35 Chapter 2
Exercise 2.6
Simultaneous Equation: A system of equation which have a common solution is called simultaneous system of equation.i.e.
Put x 2 = 4 in eq (1) 4 + y
2
=
4
ax dx + + by = c ey = f
⎫ ⎬ ⎭
→ Simultaneous Linear equation
Exp23: Solve 2 x + y
=
10
the system
y
2=
0 ⇒ y=
0 Solution of given system = { ( - 2,0 ) , ( 2,0 ) } 4 x 2 + y
2
= 68 Sol; Given system 2 x + y = 10 ..............(1)
Exercise 2.6
4 x 2 + y 2 = 68 ................(2) Form eq (1) y = 10 - 2 x ........(3) put in eq (2)
Q1i). solve 2x - y =
3
the system x 2 + y 2
=
2
4 x 2+ ( 10 - 2 x
)2 =
68
Sol: Since
4 x 2 + 100 - 40 x + 4 x
2
- 68 =
0
Form equation (1)
Putting the value of y in equation (2) we get
Either Or
Substituting these values in equation (3) When x = 1
Solution Set
Q1ii). Solve the system
Sol: Since
Form equation (1) we have
Putting the value of y in equation (2) we get ( 2 )2 4 2
32
4 2 4 2
32
8 2
32
24 2
2x - y =
3 ( 1 )x 2 + y 2 =
2 ( 2 ) 8 x 2 - 40 x + 32 = 0 divided by 8 x 2- 5 x
+ 4 =
0
x 2
- 4 x - 1 x
+ 4 =
0 x ( x - 4 ) - 1 ( x
- 4 )
=
0 ( x - 1 )( x
- 4 )
=
0 Either x- 1 = 0 or x- 4 = 0 x = 1 x = 4 put in eq (3) y= 10 -
21
( ) y= 10 -
2 ( 4 ) y= 10 -
2
y= 10 -
8 y
=
8
y
=
2 Solution of the given system = { ( 1,8 ) , ( 4,2 ) } Exp24: Solve - = 2 + + 2
= - the system Sol: Given system x - y = 7 ............(1) And x 2 + 3 xy + y 2 = - 1 ..........(2) From eq (1) x = 7 + y ...........(3) put in (2) ( ) 2 ( ) 2
2 2 2
2 2 2
x y
7 x 3 xy y
1
7 + y + 3 7 + y y + y
= -
1
49 + 14 y + y + 21 y + 3 y + y
= -
1 y + 3 y + y + 14 y + 21 y
+ 49 + 1 =
0 5 y 2 + 35 y + 50 = 0 divided by 5 y 2+ 7 y
+ 10 =
0
y 2
+ 5 y + 2 y
+ 10 =
0 y ( y + 5 ) + 2 ( y
+ 5 )
=
0 ( y + 2 )( y
+ 5 )
=
0 Either y+ 2 = 0 or y+ 5 =
0 y = - 2 y = - 5 Put in (3) x= 7 + ( -
2
) x= 7 + ( -
5 ) x
=
5 Solution of given system x
=
= 2
{ ( 5, - 2 ) , ( 2, - 5 ) } Exp25: solve
x 2 + y
2
=
4 2 x 2 - y
2
= 8
the system
Sol: Given system x 2 + y 2 = 4 .........(1)
2 x 2 - y 2 = 8 .........(2) Adding (1) & (2) 3 x 2 =
12 x 2 = 4 taking square root x = ±
2 - y + y
=
y + y
= y=y = ⇒ y
= ±
2x - y =
3 2x - 3 =
y 3
( x 2
+ 2x - 3 )
2 =
2 x 2+ ( 2x ) 2 - 2 ( 2x )( 3 ) + ( 3 )
2 =
2
x 2 + 4x 2
- 12x + 9 - 2 =
0 5x 2- 12x + 7 =
0
5x 2
- 7x - 5x + 7 =
0 x ( 5x - 7 ) - 1 ( 5x - 7 )
=
0 ( x - 1 )( 5x - 7 )
=
0
x - 1 =
0 5x x =
1
- 7 =
0 5x =7x = 7 5y 2 ( 1 ) 3
x = 7 5 y 2 3 y 1
= -
= - = -
y = 2 ( 7 5
) - 3 y = 14 5 -3 y = 14 5 - 15 =
- 5 1 = ⎧ │ ⎨ │ ⎩ ( 1, - 1 ) , ⎛ │ ⎝ 7 5 ,1 5
- ⎞ │ ⎠ ⎫ │ ⎬ │ ⎭ x + 2y =
0
x 2 + 4y 2
=32
x + 2y =
0 ( 1
)
x 2 + 4y 2
=
32 ( 2
)
x + 2y =
0 x = -
2y 3
( )
( )
36 Exercise 2.6
Substituting these values in equation (3) When
Solution Set
