Sponsor Area
Simple Equations
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 |
– |
(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) = 2
L.H.S. =
By putting m = −6,
L. H. S. = ≠ R.H.S.
∴No, the equation is not satisfied.
(x) = 2
L.H.S. =
By putting m = 0,
L.H.S. = ≠ R.H.S.
∴No, the equation is not satisfied.
(xi) = 2
L.H.S. =
By putting m = 6,
L.H.S. = = R.H.S.
∴ Yes, the equation is satisfied.
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)
(e) 4p − 3 = 13 (p = − 4) (f) 4p − 3 = 13 (p = 0)
(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.
(e) 4p − 3 = 13 (p = −4)
Putting p = −4 in L.H.S.,
4p − 3 = 4 × (−4) − 3 = − 16 − 3 = −19 ≠ 13
As L.H.S. ≠ R.H.S.,
Therefore, p = −4 is not a solution of the given equation, 4p − 3 = 13.
(f) 4p − 3 = 13 (p = 0)
Putting p = 0 in L.H.S.,
4p − 3 = (4 × 0) − 3 = −3 ≠ 13
As L.H.S. ≠ R.H.S.,
Therefore, p = 0 is not a solution of the given equation, 4p − 3 = 13.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m − 14 = 4
(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.
(ii) 3m − 14 = 4
Putting m = 4,
(3 × 4) − 14 = −2 ≠ R.H.S.
Putting m = 5,
(3 × 5) − 14 = 1 ≠ R.H.S.
Putting m = 6,
(3 × 6) − 14 = 18 − 14 = 4 = R.H.S.
Hence, m = 6 is a solution of the given equation.
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.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
(i) x + 4 = 9
(ii) y − 2 = 8
(iii) 10a = 70
(iv)
(v)
(vi) Seven times of m is 7m.
7m + 7 = 77
(vii) One-fourth of a number x is .
(viii) Six times of y is 6y.
6y − 6 = 60
(ix) One-third of z is .
Sponsor Area
Mock Test Series
Mock Test Series