(v) (1,1) Given equation: x - 2y = 4 In x - 2y = 4, putting x = 1 and y = 1, we have, 1 - 2 × 1 = -1 ≠ 4 Hence, (1, 1) is not a solution of the equation.
(v) (1,1)
Given equation: x – 2y = 4
In x – 2y = 4, putting x = 1 and y = 1, we have, 1 – 2 × 1 = -1 ≠ 4
Hence, (1, 1) is not a solution of the equation.
(i) x + y = 4 ⇒ y = 4 - x Putting x = 0, we have, y = 4 - 0 = 4 Putting x = 1, we have, y = 4 - 1 = 3 Hence, A(0, 4) and B(1, 3) are the solutions of the equation.
(i) x + y = 4
⇒ y = 4 – x
Putting x = 0, we have, y = 4 – 0 = 4
Putting x = 1, we have, y = 4 – 1 = 3
Hence, A(0, 4) and B(1, 3) are the solutions of the equation.
(ii) x - y = 2 ⇒ y = x - 2 Putting x = 0, we have, y = 0 - 2 = -2 Putting x = 1, we have, y = 1 - 2 = -1 Hence, C(O,-2) and D(1,-1) are the solutions of the equation.
(ii) x – y = 2
⇒ y = x – 2
Putting x = 0, we have, y = 0 – 2 = -2
Putting x = 1, we have, y = 1 – 2 = -1
Hence, C(O,-2) and D(1,-1) are the solutions of the equation.
Check which of the following are solutions of the equation x – 2y = 4 and which are not: ( √2, 4√2)
(v) (1,1) Given equation: x - 2y = 4 In x - 2y = 4, putting x = 1 and y = 1, we have, 1 - 2 × 1 = -1 ≠ 4 Hence, (1, 1) is not a solution of the equation.
(v) (1,1)
See lessGiven equation: x – 2y = 4
In x – 2y = 4, putting x = 1 and y = 1, we have, 1 – 2 × 1 = -1 ≠ 4
Hence, (1, 1) is not a solution of the equation.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Given equation: x = 2, y = 1 In 2x + 3y = k, putting x = 2 and y = 1, we have, 2x × 2 + 3 × 1 = k ⇒ k = 7 Hence, the value of k is 7.
Given equation: x = 2, y = 1
See lessIn 2x + 3y = k, putting x = 2 and y = 1, we have,
2x × 2 + 3 × 1 = k
⇒ k = 7
Hence, the value of k is 7.
Draw the graph of each of the following linear equations in two variables: x + y = 4
(i) x + y = 4 ⇒ y = 4 - x Putting x = 0, we have, y = 4 - 0 = 4 Putting x = 1, we have, y = 4 - 1 = 3 Hence, A(0, 4) and B(1, 3) are the solutions of the equation.
(i) x + y = 4
See less⇒ y = 4 – x
Putting x = 0, we have, y = 4 – 0 = 4
Putting x = 1, we have, y = 4 – 1 = 3
Hence, A(0, 4) and B(1, 3) are the solutions of the equation.
Draw the graph of each of the following linear equations in two variables: x – y = 2.
(ii) x - y = 2 ⇒ y = x - 2 Putting x = 0, we have, y = 0 - 2 = -2 Putting x = 1, we have, y = 1 - 2 = -1 Hence, C(O,-2) and D(1,-1) are the solutions of the equation.
(ii) x – y = 2
See less⇒ y = x – 2
Putting x = 0, we have, y = 0 – 2 = -2
Putting x = 1, we have, y = 1 – 2 = -1
Hence, C(O,-2) and D(1,-1) are the solutions of the equation.
Draw the graph of each of the following linear equations in two variables: y = 3x
(iii) y = 3x Putting x = 0, we have, y = 3 × 0 = 0 Putting x = 1, we have, y = 3 × 1 = 3 Hence, E(0, 0) and B(1,3) are the solutions of the equation.
(iii) y = 3x
See lessPutting x = 0, we have, y = 3 × 0 = 0
Putting x = 1, we have, y = 3 × 1 = 3
Hence, E(0, 0) and B(1,3) are the solutions of the equation.