Q1iii). Solve the system
Sol: Since
Form equation (1)
Putting the value of y in equation (2) we get
Either Or
Substituting these values in equation (3) When x = 2 x = - 4
Solution Set
Q1iv). Solve the system
Sol: Since
Form equation (1)
Putting the value of y in equation (2) we get
2( )2 2 2
2
Chapter 2
y = 2 y = - 2 x = -
2 ( 2
) x = - 2 ( -
2 ) x = - 4
= { ( - 4,2 ) x , ( = 4, 4 - 2 ) } 2x - y = -
8
Solution Set
x 2+ 4x =
y
2x - y = -
8 ( 1
)
Q1v). Solve the system
x 2+ 4x =
y ( 2
) Sol: Since
2x - y = -
8 2x + 8 =
y 3
From equation (2) we get
x 2+ 4x = 2x +
8 x 2+ 4x - 2x - 8 =
0 x x + 4 - 2 x + 4 =
0 Putting 4 2 5 ( the 3 x - 2 x + 4 =
0
x + - value 3 x
2
) =
of 4 y in equation (1) we get ( ) ( ) ( )( )
4 x 2 + 15 - 15 x
2
=
4
4 x 2 - 15 x
2
= 4 -
15
x - 2 =
0
x + 4 =
0
- 11 x
2
= -
11
x =
2
x = -
4
x2= - -
1111
y = 2 ( 2 ) +
8
y = 2 ( - 4 ) +
8
x
2=
1 Substituting in equation (3) we get y = 4 +
8
y = - 8 +
8 y =
12
y =
0 = { ( 2,12 ) , ( -
4,0 ) } Therefore and
2x + y =
4 x 2 - 2x + y 2
=
3
Taking square root and
2x + y =
4 ( 1
)
Solution set
x 2 - 2x + y 2
=
3 ( 2
) Q1vi). Solve the system
2x + y =
4
Sol: Since y = 4 -
2x 3
Adding equations (1) and (2) x - 2 x + 4 - 2 x
=
3
x - 2 x + 16 - 16 x + 4 x
=
3 5 x - 18 x
+ 13 =
0 5 x 2 - 13 x - 5 x
+ 13 =
0 x ( 5 x - 13 ) - 1 ( 5 x
- 13 )
=
0 ( x - 1 )( 5 x - 13 ) =
0 Substituting in equation (1) we get 5 ( 9 ) = y2
+
9 Either Or
45 - 9
=
y
2
x - 1 =
0
- y 2 = 36 x =
1
135
Thus & Taking square root
&
S.S Q1vii). Solve the system
5 x13 =
0 5 x=13 x
= Substituting these values in equation (3) When x = 1 x =
135
( )
( )
3x + y =
3 y = 3 -
3x 3
2 5x = y 2
+
9 x 2 = - y 2
+
45
6x 2
= +
54 x 2= 54 6=
9 x 2= 9 x 2= 9 y 2= 36 x = ± 3 y = ± 6 = { ( ± 3, ± 6 ) } = { ( 3,6 ) , ( - 3,6 ) , ( 3, - 6 ) , ( - 3, -
6 ) } 4x 2 + 3y 2
- 5 =
0 2x 2 + 3y 2
- 4 =
0
2x = 1 y 2= 3 - 31 ( ) y 2= 3 - 3 =
0
x 21= y 2= 0 x = ± 1 y =0 = { ( ± 1,0 ) } = { ( 1,0 ) , ( -
1,0 ) } 5x 2 = y 2
+
9 x 2 = - y 2
+
45
5x 2 = y 2
+
9 ( 1 )x 2 = - y 2
+
45 ( 2
)
( y = 4 -
21 ) y = 4 -
2 y =
2
2 2
2 2
4x + 5y =
4 1
3x + y =
3 2
⎧ = │ ⎨ │ ⎩ ⎛ │ ⎝ - ⎞ │ ⎠ ⎫ │ ⎬ │ ⎭ 4x 2 + 5y 2
=
4 3x 2 + y 2
=
3
2 2
( )
2 2
( )
( )
⎛ = - │ ⎝ ⎞ │ ⎠
= - = - ( 1,2 ) , 13 5 , 6 5
y 4 2 13 5 y 20 5 26 y 5 6
37 Chapter 2
Exercise 2.6
Sol: Since
4x 2 + 3y 2
- 5 =
0 ( 1
)
x 2+ 3 x
+ 2 =
0
2x 2 + 3y 2
- 4 =
0 ( 2
)
x 2
+ 2 x + 1 x
+ 2 =
0 Subtracting
x ( x + 2 ) + 1 ( x
+ 2 )
=
0 4x 2 + 3y 2
- 5 =
0 ± 2x 2 ± 3y 2
4 =
0
( x + 1 )( x
+ 2 )
=
0 Either x+ 1 = 0 or x+ 2 = 0 2x 2- 1 =
0 x = - 1 x = - 2 put in (3) 2x 2
=1y= 4 + ( -
1
) y= 4 + ( -
2 ) x 2= 21 Substituting x 2 =
2 1 in equation (1) we get 4 ⎛ │ ⎝ 2
1 ⎞ │ ⎠
y
=
3
y
=
