- 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/3) = 2 | m = – 6 | |
| (x) | (m/3) = 2 | m = 0 | |
| (xi) | (m/3) = 2 | m = 6 |
Solution:
(i) x + 3 = 0
Given equation:
X + 3 = 0
Value of X = 3
LHS = x + 3
By substituting the value of x
LHS = 3 + 3 = 6
By comparing LHS and RHS
LHS ≠ RHS
No, the equation is not satisfied.
(ii) x + 3 = 0
Given equation:
X +3 = 0
Value of x = 0
LHS = x + 3
By substituting the value of x
LHS = 0 + 3 = 3
By comparing LHS and RHS
LHS ≠ RHS
No, the equation is not satisfied.
(iii) x + 3 = 0
Given equation:
X + 3 = 0
Value of x = – 3
LHS = x + 3
By substituting the value of x
LHS = – 3 + 3 = 0
By comparing LHS and RHS
LHS = RHS
Yes, the equation is satisfied
(iv) x – 7 = 1
LHS = x – 7
By substituting the value of x = 7
Then,
LHS = 7 – 7 = 0
By comparing LHS and RHS
LHS ≠ RHS
No, the equation is not satisfied
(v) x – 7 = 1
Given equation:
X – 7 = 1
Value of x = 8
LHS = x – 7
By substituting the value of x
LHS = 8 – 7 = 1
By comparing LHS and RHS
LHS = RHS
Yes, the equation is satisfied.
(vi) 5x = 25
Given equation:
5x = 25
Value of x = 0
LHS = 5x
By substituting the value of x
LHS = 5 × 0 = 0
By comparing LHS and RHS
LHS ≠ RHS
No, the equation is not satisfied.
(vii) 5x = 25
Given equation:
5x = 25
Value of x = 5
LHS = 5x
By substituting the value of x
LHS = 5 × 5 = 25
By comparing LHS and RHS
LHS = RHS
Yes, the equation is satisfied.
(viii) 5x = 25
Given equation:
5x = 25
Value of x = -5
LHS = 5x
By substituting the value of x
LHS = 5 × (-5) = – 25
By comparing LHS and RHS
LHS ≠ RHS
No, the equation is not satisfied.
(ix) m/3 = 2
Given equation:
m/3 = 2
value of x = -6
LHS = m/3
By substituting the value of m
LHS = -6/3 = – 2
By comparing LHS and RHS
LHS ≠ RHS
No, the equation is not satisfied.
(x) m/3 = 2
Given equation:
m/3 = 2
value of x = 0
LHS = m/3
By substituting the value of m
LHS = 0/3 = 0
By comparing LHS and RHS
LHS ≠ RHS
No, the equation is not satisfied.
(xi) m/3 = 2
Given equation:
m/3 = 2
value of x = 6
LHS = m/3
By substituting the value of m
LHS = 6/3 = 2
By comparing LHS and RHS
LHS = RHS
Yes, the equation is satisfied.
| S. No. | Equation | Value | Say, whether the equation is satisfied. (Yes/No) |
| (i) | x + 3 = 0 | x = 3 | No |
| (ii) | x + 3 = 0 | x = 0 | No |
| (iii) | x + 3 = 0 | x = -3 | Yes |
| (iv) | x – 7 = 1 | x = 7 | No |
| (v) | x – 7 = 1 | x = 8 | Yes |
| (vi) | 5x = 25 | x = 0 | No |
| (vii) | 5x = 25 | x = 5 | Yes |
| (viii) | 5x = 25 | x = -5 | No |
| (ix) | (m/3) = 2 | m = – 6 | No |
| (x) | (m/3) = 2 | m = 0 | No |
| (xi) | (m/3) = 2 | m = 6 | Yes |
2. Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
Solution:
Given equation:
n+ 5 = 19
Where n = 1
LHS = n + 5
By substituting the value of n = 1
= n + 5
= 1 + 5
= 6
By comparing LHS and RHS
6 ≠ 19
LHS ≠ RHS
Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.
(b) 7n + 5 = 19 (n = – 2)
Solution:
Given equation:
7n + 5 = 19
Where n = -2
LHS = 7n + 5
By substituting the value of n = -2
LHS = 7n + 5
= (7 × (-2)) + 5
= – 14 + 5
= – 9
By comparing LHS and RHS
-9 ≠ 19
LHS ≠ RHS
Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.
(c) 7n + 5 = 19 (n = 2)
Solution:
Given equation:
7n + 5 = 19
Where n = 2
LHS = 7n + 5
By substituting the value of n = 2
LHS = 7n + 5
= (7 × (2)) + 5
= 14 + 5
= 19
By comparing LHS and RHS
19 = 19
LHS = RHS
Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.
(d) 4p – 3 = 13 (p = 1)
Solution:
Given equation:
4p – 3 = 13
Where p = 1
LHS = 4p – 3
By substituting the value of p = 1
LHS = 4p – 3
= (4 × 1) – 3
= 4 – 3
= 1
By comparing LHS and RHS
1 ≠ 13
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.
