-2) Verify whether the value in the brackets is the solution of the given equation.
x + 5 = 18 ( x = 1)
7x + 5 = 17 ( x = – 2)
7x + 5 = 19 ( x = 2)
4y – 3 = 11 (y = 1)
4y – 3 = 12 ( y = – 4)
4y – 3 = 14 ( y = 0)
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Answers
Answer:
) Here, x + 5 is H.S, 18 is R.H.S and x = 1 ( given data)
L.H.S = x + 5,
By putting, x = 1,
L.H.S = 1 + 5 = 6 \neq R.H.S
As, L.H.S \neq R.H.S , so x = 1 is not a solution.
2) Here, 7x + 5 is H.S, 17 is R.H.S and x = – 2 ( given data)
L.H.S = 7x + 5,
By putting, x = – 2,
L.H.S = 7(-2) + 5 = -9 \neq L.H.S
As, L.H.S \neq R.H.S , so x = – 2 is not a solution.
3) Here, 7x + 5 is H.S, 19 is R.H.S and x = 2 ( given data)
L.H.S = 7x + 5,
By putting, x = 2,
L.H.S = 7(2) + 5 = 19 = R.H.S
As, L.H.S = R.H.S, so x = 2 is a solution.
4) Here, 4y – 3 is H.S, 11 is R.H.S and y = 1 ( given data)
L.H.S = 4y – 3,
By putting, y = 1,
L.H.S = 4(1) – 3 = 1 \neq L.H.S
As, L.H.S \neq R.H.S, so y = 1 is not a solution.
5) Here, 4y – 3 is H.S, 12 is R.H.S and y = – 4 ( given data)
L.H.S = 4y – 3,
By putting, x = – 4,
L.H.S = 4(-4) – 3 = -19 \neq R.H.S
As, L.H.S \neq R.H.S, so y = – 4 is not a solution.
6) Here, 4y – 3 is H.S, 14 is R.H.S and y = 0 ( given data)
L.H.S = 4y – 3,
By putting, x = 0,
L.H.S = 4(0) – 3 = -3 \neq R.H.S
As, L.H.S \neq R.H.S, so y = 0 is not a solution.
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Here, x + 5 is H.S, 18 is R.H.S and x = 1 ( given data)
L.H.S = x + 5,
By putting, x = 1,
L.H.S = 1 + 5 = 6 \neq R.H.S
As, L.H.S \neq R.H.S , so x = 1 is not a solution.
2) Here, 7x + 5 is H.S, 17 is R.H.S and x = – 2 ( given data)
L.H.S = 7x + 5,
By putting, x = – 2,
L.H.S = 7(-2) + 5 = -9 \neq L.H.S
As, L.H.S \neq R.H.S , so x = – 2 is not a solution.
3) Here, 7x + 5 is H.S, 19 is R.H.S and x = 2 ( given data)
L.H.S = 7x + 5,
By putting, x = 2,
L.H.S = 7(2) + 5 = 19 = R.H.S
As, L.H.S = R.H.S, so x = 2 is a solution.
4) Here, 4y – 3 is H.S, 11 is R.H.S and y = 1 ( given data)
L.H.S = 4y – 3,
By putting, y = 1,
L.H.S = 4(1) – 3 = 1 \neq L.H.S
As, L.H.S \neq R.H.S, so y = 1 is not a solution.
5) Here, 4y – 3 is H.S, 12 is R.H.S and y = – 4 ( given data)
L.H.S = 4y – 3,
By putting, x = – 4,
L.H.S = 4(-4) – 3 = -19 \neq R.H.S
As, L.H.S \neq R.H.S, so y = – 4 is not a solution.
6) Here, 4y – 3 is H.S, 14 is R.H.S and y = 0 ( given data)
L.H.S = 4y – 3,
By putting, x = 0,
L.H.S = 4(0) – 3 = -3 \neq R.H.S
As, L.H.S \neq R.H.S, so y = 0 is not a solution.