2 Solution set = { ( - 1,3 ) , ( - 2,2 ) } Exp26: A rectangular shed is being build that has an area of 120 square feet and is 7 + 3y 2
- 5 = 0 Feet longer than it is wide. Determine its
2 + 3y 2
- 5 =
0
dimensions. Sol: Let width = x feet 3y 2- 3 =
0
Length = x+ 7 feet
3y 2 =3y 2 = 3 3
=
1 Given that area = 120 square feet
x ( x + 7 )
=
120
Therefore x 2 = 2 1 and
y 2= 1 Taking square root
x 2+ 7 x
- 120 =
0
x 2
+ 15 x - 8 x
- 120 =
0 x ( x + 15 ) - 8 ( x
+ 15 )
=
0 x = ± 1 2 and
y = ±1 ( x - 8 )( x
+ 15 )
=
0
Solution set
= ⎧ │ ⎨ │ ⎩ ⎛ │ ⎝ ± 2
1 , ± 1 ⎞ │ ⎠ ⎫ │ ⎬ │ ⎭
Either x- 8 = 0 or x+ 15 =
0 x = 8 x = - 15 which is impossible Therefore width of rectangle = 8 feets
Q2i) Solve x + y
=
9 x 2 + 3 xy + 2 y
2
= 0
The system
And Length of rectangle = 8 + 7 = 15 feets Exp27: A men purchased a number of shares of stock for an amount of Rs. 6000 if he had Sol: Given system x + y = 9 ..........(1)
x 2 + 3 xy + 2 y 2 = 0 ............(2)
paid Rs. 20 less per share, number of share that could have been purchased for amount of money would have increased by 10. How From (1) y = 9 - x ............(3) put in (2)
many share did he buy? x 2+ 3 x ( 9 - x ) + 2 ( 9 - x
)
2 =
0 x 2 + 27 x - 3 x 2 + 2 ( 81 - 18 x + x
2
)
=
0
x 2 - 3 x 2 + 27 x + 162 - 36 x + 2 x
2
=
0
x 2 - 3 x 2 + 2 x 2
+ 27 x - 36 x
+ 162 =
0 - 9 x+ 162 =
0
Sol: Let number of share = x The amount paid per share = y Total amount = 6000
xy = 6000 ..................(1) y = 6000 x ...............(2) According to condition - 9 x
= -
162
( x + 10 )( y - 20 ) = 6000 x=
18
xy - 20 x + 10 y - 200 = 6000 put (1) Put in eq (3) y = 9 - 18 = - 9 Solution set = { ( 18, -
9 ) } 6000 - 20 x + 10 y - 200 =
6000 - 20 x + 10 y - 200 = 0 put value of x
Q2ii). Solve 2 2
y - x
=4 2 x + xy + y
= 8
the system
- 20 x + 10 ⎛ │ ⎝ 6000 x
Sol: Given system y - x = 4 ...........(1)
2 x 2 + xy + y 2 = 8 ..........(2)
⎞ │ ⎠ - 200 = 0 multiply by x - 20 x 2 + 60000 - 200 x 0 divided by -20 x 2 + 10 x - 3000 = 0 From (1) y = 4 + x ........(3) Put in (2)
2( ) ( )2 2 2 2
( ) ( ) ( )( )
2 2 2
x 2 + 60 x - 50 x
- 3000 =
0
2 x + x 4 + x + 4 + x
=
8
x x + 60 - 50 x
+ 60 =
0
2 x + 4 x + x + 16 + 8 x + x
=
8 2 x + x + x + 4 x + 8 x
+ 16 - 8 =
0 4 x 2 + 12 x + 8 = 0 divided by 4
x - 50 x
+ 60 =
0 Either x- 50 = 0 or x+ 60 =
0 x = 50 x = - 60 is not admissible Thus number of share purchased is 50
38 Exercise 2.7
Exercise 2.7
Q1. Find two consecutive positive integers whose product is 72. Sol: Consider consecutive integers First positive integer = x Second positive integer = x + 1 Then according to the given condition
Either Or
We take only positive integer as given in question First integer = 8 Second integer = 8 + 1 = 9 Q2.
