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)
(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)
Solution and Explanation
(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 x (-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 x (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 x 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 x (-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 x 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.
Top CBSE Class VII Mathematics Questions
- Find the area of each of the following parallelograms:
- Find the area of each of the following triangles:
- Find the missing values:
So No Base Height Area of parallelogram a. 20 cm - 246 \(cm^2\) b. - 15 cm 154.5 \(cm^2\) c. - 8.4 cm 48.72 \(cm^2\) d. 15.6 cm - 16.38 \(cm^2\) - Find the missing values:
Base Height Area of triangle 15 cm - 87 \(cm^2\) - 31.4 mm 1256 \(mm^2\) 22 cm - 170.5 \(cm^2\) - PQRS is a parallelogram (see the given figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = \(12\) cm and QM = \(7.6\) cm. Find:
- the area of the parallelogram PQRS
- QN, if PS = 8 cm
Top CBSE Class VII A Mind-Reading Game Questions
- Complete the last column of the table
S. No. Equation Value Say, whether the Equation No. 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) \(\frac{m}{3}=2\) m = – 6 (x) \(\frac{m}{3}=2\) m = 0 (xi) \(\frac{m}{3}=2\) m = 6 Solve the following equations by trial and error method: (i) 5p + 2 = 17 (ii) 3m – 14 = 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.
(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- Write the following equations in statement forms:
(i)p+4=15
(ii)m-7=3
(iii)2m=7
(iv)\(\frac{m}{5}\)=3
(v)\(\frac{3m}{5}\)=6
(vi)3p+4=25
(vii)4p-2=18
(viii)\(\frac{p}{2}\)+2=8 Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
Top CBSE Class VII Questions
- Find the area of each of the following parallelograms:
- Find the area of each of the following triangles:
- Find the missing values:
So No Base Height Area of parallelogram a. 20 cm - 246 \(cm^2\) b. - 15 cm 154.5 \(cm^2\) c. - 8.4 cm 48.72 \(cm^2\) d. 15.6 cm - 16.38 \(cm^2\) - Find the missing values:
Base Height Area of triangle 15 cm - 87 \(cm^2\) - 31.4 mm 1256 \(mm^2\) 22 cm - 170.5 \(cm^2\) - PQRS is a parallelogram (see the given figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = \(12\) cm and QM = \(7.6\) cm. Find:
- the area of the parallelogram PQRS
- QN, if PS = 8 cm