(e) 4p – 3 = 13 (p = – 4)
Solution:
Given equation:
4p – 3 = 13
Where p = -4
LHS = 4p – 3
By substituting the value of p = – 4
LHS = 4p – 3
= (4 × (-4)) – 3
= -16 – 3
= -19
By comparing LHS and RHS
-19 ≠ 13
LHS ≠ RHS
Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.
(f) 4p – 3 = 13 (p = 0)
Solution:
Given equation:
4p – 3 = 13
Where p = 0
LHS = 4p – 3
By substituting the value of p = 0
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= -3
By comparing LHS and RHS
– 3 ≠ 13
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.
3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
Solution:
Given,
LHS = 5p + 2
Now,
Let p = 0
By substituting the value of p = 0
We get
LHS = 5p + 2
= (5 × 0) + 2
= 0 + 2
= 2
By comparing LHS and RHS
2 ≠ 17
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation.
Now,
Let, p = 1
By substituting the value of p = 1
LHS = 5p + 2
= (5 × 1) + 2
= 5 + 2
= 7
By comparing LHS and RHS
7 ≠ 17
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation.
Now,
Let, p = 2
LHS = 5p + 2
= (5 × 2) + 2
= 10 + 2
= 12
By comparing LHS and RHS
12 ≠ 17
LHS ≠ RHS
Hence, the value of p = 2 is not a solution to the given equation.
Now,
Let p = 3
LHS = 5p + 2
= (5 × 3) + 2
= 15 + 2
= 17
By comparing LHS and RHS
17 = 17
LHS = RHS
Hence, the value of p = 3 is a solution to the given equation.
(ii) 3m – 14 = 4
Solution:
Given,
LHS = 3m – 14
By substituting the value of m = 2
LHS = 3m – 14
= (3 × 2) – 14
= 6 – 14
= – 8
By comparing LHS and RHS
-8 ≠ 4
LHS ≠ RHS
Hence, the value of m = 2 is not a solution to the given equation.
Now,
Let, m = 3
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing LHS and RHS
–5 ≠ 4
LHS ≠ RHS
Hence, the value of m = 4 is not a solution to the given equation.
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing LHS and RHS
1 ≠ 4
LHS ≠ RHS
Hence, the value of m = 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing LHS and RHS
4 = 4
LHS = RHS
Hence, the value of m = 6 is a solution to the given equation.
4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
Solution:
The above statement can be written as,
= x + 4 = 9
(ii) 2 subtracted from y is 8.
Solution:
The above statement can be written as,
= y – 2 = 8
(iii) Ten times a is 70.
Solution:
The above statement can be written as,
= 10a = 70
(iv) The number b divided by 5 gives 6.
Solution:
The above statement can be written as,
= (b/5) = 6
(v) Three-fourth of t is 15.
Solution:
The above statement can be written as,
= ¾t = 15
(vi) Seven times m plus 7 gets you 77.
Solution:
The above statement can be written as,
= 7m + 7 = 77
(vii) One-fourth of a number x minus 4 gives 4.
Solution:
The above statement can be written as,
One-fourth of a number x is x/4
Then,
= x/4 – 4 = 4
(viii) If you take away 6 from 6 times y, you get 60.
Solution:
The above statement can be written as,
6 times of y is 6y
Now,
= 6y – 6 = 60
(ix) If you add 3 to one-third of z, you get 30.
Solution:
The above statement can be written as,
One-third of z is z/3
= 3 + z/3 = 30
5. Write the following equations in statement forms:
(i) p + 4 = 15
Solution:
The sum of numbers p and 4 gives 15.
(ii) m – 7 = 3
Solution:
7 subtracted from m gives 3.
(iii) 2m = 7
Solution:
Twice of number m gives 7.
(iv) m/5 = 3
Solution:
The number m divided by 5 gives 3.
(v) (3m)/5 = 6
Solution:
Three times of m divided by 5 gives 6.
(vi) 3p + 4 = 25
Solution:
Three times p plus 4 gives us 25.
(vii) 4p – 2 = 18
Solution:
Four times p minus 2 gives us 18.
(viii) p/2 + 2 = 8
Solution:
When we add half of a number p to 2, we get 8.
6. Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)
Solution:
Let the Parmit’s marble be m
Then, Irfan’s marble = 5m + 7
The total number of marble = 37
Hence, the required equation is:
5m + 7 = 37
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
Solution:
Let the age of Laxmi is y
Age of Laxmi’s Father = 3y + 4
but the age of Laxmi’s father is given by 49
Hence the required equation:
3y + 4 = 49
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
Solution:
Let the lowest score be l
then, the highest score = 2l + 1
but the given highest score = 87
Hence, the required equation:
2l + 1 = 87
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Solution:
Let the base of each angle be ‘b’ degree
then, the vertex angle of the triangle = 2b
We know that the sum of the angles of a triangle is 180o
Hence the required equation is:
b + b + 2b = 180o or 4b
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