The sum of the square of three consecutive integer is 50. Find the integers. Sol: Consider consecutive integers First integer = x Second integer = x + 1 Third integer = x + 2 Then according to the given condition
Either Or
Take x = 3 Take x = - 5 1st integer = 3 1st integer = - 5 2nd integer = 3 + 1 = 4 2nd integer =- 5+1= - 4 3rd integer =3+2=5 3rd integer =-5+2= - 3 Q3. The length of prayer hall is 5 meter more than its width. If the area of the hall is 36 square meter. Find length and width of hall. Sol: Consider dimension of the hall Width of the hall = x Length of the hall = x + 5 Then ( according )
to condition Area = 36 m2
2
Chapter 2
between any two points should be positive Width of the hall = 4 meter Length of the hall = 4 + 5 = 9 meter Q4. The sum of two numbers is 11 and sum of their squares is 65. Find the numbers. Sol: Consider two numbers First number = x x ( x + 1 )
=
72
Second number = y
x 2+ x - 72 =
0
Sum x + y of =
11 two numbers ( 1 ) is 11
x 2+ 9x - 8x - 72 =
0
Sum of square of two numbers is 65 x ( x + 9 ) - 8 ( x + 9 )
=
0 ( x - 8 )( x + 9 )
=
0
From equation (1)
x - 8 =
0
x + 9 =
0 x =
8
x = -
9
Putting the value of y in equation (2)
x 2+ ( x + 1 ) 2 + ( x + 2 )
2 =
50
Either Or
x 2 + x 2 + 2.x.1 + 1 2 + x 2 + 2.x.2 + 2 2
- 50 =
0 x 2 + x 2 + x 2
+ 2x + 4x + 1 + 4 - 50 =
0
Putting the value of x in equation (3) 3x 2
+ 6x - 45 = 0 ÷
by 3 x 2+ 2x - 15 =
0 x 2+ 5x - 3x - 15 =
0 x ( x + 5 ) - 3 ( x + 5 )
=
0 ( x - 3 )( x + 5 )
=
0
When x = 4 When x = 7 1st number = 4 1st number = 7 2nd number = 7 2nd number = 4 Q5. The sum of square of two numbers is 100. One number is two more than the other.
x - 3 =
0
x + 5 =
0
Find the numbers. Sol: Consider two numbers according to x =
3
x = -
5
condition ( one number is 2 more than other ) First number = x Second number = x + 2 Sum of square of two numbers is 100
x x
+ 5 =
36
x + 5 x- 36 =
0 x 2 + 9 x - 4 x
- 36 =
0
Either Or
x ( x + 9 ) - 4 ( x
+ 9 )
=
0 ( x - 4 )( x
+ 9 )
=
0 Either x x When x = 6 When x = - 8
1st number = 6 1st number = -8 2nd number=6+2= 8 2nd number= -8+2= -6
- 4 =
0 =
4
Or
x + 9 =
0 x = -
9 We take only positive integer because distance
( x 2+ 11 - x )
2 =
65
x 2 + 11 2 - 2.11.x + x 2
- 65 =
0
x 2 + x 2
- 22x + 121 - 65 =
0
2x 2
- 22x + 56 = 0 ÷
by 2 x 2- 11x + 28 =
0 x 2- 7x - 4x + 28 =
0 x ( x - 7 ) - 4 ( x - 7 )
=
0 ( x - 4 )( x - 7 )
=
0
x - 4 =
0 x =
4
x - 7 =
0 x =
7
y = 11 -
4
y = 11 -
7 y =
7
y =
4
x 2+ ( x + 2 )
2 =
100
x 2 + x 2 + 2.x.2 + 2 2
- 100 =
0 2x 2+ 4x + 4 - 100
=
2x 2
+ 4x - 96 = 0 ÷
by 2 x 2+ 2x - 48 =
0 x 2+ 8x - 6x - 48 =
0 x ( x + 8 ) - 6 ( x + 8 )
=
0 ( x - 6 )( x + 8 )
=
0
x - 6 =
0 x =
6
x + 8 =
0 x = -
8
( ) 2 2 x y 65 2 + =
x + y =
11 y = 11 -
x 3
( )
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