Practice Set 1.3
Q 1 Fill in the blanks with correct number
Q 2 Find the value of following determinants .
Sunday, September 12, 2021
Sunday, September 5, 2021
Maths 1 Practice Set 1.2
Practice Set 1.2
x
3
2
0
y
0
5
3
(x,y)
(3,0)
(-2,5)
(0,3)
2] x - y = 4
x
4
-1
0
y
0
-5
-4
(x,y)
(4,0)
(-1,-5)
(0,-4)
1]Q 2 Solve the following equations graphically.
1] x + y = 6 ; x - y = 4
x + y = 6
.`. y = 6 - x
x
0
6
3
4
y
6
0
3
2
(x,y)
(0,6)
(6,0)
(3,3)
(4,3)
x - y = 4
.`. y = x - 4
x
0
4
2
5
y
-4
0
-2
1
(x,y)
(0,-4)
(4,0)
(2,-2)
(5,1)
The two lines intersect at point (5,1).
.`. x = 5 and y = 1 is the
solution
of the equation
x + y = 6 and x - y = 4 .
2] x + y = 5 ; x - y = 3
x + y = 5
.`. y = 5 - x
x
0
5
2
3
y
5
0
3
2
(x,y)
(0,5)
(5,0)
(2,3)
(3,2)
x - y = 3
.`. y = x - 3
x
0
3
2
1
y
-3
0
-1
-2
(x,y)
(0,-3)
(3,0)
(2,-1)
(1,-2)
The two lines intersect at point (4,1).
.`. x = 4 and y = 1 is the solution of
simultaneous equation
x + y = 5 and x - y = 3 .
3] x + y = 0 ; 2x - y = 9
x + y = 0
.`. y = -x
x
0
2
-2
-1
y
0
-2
2
1
(x,y)
(0,0)
(2,-2)
(-2,2)
(-1,1)
2x - y = 9
.`. y = 2x - 9
x
0
2
5
4
y
-9
-5
1
-1
(x,y)
(0,-9)
(2,-5)
(5,1)
(4,-1)
The two lines intersect at point (3,-3)
.`. x = 3 and y = -3 is the solution of
the simultaneous equation
x + y = 0 and 2x - y = 9 .
4] 3x - y = 2 ; 2x - y = 3
3x - y = 2
.`. y = 3x - 2
x
0
1
-1
2
y
-2
1
-5
4
(x,y)
(0,-2)
(1,1)
(-1,-5)
(2,4)
2x - y = 3
.`. y = 2x - 3
x
0
1
2
3
y
-3
-1
1
3
(x,y)
(0,-3)
(1,-1)
(2,1)
(3,3)
The two lines intersect at point (-1,-5)
.`. x = -1 and y = -5 is the solution of
the simultaneous equation
3x - y = 2 and 2x - y = 3 .
5] 3x - 4y = -7 ; 5x -2y = 7
3x - 4y = -7
.`. 4y = 3x + 7
.`. y = 3x + 7
4
x
-1
-5
3
5
y
1
-2
4
5.5
(x,y)
(-1,1)
(-5,-2)
(3,4)
(5,5.5)
5x - 2y = 0
.`. 2y = 5x
.`. y = 5 x
2
x
0
2
-2
1
y
0
5
-5
2.5
(x,y)
(0,0)
(2,5)
(-2,-5)
(1,2.5)
The two lines intersect at point (1,2.5)
.`. x = 1 and y = 2.5 is the solution of
the simultaneous equation
3x - 4y = -7 and 5x -2y = 7 .
6] 2x - 3y = 4 ; 3y - x = 4
2x - 3y = 4
.`. 3y = 2x - 4
.`. y = 2x - 4
3
x
2
-1
5
8
y
0
-2
2
4
(x,y)
(2,0)
(-1,-2)
(5,2)
(8,4)
3y - x = 4
.`. 3y = x + 4
.`. y = x + 4
3
x
2
-4
5
-1
y
2
0
3
1
(x,y)
(2,2)
(-4,0)
(5,3)
(-1,1)
The two lines intersect at point (8.4)
.`. x = 8 and y = 4 is the solution of
the simultaneous equation
2x - 3y = 4 and 3y - x = 4 .
Thursday, September 2, 2021
Maths 1 practice set 1.1
* Linear Equation in two variables *
_____________(Practice Set 1.1)________________________________
Q 1 complete the following activity to solve the simultaneous equation.
5x + 3y = 9 ___(i)
2x - 3y = 12 ___(ii)
Let's add equation (i) and (ii) .
5x + 3y = 9
+ 2x - 3y = 12
________________________
7 x = 21
x = 21
7
x = 3
Place x = 3 in equation ---(i) , we get ;
5 (3) + 3y = 9
15 + 3y = 9
3y = 9 - 15
y = -6
3
y = -2
. ` . Solution is (x , y) = (3 , -2)
Q 2 Solve the following simultenious equation .
1] 3a + 5b = 26 ; a + 5b = 22
---:-
3a + 5b = 26 -----(i)
a + 5b = 22 -----(ii)
Let's subract equation --(i) and --(ii) , we get ;
3a + 5b = 26
a + 5b = 22
(-) (-) (-)
2a = 4
a = 4
2
[a = 2]
Substituting a = 2 in equation ---(ii) , we get ;
2 + 5b = 22
5b = 22 - 2 = 20
b = 20
5
[b = 4]
. ` . (a , b) = (2 , 4) is the solution .
