maths1 practice set 1.2 10 std
Answers
Answer:
Complete the following table to draw graph of the equations
(I) x + y = 3
(II) x – y = 4
Solution:
(I) Given
x + y = 3 …. (i)
(i) Put value x=3 in equation (i)
We get, y = 3 – 3
⇒ y = 0
ii. Put value y = 5 in equation (i)
We get, x = 3 – 5
⇒ x = -2
iii. Put value y = 3 in equation (i)
We get, x = 3 – 3
⇒ x = 0
Now the table becomes,
x 3 -2 0
y 0 5 3
(x, y) (3, 0) (-2, 5) (0, 3)
(2) Given
x – y = 4 ……. (ii)
i. Put value y = 0 in equation (ii)
we get, x = 4 – 0
⇒ x = 4
ii. Put value x = –1 in equation (ii)
we get, – y = 5
iii. Put value y = –4 in equation (ii)
we get, x = 4 + 4
⇒ x = 8
x 4 -1 0
y 0 -5 -4
(x, y) (4, 0) (-1, -5) (0, -4)
2. Solve the following simultaneous equations graphically.
(1) x + y = 6; x – y = 4
Solution:
Given x + y = 6 …. (i)
x 0 6 5
y 6 0 1
(x, y) (0, 6) (6, 0) (5, 1)
x – y = 4 …… (ii)
x 0 2 5
y -4 -2 1
(x, y) (0, -4) (2, -2) (5, 1)
Calculating intersecting point
x + y = 6
x – y = 4
2x = 10
x = 10/2
x = 5
Putting x= 5 in equation (i)
5 + y = 6
y = 6 – 5
y = 1
(2) x + y = 5; x – y = 3
Solution:
x + y = 5 …. (i)
x 0 2 4
y 5 3 1
(x, y) (0, 5) (2, 3) (4, 1)
x – y = 3 …… (ii)
x 0 2 4
y -3 -1 1
(x, y) (0, -3) (2, -1) (4, 1)
Calculating intersecting point
x + y = 5
x – y = 3
which implies
2x = 8
x = 8/2
x = 4
Putting x= 4 in equation (i)
4 + y = 5
y = 5 – 4
y = 1
Intersection Point (4,1)
(3) x + y = 0; 2x – y = 9
Solution:
x + y = 0 …. (i)
x 1 3 5
y -1 -3 -5
(x, y) (1, -1) (3, -3) (5, -5)
2x – y = 9 …. (ii)
x 2 3 4
y -5 -3 -1
(x, y) (2, -5) (3, -3) (4, -1)
Calculating intersecting point
x + y = 0
2x – y = 9
3x = 9
x = 9/3
x = 3
Putting x= 3 in equation (i)
3 + y = 0
y = 0 – 3
y = -3
Intersection point (3, –3)
(4) 3x – y = 2; 2x – y = 3
Solution:
3x – y = 2 …… (i)
x 0 1 -1
Y -2 1 -5
(x, y) (0, -2) (1, 1) (-1, -5)
2x – y = 3 …… (ii)
x 3 2 -1
y 3 1 -5
(x, y) (3, 3) (2, 1) (-1, -5)
Calculating intersecting point
3x – y = 2
– 2x + y = -3
x = -1
Putting x= –1 in equation (i)
3× –1 – y = 2
– 3 – y = 2
– y = 2 + 3
y = -5
Intersection point (–1, –5)
(5) 3x – 4y = –7; 5x – 2y = 0
Solution:
3x – 4y = 7 …. (i)
When x = 0, 4y = 7, y = 7/4
When y = 0, 3x = -7, x = -7/3
5x – 2y = 0 …… (ii)
When x = 0, y = 0
When x = 1, y = 5/2
Plotting both the graphs we get,
Calculating intersecting point
3x – 4y = -7
5x – 2y = 0
x = -1
Putting x= –1 in equation (i)
3 × –1 – y = 2
– 3 – y = 2
– y = 2 + 3
y = – 5
Intersection point (–1, –5)
hope it helps..
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