or, ab(xa + yb) = 2ab [multiplying both sides by ab]
or, xb + ya = 2ab
or, xb = 2ab - ya
or, x = 2a - ya b.....(iii) [divide both sides by b]

Now putting the value of x into equation (ii),
     a(2a - ya b) + by = a² + b²
or, a.2a - (ya b).a + by = a² + b²
or, - (ya b).a = a² + b² – a.2a – by
or, - (ya b).a = a² + b² – 2a² - by
or, - (ya b).a = b² – a² - by
or, - ya.a = b(b² – a² – by)
or, - ya² = b³ – a²b – b²y
or, - ya² + b²y = b(b² - a²)
or, y(b² - a²) = b(b² - a²)
or, y = b    [divide both sides by (b² - a²)]

Here, substituting this value of b into equation (iii),
x = 2a - ab b
or, x = 2a – a = a

Hence, the solution determined by substitution method is: (x, y) = (a, b)   (Ans.)


(iv)  x18 +  y18 = 1
      x + y    2 + 3x + 5y      2 = 2

Given,
 x18 +  y18 = 1......(i)
x + y    2 + 3x + 5y      2 = 2.....(ii)

(ii) Multiplying both sides of the equation by 2, we get
    2(x + y   2 + 3x + 5y     2) = 2 ✕ 2
or, x + y + 3x + 5y = 4
or, 4x + 6y = 4
or, 2(2x + 3y) = 4
or, 2x + 3y = 2
or, 2x = 2 - 3y
or, x = 2 - 3y   2.....(iii)

Now substituting this value of x into equation (i),

      x ✕  114 +  y18 = 1
or,  2 - 3y   2 ✕  114 +  y18 = 1
or, (2 - 3y)    28 +  y18 = 1
or, (9(2 - 3y) + 14y)          252 = 1
or, 9(2 - 3y) + 14y = 252 [multiply both sides by 252]
or, 18 - 27y + 14y = 252
or, -13y = 252 - 18
or, - 13y = 234
or, y = -18

Here, substituting this value of y into equation (iii),
x = 2 - 3(-18)      2
or, x = 2 + 54   2
or, x = 56 2
or, x = 28

Hence, the solution to be determined by substitution method, (x, y) = (28, - 18)  (Ans.)

v) p(x + y) = q(x - y) = 2pq

Given,
p(x + y) = 2pq......(i)
q(x - y) = 2pq......(ii)

From equation no. (i),
x + y = 2q
or, x = 2q - y......(iii)

Here, putting this value of x into equation (ii),
    q(2q - y - y) = 2pq
or, q(2q - 2y) = 2pq
or, q.2(q - y) = 2pq
or, 2q(q - y) = 2q.p
or, (q - y) = p
or, -y = p - q
or, y = q - p

Here, substituting this value of y into equation (iii),
     x = 2q - (q - p) 

or, x = 2q - q + p = q + p

Hence, the solution to be determined by substitution method is, (x, y) = (q+p, q-p) (Ans.)

E.5.4 Solve using elimination method:

i)  3x - 5y = -9 

     5x - 3y = 1

ii)  x + 1y + 1 = 45 

      x - 5y - 5 = 12

iii)  2x + 3y = 5 

       5x - 2y = 3

iv)  ax + by = 1 

       bx + ay =    2aba² + b²

Solution:

Solve using elimination method-

(i) 3x - 5y = - 9
     5x - 3y = 1

Given,
3x - 5y = - 9
or, 9x - 15y = - 27.....(i) [multiply both sides by 3]
And,
5x - 3y = 1
or, 25x - 15y = 5.....(ii) [multiply both sides by 5]

Here, by subtracting (ii) from (i), we get,
25x - 15y - (9x - 15y) = 5 – ( - 27)
or, 25x - 15y - 9x + 15y = 5 + 27
or, 16x = 32
or, x = 2

