Solve the following equations by trial and error method: (i) 5p + 2 = 17 (ii) 3m – 14 = 4
Solve the following equations by trial and error method: (i) 5p + 2 = 17 (ii) 3m – 14 = 4
Solution and Explanation
(i)5p+2=17.
L.H.S=5p+2
By putting, p=0,
we get
5(0)+2=2≠17
By putting, p=1,
we get
5(1)+2=7≠17
By putting,p=2,
we get
5(2)+2=12 L.H.S≠ R.H.S
By putting, p=3,
we get
5(3)+2=17
L.H.S = R.H.S
So, p = 3is a solution of the equation.
ii) 3m–14=4
3m–14= L.H.S
By putting,m=5,
we get
3(5)–14=1≠6
By putting, m=6,
we get
3(6)–14=4
L.H.S= R.H.S. Therefore, m = 6 is a solution of the equation.
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 - 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) 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