2] x + 7y = 10 ; 3x - 2y = 7
---:-
x + 7y = 10 ----(i)
3x - 2y = 7 ----(ii)
Multiplying eq ---(i) by 3 , we get ;
3x + 22y = 30 ----(iii)
Subracting eq ---(ii) from ---(iii) , we get ;
3x + 21y = 30
3x - 2y = 7
(-) (+) (-)
23y = 23
y = 23
23
[y = 1]
Substituting y=1 in equation --(i) , we get ;
x + 7(1) = 10
x + 7 = 10
x = 10 - 7
[x = 3]
. ` . (x , y) = (3, 1) is the Solution .
3] 2x - 3y = 9 ; 2x + y = 13
---:-
2x - 3y = 9 ---(i)
2x + y = 13 ---(ii)
Subracting eq ---(i) from ---(ii) , we get ;
2x - 3y = 9
2x + y = 13
(-) (+) (-)
4y = 4
y = 4
4
[y = 1]
Substituting y=1 in equation --(ii) , we get ;
2x + 1 = 13
2x = 13 - 1
x = 12
2
[ x = 6]
. ` . (x , y) = (6 , 1) is the Solution .
4] 5m - 3n = 19 ; m - 6n = -7
---:-
5m - 3n = 19 ---(i)
m - 6n = -7 ---(ii)
Multiplying equation --(ii) by 5 , we get ;
5m - 3n = -35 ----(iii)
Subracting equation ---(i) from --(iii) ,we get ;
5m - 3n = -35
5m - 3n = 19
(-) (+) (-)
- 27n = -54
n = -54
-27
[n = 2]
Substituting n = 2 in equation ---(i) , we get ;
5m - 3(2) = 19
5m - 6 = 19
5m = 19 + 6
5m = 25
m = 25
5
[m = 5]
. ` . (m , n) = (5 , 2) is the Solution .
5] 5x + 2y = -3 ; x + 5y = 4
---:-
5x + 2y = -3 ---(i)
x + 5y = 4 ---(ii)
Multiplying equation ---(ii) by 5 , we get ;
5x + 25y = 20
5x + 2y = -3
(-) (-) (+)
23y = 23
y = 23
23
[y = 1]
Substituting y = 1 in eq ---(ii) , we get ;
x + 5(1) = 4
x + 5 = 4
x = 4 - 5
[x = -1]
. ` . (x , y) = (-1 , 1) is the Solution .
6] 1 x + y = 10 ; 2x + 1 y = 11
3 3 4 4
---:-
1 x + y = 10
3 3
Multiply by 3 , we get ;
3 * 1 x + 3 * y = 10 * 3
3 3
x + 3y = 10 ---(i)
Mutiply by 4 , we get ;
4 * 2x + 4 * 1 y = 11 * 4 ----(ii)
4 4
8x + y = 11 ---(ii)
Multiplying equation --(i) by 8 , we get ;
8x + 24y = 80 ---(iii)
Subracting equation ---(iii) by ---(ii) , we get ;
8x + 24y = 80
8x + y = 11
(-) (-) (-)
23y = 69
y = 69
23
[y = 3]
Substituting y = 3 in equation ---(i) , we get ;
x + 3(3) = 10
x + 9 = 10
x = 10 - 9
[x = 1]
. ` . (x , y) = (1 , 3) is the Solution .
7] 99x + 101y = 499 ; 101x + 99y = 501
---:-
99x + 101y = 499 ----(i)
101x + 99y = 501 ----(ii)
Adding equation ---(i) and ---(ii) , we get ;
99x + 101y = 499
101x + 99y = 501
200x + 200y = 1000
200 (x + y) = 1000
(x + y) = 1000
200
x + y = 5 ----(iii)
Subracting equation ---(i) from ---(ii) , we get ;
99x + 101y = 499
101x + 99y = 501
(-) (-) (-)
2x - 2y = 2
2 (x - y) = 2
x - y = 1 ----(iv)
Adding equation ---(iii) and ---(iv) , we get ;
x + y = 5
x - y = 1
2x = 6
x = 6
2
[x = 3]
Substituting x = 3 in equation ---(iii) , we get ;
3 + y = 5
y = 5 - 3
[ y = 2 ]
. ` . (x , y) = (3 , 2) is the Solution
8] 49x - 57y = 172 ; 57x - 49y = 252
---:-
49x - 57y = 172 ----(i)
57x - 49y = 252 ----(ii)
Adding equation ---(i) and ---(ii) , we get ;
49x - 57y = 172
57x - 49y = 252
106x - 106y = 424
106 (x - y) = 424
x - y = 424
106
x - y = 4 ----(iii)
Subracting equation ---(i) and ---(ii)
49x - 57y = 172
57x - 49y = 252
(-) (+) (-)
8x + 8x = 80
8 (x + y) = 80
x + y = 80
8
x + y = 10 ----(iv)
Adding equation ---(iii) and ---(iv) , we get ;
x - y = 4
x + y = 10
2x = 14
x = 14
2
| x = 7 |
Substituting x = 7 in equation ---(iv) , we get ;
7 + y = 10
y = 10 - 7
| y = 3 |
. ` . (x , y) = (7 , 3) is the Solution .
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Maths 1 Practice Set 1.3
Practice Set 1.3 Q 1 Fill in the blanks with correct number Q 2 Find the value of following determinants ....