Here, putting x = 2 in equation (ii) we get,
25x - 15y = 5
or, 25 ✕ 2 – 15y = 5
or, 50 – 15y = 5
or, -15y = 5 - 50
or, -15y = - 45
or, y = 3

Therefore, the solvable solution by elimination method is (x, y) = (2, 3)

(ii) x + 1y + 1 = 45
      x - 5y - 5 = 12

Given,
x + 1y + 1 = 45
or, 5(x + 1) = 4(y + 1)
or, 5x + 5 = 4y + 4
or, 5x - 4y = 4 - 5
or, 5x - 4y = - 1.....(i)
And,
x - 5y - 5 = 12
or, 2(x - 5) = 1(y - 5)
or, 2x - 10 = y - 5
or, 2x - y = -5 + 10
or, 2x - y = 5
or, 8x - 4y = 20.....(ii) [multiply both sides by 4]

Here, by subtracting (ii) from (i) we get,
8x - 4y - (5x - 4y) = 20 - ( - 1)
or, 8x - 4y - 5x + 4y = 20 + 1
or, 3x = 21
or, x = 21 3
or, x = 7

Here, the value of x = 7, put in equation (ii) and we get,
8 ✕ 7 - 4y = 20
or, 56 - 4y = 20
or, -4y = 20 – 56
or, -4y = -36
or, y = 9

Therefore, the solvable solution by elimination method is (x, y) = (7, 9)


(iii) 2x + 3y = 5
       5x - 2y = 3

Given,
2x + 6y = 5
or, 4x + 3y = 10.....(i) [multiply both sides by 2]
And,
5x - 2y = 3
or, 15x - 6y = 9.....(ii) [multiply both sides by 3]

Here, (i) and (ii) add,
4x + 6y + 15x - 6y = 10 + 9
or, 19x = 19
or, x = 1

Here, x = 1, Substituting this value into equation (ii), we get,
15 ✕ 1 - 6y = 9
or, - 6y = 9 - 15
or, - 6y = - 6
or, -6y = - 6
or, y = 1

Therefore, the solvable solution by elimination method isঃ (x, y) = (1, 1)


(iv) ax + by = 1
       bx + ay = 2ab(a² + b²)

Given,
ax + by = 1
or, b(ax + by) = 1.b [multiply both sides by b]
or, abx + b²y = b.....(i)
And,
bx + ay =      2ab(a² + b²)
or, a(bx + ay) =  a ✕ 2ab(a² + b²)
or, abx + a²y =    2a²b(a² + b²).....(ii) [multiply both sides by a]

Here, subtracting (ii) from (i) we get,
abx + a²y - ( abx + b²y) = 2a²b/(a² + b²) - b
or, a²y - b²y =     2a²b(a² + b²) – b
or, y(a² - b²) = 2a²b - b(a² + b²)        a² + b²
or, y(a² - b²) = 2a²b - a²b - b³      a² + b²
or, y(a² - b²) = a²b - b³a² + b²
or, y(a² - b²) = b(a² - b²)  a² + b²
or, y =      b a² + b²

Here, ax + by = 1 Substitute the obtained value of y into the equation,
ax + b.      b(a² + b²) = 1
or, ax(a² + b²)+ b²       a² + b² = 1
or, ax(a² + b²) + b² = a² + b²
or, ax(a² + b²) = a² + b² - b²
or, ax(a² + b²) = a²
or, x(a² + b²) = a
or, x =     aa² + b²

Therefore, the solvable solution by elimination method is
x =      aa² + b²    And
y =     aa² + b²

E.5.5 Solve using cross multiplication method:

i)  3x - 2y = 2 

    7x + 3y = 43

ii)  x2 + y3 = 8 

     5x 4 - 3y = -3

iii)  px + qy = p² + q² 

      2px - py = pq

iv)  ax - by = ab 

       bx - ay = ab

Solution:

Solution by cross multiplication method-

(i) 3x - 2y = 2
    7x + 3y = 43

The given equation can be written as-
3x - 2y = 2
or, 3x - 2y - 2 = 0
and,
7x + 3y = 43
7x + 3y - 43 = 0

