Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th
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Given, x = 2 and y = 1 (i) Given, linear equation is 2x + 5y = 9. On putting x = 2 and y= 1 in LHS, we get  LHS = 2x + 5y =2(2) + 5(1) = 4 + 5 = 9 = RHS  So, x = 2 and y=1 is a solution of given equation. (ii) Given, linear equation is  5 / 2 x + 3y = 14. On putting the values of x and y in LHS, we get LHS = 5 / 2 x + 3 y = 5 / 2 (2) + 3 (1) = 5 + 3 = 8 ≠ 14  ⇒ LHS  ≠  RHS  So, x = 2 and y = 1 is not a solution of given equation.
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Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

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Last Answer : Given, x = 2 and y = 1 (i) Given, linear equation is 2x + 5y = 9. On putting x = 2 and y= 1 in LHS, we get LHS = 2x + 5y =2(2) + 5(1) = 4 + 5 = 9 = RHS So, x = 2 and y=1 is a solution of given ... + 3 (1) = 5 + 3 = 8 ≠ 14 ⇒ LHS ≠ RHS So, x = 2 and y = 1 is not a solution of given equation.

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Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

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Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

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Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

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Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

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