Problem based on 'Linear equation in two variable'
1. Find the given co-ordinates is a solution of equation (x –
2y) = 0 ?
a). (0,1)
b). (0,-1)
c). (2,1)
d). (-2,-1)
e). (4,2)
f). (1,2)
Sol. 1
a). (0,1)
(x – 2y) = 0
Putting x = 0 and y = 1
0 – 2 * 1 # 0
-2 # 0
L.H.S. is not equal to R.H.S.
So (0,1) is not a solution of this equation.
b). (0,-1)
(x – 2y) = 0
Putting x = 0 and y = -1
0 – 2 * (-1) # 0
2 # 0
L.H.S. is not equal to R.H.S.
So (0,-1) is not a solution of this equation.
c). (2,1)
(x – 2y) = 0
Putting x = 2 and y = 1
2 – 2 * 1 = 0
0 = 0
L.H.S. is equal to R.H.S.
So (2,1) is a solution of this equation.
d). (-2,-1)
(x – 2y) = 0
Putting x = -2 and y = -1
-2 – 2 * -1 = 0 => -2 + 2 = 0
0 = 0
L.H.S. is equal to R.H.S.
So (-2,-1) is a solution of this equation.
e). (4,2)
(x – 2y) = 0
Putting x = 4 and y = 2
4 – 2 * 2 = 0 => 4 – 4 = 0
0 = 0
L.H.S. is equal to R.H.S.
So (4,2) is a solution of this equation.
f). (1,2)
(x – 2y) = 0
Putting x = 1 and y = 2
1 – 2 * 2 # 0 => 1 – 4 # 0
-3 # 0
L.H.S. is not equal to R.H.S.
So (1,2) is not a solution of this equation.
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| maths ncert solutions |
2. Find the value of K for given equation 2 x + 2 y = K for
co-ordinates are
a). (0,1)
b). (0,-1)
c). (2,1)
d). (-2,-1)
e). (4,2)
f). (1,2)
a). Given 2 x + 2 y = K
Put x = 0 y = 1
2 * 0 + 2 * 1 = K
K = 0 + 2
K = 2
b). Given 2 x + 2 y = K
Put x = 0 y = -1
2 * 0 + 2 * (-1) = K
K = 0 - 2
K = -2
c). Given 2 x + 2 y = K
Put x = 2 y = 1
2 * 2 + 2 * 1 = K
K = 4 + 2
K = 6
d). Given 2 x + 2 y = K
Put x = -2 y = -1
2 * (-2) + 2 * (-1) = K
K = -4 - 2
K = -6
e). Given 2 x + 2 y = K
Put x = 4 y = 2
2 * 4 + 2 * 2 = K
K = 8 + 4
K = 12
f). Given 2 x + 2 y = K
Put x = 1 y = 2
2 * 1 + 2 * 2 = K
K = 2 + 4
K = 6
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