Exercise-4.1
Question 1:
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 |
Answer:
(i) The equation we have is: x + 3 = 0
Note: To check whether x =3 satisfies the given equation, we put x = 3 in the equation.
L.H.S. = 3 + 3 = 6 ≠ RHS
Thus, the answer is No i.e. the equation is not satisfied.
(ii) The equation we have is: x + 3 = 0
Note: To check whether the x = 0 satisfies the given equation, we put x = 0 in the given equation.
L.H.S. = 0 + 3 = 3 ≠ RHS
Thus, the answer is No i.e. the equation is not satisfied.
(iii) The equation we have is: x + 3 = 0
Note: To check whether the x = -3 satisfies the given equation, we put x = -3 in the given equation.
L.H.S. = -3 + 3 = 0 = RHS
Thus, the answer is Yes i.e. the equation is satisfied.
(iv) The equation we have is: x – 7 = 1
Note: To check whether the x = 7 satisfies the given equation, we put x = 7 in the given equation.
L.H.S. = 7 – 7 = 0 ≠ RHS
Thus, the answer is No i.e. the equation is not satisfied.
(v) The equation we have is: x – 7 = 1
Note: To check whether the x = 8 satisfies the given equation, we put x = 8 in the given equation.
L.H.S. = 8 – 7 = 1 = RHS
Thus, the answer is Yes i.e. the equation is satisfied.
(vi) The equation we have is: 5x = 25
Note: To check whether the x = 0 satisfies the given equation, we put x = 0 in the given equation.
L.H.S. = 5(0) = 0 ≠ RHS
Thus, the answer is No i.e. the equation is not satisfied.
(vii) The equation we have is: 5x = 25
Note: To check whether the x = 5 satisfies the given equation, we put x = 5 in the given equation.
L.H.S. = 5(5) = 25 = RHS
Thus, the answer is Yes i.e. the equation is satisfied.
(viii) The equation we have is: 5x = 25
Note: To check whether the x = -5 satisfies the given equation, we put x = -25 in the given equation.
L.H.S. = 5(-5) = -25 = RHS
Thus, the answer is No i.e. the equation is not satisfied.
Exercise-4.2
Question 1:
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
Answer: (a) x – 1 = 0
Solution: We have to separate the variables
Thus, x – 1 = 0
Now,
Adding 1 on both side of the equation, We get,
x – 1 + 1 = 0 + 1
x = 1
(b) x + 1 = 0
Solution: Here, we have to separate the variables
We have,
x + 1 = 0
Now,
Subtracting 1 on both side of the equation,We get,
x – 1 + 1 = 0 – 1
x = -1
(c) x – 1 = 5
Solution: Here, we have to separate the variables
We have,
x – 1 = 5
Now,Adding 1 on both side of the equation,We get,
x – 1 + 1 = 5 + 1
x = 6
(d) x + 6 = 2
Solution: Here, we have to separate the variables
We have,
x + 6 = 2
Now, subtracting 6 on both side of the equation,
We get,
x + 6 – 6 = 2 – 6
x = -4
(e)y – 4 = – 7
Solution: Here, we have to separate the variables
We have,
y – 4 = -7
Now,
Adding 4 on both side of the equation,
We get,
x – 4 + 4 = -7 + 4
x = -3
(f) y – 4 = 4
Solution: Here, we have to separate the variables
We have,
y – 4 = 4
Now,Adding 4 on both side of the equation,We get,
x – 4 + 4 = 4 + 4
x = 8
(g) y + 4 = 4
Solution: Here, we have to separate the variables
Thus,
We have, y + 4 = 4
Now,
Subtracting 4 on both side of the equation, We get,
x + 4 – 4 = 4 – 4
x = 0
(h )y + 4 = – 4
Solution: Here,We have to separate the variables
Thus,We have,
y + 4 = -4
Now,
Subtracting 4 on both side of the equation,We get,
x + 4 – 4 = -4 – 4
x = -8
Question 2:
Check whether the value given in the brackets is a solution to the given equation or not:
(a)3l = 42
(b )b/2 = 6
(c)p/7 = 4
(f)
(h)20t = -10
The given parts of the question are solved below:
Here,
Thus,
Dividing both sides by 3, we get, l=423
We have to separate the variables
Multiplying both sides by 2, we get,
b = 12
(c) p/7 = 4
We have, p7=4
p/7×7 = 4 × 7
Exercise-4.3
Exercise-4.4
Question 1:
Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of
(f) Notebooks he has from 50, he finds the result to be 8.
(g) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(h) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
Answer:
The parts of this given questions are solved below:
(a) Add 4 to eight times a number you get 60.We have to frame the equation on the basis of given statement
And then solve the so framed equation
Thus,
Let the number be a
We know that,
Eight times of a is 8a
Therefore,
As per the statement,
We can write the equation as,
8a + 4 = 60
8a = 60 – 48a = 56
Now,
Dividing 8 on both the sides, we get
8a/8=56/8
Therefore,
a = 7
(b) One-fifth of a number minus 4 gives 3.
We have to frame the equation on the basis of given statement
And then solve the so framed equation
Thus,
Let the number be a
We know that,
One fifth of a is x/5
Therefore,
As per the statement,
We can write the equation as,
x/5-4=3
x/5 = 3 + 4
x/5 = 7
Now,
Multiplying 5 to both the sides, we get
a/5×5=7×5