Here, applying the cross multiplication method and we get,
               x(-2)(-43) - (-2)(3) =           1(3)(3) - (7)(-2)
or,       x86 - (-6) =      19 - (-14)
or,  x92 =  123
or, 23x = 92 [divide both sides by 23]
or, x = 4

Again,
            y(-2)(7) - (-43)(3) =           1(3)(3) - (7)(-2)
or,           y- 14 - (-129) =      19 - (-14)
or,   y115 =  123
or, 23y = 115 [divide both sides by 23]
or, y = 5

Therefore, the solution, solved by cross multiplication, (x, y) = (4, 5)

(ii) x2 + y3 = 8
     5x 4 - 3y = -3

The given equation can be written as -
x2 + y3 – 8 = 0
and,
5x 4 - 3y + 3 = 0

Here, applying cross multiplication method, and we get-
           x(13)(3) - (-3)(-8) =             1(12)(-3) - (54)(13)
or,      x1 - (24) =     1-32 - 512
or,   x-23 =      1-23/12
or,   x-23 = 1 ✕ (-1223)

or, -23-23.x = (-23) ✕ 12      -23     [divide both sides by - 23]
or, 1.x = 1 ✕ 12 [divide both sides by -23]
or, x = 12

Again,
            y(-8)(54) - (3)(12) =             1(12)(-3) - (54)(13)
or,       y- 10 - 32 =      1- 32 - 512
or,       y  -20-3        2 =     1-18-5  12
or,    y -23    2 =  1-23 12
or, 2y23 = 1223
or, 2y = 12 [multiply both sides by -23]
or, y = 12 2
or, y = 6

Therefore, the solution, solved by cross multiplication, (x, y) = (12, 6)

(iii) px + qy = p² + q²
      2qx - py = pq

The given equation can be written as-
px + qy = p² + q²
or, px + qy - p² - q² = 0
and,
2qx - py = pq
or, 2qx - py - pq = 0

Here, the method of cross multiplication applied,
                 x(q)(-pq) - (-p)(-p² - q²) =             1(p)(-p) - (2q)(q)
or,              x- pq² - p³ - pq² =        1- p² - 2q²
or,          xp(-p² - 2q²) =        1- p² - 2q²
or,          xp(-p² - 2q²) =        1- p² - 2q²
or, xp = 1 [multiply both sides by (-p² - 2q²)]
or, x = p

Again,
                    y(-p² - q²)(2q) - (-pq)(p) =             1(p)(-p) - (2q)(q)
or,                 y- 2p²q - 2q³ + p²q =        1- p² - 2q²
or,          y- p²q - 2q³ =        1- p² - 2q²
or,          yq(-p² - 2q²) =        1- p² - 2q²
or,  y   q = 1 [multiply both sides by (-p² - 2q²)]
or, y = q

Therefore, the solution, solved by cross multiplication, (x,y) = (p,q)


(iv) ax - by = ab
      bx - ay = ab

The given equation can be written as-
ax - by = ab
or, ax - by - ab = 0
and,
bx - ay = ab
bx - ay - ab = 0

Here, applying the cross multiplication method, we get,
               x(-b)(-ab) - (-a)(-ab) =            1(a)(-a) - (b)(-b)
or,        xab² - a²b =        1- a² + b²
or,        xab(b - a) =          1(b - a)(b + a)
or, x(b - a)(b + a) = ab(b - a)
or, x(b + a) = ab [divide both sides by (b - a)]
or, x =   aba + b

Again,

           y(-ab)b - (-ab)a =            1(a)(-a) - (b)(-b)
or,          y- ab² + a²b =       1- a² + b²
or,       ya²b - ab² =        1- a² + b²
or,        yab(a - b) =          1(b - a)(b + a)
or, y(b - a)(b + a) = ab(a - b)
or, y(b - a)(b + a) = -ab(b - a)
or, y(b + a) = -ab [divide both sides by (b - a)]
or, y =  -aba + b

