# In

(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 – 7 = -4
and R.H.S = 3(3) + 8 = 9 + 8 = 17
... L.H.S ≠ R.H.S.
Hence, 3 is not a solution of y – 7 = 3y + 8

(iii) If we substitute u = 5, we get
L.H.S = 3u + 2 = 3(5) + 2 = 15 + 2 = 17
and R.H.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

(iv) If we substitute x = √2, we get
L.H.S = 2 x – 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

(v) If we substitute x = 3, we get
L.H.S = (5/2)x + 3 = (5/2)(3) + 3 = 15/2 + 3 = (15 + 6)/2 = 21/2
and R.H.S = 21/2
... L.H.S = R.H.S.
Hence, 3 is a solution of (5/2)x + 3 = 21/2.

(vi) If we substitute u = -1, we get
L.H.S = 24 – 3{(-1) – 2}= 24 – 3(-1 –2) = 24 – (-9) = 24 + 9 = 33
and R.H.S = u + 8 = -1 + 8 = 7
... L.H.S ≠ R.H.S.
Hence, -1 is not a solution of 24 – 3(u – 2) = u + 8.

(vii) If we substitute x = 0, we get
L.H.S = (x – 2) + (x + 3) = (0 – 2) + (0 + 3) = -2 + 3 = 1
and R.H.S = x + 8 = 0 + 8 = 8
... L.H.S ≠ R.H.S.
Hence, 0 is not a solution of (x – 2) + (x + 3) = x + 8.

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