EVENTS CONVENT HIGH SCHOOL
12/01/2021 CLASS-7 SLOT-2
MATHS
Chapter-4 SIMPLE EQUATIONS
_________________________________________________________________
Question 1:Complete
the last column of the table.
|
S.
No. |
Equation |
Value |
Say,
whether the equation is satisfied. (Yes/No) |
|
(i) |
x + 3 = 0 |
x = 3 |
– |
|
(ii) |
x + 3 = 0 |
x = 0 |
– |
|
(iii) |
x + 3 = 0 |
x = − 3 |
– |
|
(iv) |
x − 7 = 1 |
x = 7 |
– |
|
(v) |
x − 7 = 1 |
x = 8 |
– |
|
(vi) |
5x =
25 |
x = 0 |
– |
|
(vii) |
5x =
25 |
x = 5 |
– |
|
(viii) |
5x =
25 |
x = − 5 |
– |
|
(ix) |
|
m = − 6 |
– |
|
(x) |
|
m = 0 |
– |
|
(xi) |
|
m = 6 |
– |
Answer:
(i) x + 3 = 0
L.H.S. = x + 3
By putting x = 3,
L.H.S. = 3 + 3 = 6 ≠ R.H.S.
∴ No, the equation is not satisfied.
(ii) x + 3 = 0
L.H.S. = x + 3
By putting x = 0,
L.H.S. = 0 + 3 = 3 ≠ R.H.S.
∴ No, the equation is not satisfied.
(iii) x + 3 = 0
L.H.S. = x + 3
By putting x = −3,
L.H.S. = − 3 + 3 = 0 = R.H.S.
∴ Yes, the equation is satisfied.
(iv) x − 7 = 1
L.H.S. = x − 7
By putting x = 7,
L.H.S. = 7 − 7 = 0 ≠ R.H.S.
∴ No, the equation is not satisfied.
(v) x − 7 = 1
L.H.S. = x − 7
By putting x = 8,
L.H.S. = 8 − 7 = 1 = R.H.S.
∴ Yes, the equation is satisfied.
(vi) 5x = 25
L.H.S. = 5x
By putting x = 0,
L.H.S. = 5 × 0 = 0 ≠ R.H.S.
∴ No, the equation is not satisfied.
(vii) 5x = 25
L.H.S. = 5x
By putting x = 5,
L.H.S. = 5 × 5 = 25 = R.H.S.
∴ Yes, the equation is satisfied.
(viii) 5x = 25
L.H.S. = 5x
By putting x = −5,
L.H.S. = 5 × (−5) = −25 ≠ R.H.S.
∴ No, the equation is not satisfied.
(ix)
L.H.S. =
By putting m = −6,
L. H. S. =
∴No, the equation is not satisfied.
(x)
L.H.S. =
By putting m = 0,
L.H.S. =
∴No, the equation is not satisfied.
(xi)
L.H.S. =
By putting m = 6,
L.H.S. =
∴ Yes, the equation is satisfied.
Question 2:Check
whether the value given in the brackets is a solution to the given equation or
not:
(a) n + 5 = 19 (n =
1) (b) 7n + 5 = 19 (n =
− 2)
(c) 7n + 5 = 19 (n =
2) (d) 4p − 3 = 13 (p =
1)
Answer:(a) n + 5 = 19 (n = 1)
Putting n = 1 in L.H.S.,
n + 5 =
1 + 5 = 6 ≠ 19
As L.H.S. ≠ R.H.S.,
Therefore, n = 1 is not a
solution of the given equation, n + 5 = 19.
(b) 7n + 5 = 19 (n = −2)
Putting n = −2 in L.H.S.,
7n + 5 = 7 × (−2) + 5 = −14 + 5
= −9 ≠ 19
As L.H.S. ≠ R.H.S.,
Therefore, n = −2 is not a
solution of the given equation, 7n + 5 = 19.
(c) 7n + 5 = 19 (n = 2)
Putting n = 2 in L.H.S.,
7n + 5 = 7 × (2) + 5 = 14 + 5 =
19 = R.H.S.
As L.H.S. = R.H.S.,
Therefore, n = 2 is a
solution of the given equation, 7n + 5 = 19.
(d) 4p − 3 = 13 (p = 1)
Putting p = 1 in L.H.S.,
4p − 3 = (4 × 1) − 3 = 1 ≠ 13
As L.H.S ≠ R.H.S.,
Therefore, p = 1 is not a
solution of the given equation, 4p − 3 = 13.
Question 3:Solve
the following equations by trial and error method:
(i) 5p + 2 = 17
Answer:
(i) 5p + 2 = 17
Putting p = 1 in L.H.S.,
(5 × 1) + 2 = 7 ≠ R.H.S.
Putting p = 2 in L.H.S.,
(5 × 2) + 2 = 10 + 2 = 12 ≠ R.H.S.
Putting p = 3 in L.H.S.,
(5 × 3) + 2 = 17 = R.H.S.
Hence, p = 3 is a solution
of the given equation.
Question 4:Write
equations for the following statements:
(i) The sum of numbers x and
4 is 9.
(ii) 2 subtracted from y is
8.
(iii) Ten times a is 70.
Answer:(i) x + 4 = 9
(ii) y − 2 = 8
(iii) 10a = 70
Question 5:Write
the following equations in statement forms:
(i) m − 7 = 3 (ii) 2m = 7
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Thank your for your valuable responce.
Mrfarooqui