Therefore, the solution, solved by cross multiplication,
x =   aba + b and
y =  -aba + b

E.5.6 Apu has a rectangular vegetable garden. The perimeter of the garden is 120 meters. Doubling the width and subtracting 3 meters from the length the perimeter will 150 meters. 

a) The garden is enclosed on 3 sides and is open on one side along its length. How much money will it cost to enclose the empty side with a fence of tk. 10 Per meter? 

b) If organic fertilizer costs tk.7 per square meter, how much will Apu have to spend on fertilizer in total?

Solution:

a) To determine the cost of enclosing the remaining empty area with fencing-

Let,
The rectangular garden, 

length = x meter and
width = y meter।

According to question,
Perimeter, 2(x + y) = 120......(i)
and 2{2y + (x - 3)} = 150.....(ii)

Now, from equation (i) we get,
2(x + y) = 120
or, x + y = 60
or, x = 60 - y......(iii)

From equation (ii) we get,

2{2y + (x - 3)} = 150

or, {2y + (x - 3)} = 150  2

or, 2y + x - 3 = 75

or, 2y + x = 75 + 3

or, 2y + x = 78

or, 2y + (60 - y) = 78   [substituting the value of equation (iii)]

or, y = 78 - 60

∴ y = 18

Substituting the value of y into equation (iii),
    x = 60 - y
or, x = 60 – 18

∴ x = 42

 
From the above solution find the length of the rectangular vegetable garden, x = 42 m.
According to the question one side of the garden is free along the length 42 meters.

Given, to fence around, 

1 meter cost   = 10 Taka
42 meter cost = (42 ✕ 10) Taka
                             = 420 Taka

So, the cost of fencing is 420 Taka.  (Answer)




b) How much will be spent on fertilizers to determine-

From the above solution, length of rectangular vegetable garden, x = 42 meter and width y = 18 meter

Therefore, area of rectangular vegetable garden = (42 ✕ 18) sq m = 756 sq m।

Given, cost per 1 square meter of organic fertilizer = 7 Taka
                                   ∴ Cost per 756 square meter = (7 ✕ 756) Taka
                                                                                 = 5292 Taka।

So, the cost for organic fertilizer is 5292 Taka. (Answer)

E.5.7 Find the nature of the roots of x² - 3 = 0 and solve it.

Solution:

Determine and solve the nature of the roots of the equation x² – 3 = 0

Given,
x² – 3 = 0

We know,
standard form of quadratic equation- ax² + bx + c = 0
Here, a = 1, b = 0, c = -3

Arrange the given equation according to standard form of quadratic equation- 1.x² + 0.x + (-3) = 0
Then, the determinant of the given equation is  b²-4ac = 0²-4.1.(-3) = 12
Now, the determinant is 12 > 0 and is not an integer.

Then, the nature of the roots of this equation is found, its roots are real, unequal and irrational.

This equation expresses the quadratic equation-
      
x = -b ± √ (b²-4ac)     2a        

or, x = -0 ± √ {0² - 4 ✕ 1 ✕ (-3)}               2 ✕ 1            

or, x = ±√ 12    2          

or, x = ±√ (4 ✕ 3)        2           

or, x = ±2√ 3     2
              
or, x = ±√ 3 

So, the two roots of the equation are x₁ = √ 3  and x₂ = -√ 3  (Answer)

E.5.8 Solve 3x² - 2x -1 = 0 using formula. Again solve it with the help of graph and show that there is the same solution in both the method.

Solution:

The equation 3x² - 2x - 1 = 0 is to be solved with the help of formulas and graphs.

