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प्रश्न
Solve the following equation by the trial-and-error method: z − 3 = 2z − 5
उत्तर
z − 3 = 2z − 5
We try several values of z till we get the L.H.S. equal to the R.H.S.
| z | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |
| 1 | 1 − 3 = −2 | 2 ×× 1 − 5 = −3 | No |
| 2 | 2 − 3 = −1 | 2 ×× 2 − 5 = −1 | Yes |
∴ z = 2
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Complete the entries in the third column of the table.
|
S.No. |
Equation |
Value of variable |
Equation satisfied Yes/No |
|
(a) |
10y = 80 |
y = 10 |
______ |
|
(b) |
10y = 80 |
y = 8 |
______ |
|
(c) |
10y = 80 |
y = 5 |
______ |
|
(d) |
4l = 20 |
l = 20 |
______ |
|
(e) |
4l = 20 |
l = 80 |
______ |
|
(f) |
4l = 20 |
l = 5 |
______ |
|
(g) |
b + 5 = 9 |
b = 5 |
______ |
|
(h) |
b + 5 = 9 |
b = 9 |
______ |
|
(i) |
b + 5 = 9 |
b = 4 |
______ |
|
(j) |
h − 8 = 5 |
h = 13 |
______ |
|
(k) |
h − 8 = 5 |
h = 8 |
______ |
|
(l) |
h − 8 = 5 |
h = 0 |
______ |
|
(m) |
p + 3 = 1 |
p = 3 |
______ |
|
(n) |
p + 3 = 1 |
p = 1 |
______ |
|
(o) |
p + 3 = 1 |
p = 0 |
______ |
|
(p) |
p + 3 = 1 |
p = − 1 |
______ |
|
(q) |
p + 3 = 1 |
p = − 2 |
______ |
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