Given,
3x² - 2x - 1 = 0

Comparing the given equation with quadratic equation,
ax²+bx+c=0, where, a = 3, b = -2, c = -1

Therefore,

x = -b ± √ (b²-4ac)      2a

or, x = -(-2) ± √ {(-2)² - 4 ✕ 3 ✕ (-1)}                2 ✕ 3

or, x = 2 ± √ (4+12)        6

or, x = 2 ± √ 16     6

or, x = 2 ± 4   6

So, x₁ = 2 + 4   6 = 1 and, x₂ = 2 - 4   6 = -  2  6 = -  1  3

Solution with the help of diagram-

Let,
y = 3x² - 2x – 1

Determine the value of y for different values of x

On graph paper, the length of one side of the least square is one unit and the points x and y are plotted.

Plotting the value of (x,y) on the graph shows that (- 13,0) and intersect at the points (1,0). Therefore, these point values are the solutions of the given equation.
So, x₁ = 1 and, x₂ = - 13

Therefore, both formula and diagram methods yield the same solution. (shown)

E.5.9 Setu’s mother keeps ducks and chickens at home. She bought 25 ducklings and 30 chickens for tk. 5000. If she had bought 20 ducklings and 40 chickens at the same rate, he would have spent tk. 500 less. 

a) What is the price of a duckling and a chicken? 

b) How much total profit will she make if she sells each duck at tk. 250 and each chicken at tk.160 after rearing for a few days?

Solution:

(a) The price of one duckling and one chicken is to be determined by—

Suppose,
each duckling is worth x Taka
each chicken is worth y Taka

Here, as per the question,
25x + 30y = 5000

or, 5(5x + 6y) = 5000

or, (5x + 6y) = 5000   5

or, 5x + 6y = 1000.....(i)


and
20x + 40y = 5000 – 500

or, 20x + 40y = 4500

or, 20(x + 2y) = 4500

or, (x + 2y) = 4500  20

or, x + 2y = 225

or, x = 225 - 2y......(ii)


Here, putting x = 225 - 2y, in equation (i),
    5x + 6y = 1000

or, 5(225 - 2y) + 6y = 1000

or, 1125 – 10y + 6y = 1000

or, -4y = 1000 – 1125

or, -4y = -125

or, y = -125  -4

∴ y = 31.25


Substituting the value of x into equation (ii), we get,
 x = 225 - 2y
    = 225 – 2 ✕ 31.25
    = 225 – 62.50
    = 162.50

Therefore, a duckling costs 162.50 Taka and a chicken costs 31.25 Taka. (Answer)



b) How much total profit should be determined-

The number of ducklings purchased is 25 and the number of chickens purchased is 30.

After rearing,
  selling price of 1 duck  = 250 Taka
selling price of 25 ducks = (250 ✕ 25) Taka
                                        = 6250 Taka

Again, after rearing,
  selling price of 1 chicken = 160 Taka
   selling price of 30 ducks = (160 ✕ 30) Taka
                                           = 4800 Taka

Therefore, total selling price = (6250 + 4800) Taka
                                              = 11050 Taka

Given, the total purchase price of duck and chicken was = 5000 Taka

Therefore, Setu's mother's profit is = Selling price - Purchase price
                                                        = (11050 - 5000) Taka 
                                                        = 5050 Taka

So, total profit will be 5050 Taka.      (Answer)

E.5.10 Solve the following system of equations:

            y = x² - 2x -3 

            x - 3y + 1 = 0

Solution:

Solve the following equations-

Given,
y = x² - 2x – 3.....(i)
x - 3y + 1 = 0......(ii)

Substituting the value of y from equation (i) into equation (ii),
x – 3(x² - 2x - 3) + 1 =0

or, x – 3x² + 6x + 9 + 1 = 0

or, -3x² + 7x + 10 = 0

or, 3x² – 7x – 10 = 0

or, 3x² + 3x - 10x – 10 = 0

or, 3x(x + 1) - 10(x + 1) = 0

or, (x + 1)(3x - 10) = 0

Therefore, x + 1 = 0
or, x = -1

And, 3x - 10 = 0
or, 3x = 10
or, x = 10 3 

Now, putting the value of x = -1 in equation (i),
 y = (-1)² – 2.(-1) – 3
    = 1 + 2 - 3
    = 0
and x = 10 3  the value putting in equation (i)
y = (10 3)² – 2.(10 3) – 3
    = 100  9 - 20 3 – 3
    =  100 - 20 ✕ 3 - 3 ✕ 9               9
    =  100 - 60 - 27          9
    =    13   9

Therefore, the deterministic solution, (x,y) = (-1,0),(10 3, 13 9)    (Answer)

E.5.11 Construct 3 sets system of equations of two variables (one is linear and another is quadratic) as you like and solve it.

Solution:

Form and solve 3 sets of equations in two variables (one linear and one quadratic) on your own.

1st set of equations formed-
y = x² - x - 2.....(i)
x - 2y + 5 = 0.....(ii)

From equation no.(i), value of y substituting into equation no. (ii),

x – 2(x² - x - 2) + 5 = 0

or, x – 2x² + 2x + 4 + 5 = 0

or, -2x² + 3x + 9 = 0

or, 2x² - 3x - 9 = 0

or, 2x² - 6x + 3x - 9 = 0

or, 2x(x - 3) + 3(x - 3) = 0

or, (2x + 3)(x - 3) = 0

Therefore, 

    2x + 3 = 0
or, 2x = -3   
or, x = -  3  2
and,
     x - 3 = 0
or, x = 3

Here, x = 10 3 substituting the value into equation (i),
     y = (-  3  2)² – (-  3  2) – 2
or, y =  9  4  +  3  2 – 2
  ∴ y =  7  4

Again, putting the value of x = 3 into equation (i),
     y = 3² – 3 – 2
or, y = 9 – 3 - 2
  ∴ y = 4

So, the deterministic solution, (x,y) = (3,4),(-  3  2,  7  4)    (Answer)

2nd set of structural equations- 

y = x² - 3x + 2.....(i)
x - y - 1 = 0......(ii)

From equation no.(i), value of y substituting into equation no. (ii),

x – (x² - 3x + 2) - 1 = 0

or, x – x² + 3x - 2 - 1 = 0

or, -x² + 4x - 3 = 0

or, x² - 4x + 3 = 0

or, x² – 3x – x + 3 = 0

or, x(x - 3) - 1(x - 3) = 0

or, (x - 1)(x - 3) = 0

Therefore, 

    x - 3 = 0
or, x = 3   

and

     x - 1 = 0
or, x = 1

Here, substituting the value of x = 3 in equation (i),
    y = 3² – 3.3 + 2
or, y = 9 – 9 + 2
or, y = 2

And substituting the value of x =1 into equation (i),
    y = 1² – 3.1 + 2
or, y = 1 – 3 + 2
or, y = 0

Therefore, the deterministic solution is (x,y) = (3,2),(1,0)   (Answer)

3rd set of equations formed-
y = 2x² - 2x - 3......(i)
x - y - 4 = 0......(ii)

From equation no.(i), value of y substituting into equation no. (ii),

    x – (2x² - 2x - 3) - 4 = 0

or, x – 2x² + 2x + 3 - 4 = 0

or, -2x² + 3x - 1 = 0

or, 2x² - 3x + 1 = 0

or, 2x² - x - 2x + 1 = 0

or, x(2x - 1) - 1(2x - 1)

or, (x - 1)(2x - 1) = 0

therefore, 

    x - 1 = 0
or, x = 1

and

    2x - 1 = 0
or, 2x = 1   
or, x =  1  2

Now, putting the value of x = 1 into equation (i),
    y = 2.1² - 2.1 – 3
or, y = 2 – 2 – 3
or, y = -3

and x =  1  2 substituting the value in equation (i),
 

  y = 2.(12)² - 2. 1  2 - 3

or, y =  1  2 - 1 - 3

or, y =  1  2 - 4

or, y = 1 -  8  2

or, y = -  7  2

Therefore, the deterministic solution is (x,y) = (1,-3),( 1  2,-  7  2)    (Answer)