Students can Download Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka State Syllabus Class 7 Maths Chapter 4 Simple Equations Ex 4.2
Question 1.
Given first the step you will use to separate the variable and then solve the equation:
a) x – 1 = 0
Solution:
The given equation = x – 1 = 0
Add 1 to both the sides
x – 1 + 1 = 0 + 1 ⇒ x = 1
Checking
x – 1 = 0
put the value of x = 1
1 – 1 = 0
∴ LHS = RHS (checked)
b) x + 1 = 0
The given equation = x + 1 = 0
Subtract 1 from both the sides
x – 1 + 1 = 0 – 1 ⇒ x = – 1
Verification
x+ 1 = 0
substitute the value of x = – 1 then x + 1 = 0
(-1) + 1 = 0
0 = 0
LHS = RHS (verified)
c) x – 1 = 5
The given equation is x – 1 = 5
Add 1 to both sides
x – 1 + 1 = 5 + 1 ⇒ x = 6
Substitute the value in the equation x = 6.
x – 1 = 5
6 – 1 = 5
5 = 5
∴ LHS = RHS (verified)
d) x + 6 = 2
The given equation is x + 6 = 2
Subtract 6 from both the sides
x + 6 – 6 = 2 – 6 ⇒ x = – 4
Substitute the value of x = – 4 in the given equation,
x + 6 = 2
– 4 + 6 = 2
2 = 2
∴ LHS = RHS (verified)
e) y – 4 = -7
The given equation is y – 4 = – 7
Add 4 to both the sides
y – 4 + 4 = – 7 + 4 ⇒ y = – 3
substitute the value of y = – 3 in the given equation
y – 4 = -7
– 3 – 4 = – 7
– 7 = – 7
∴ LHS = RHS (verified)
f) y – 4 = 4
Solution:
The given equation is y – 4 = 4
Add 4 to both the sides
y – 4 + 4 = 4 + 4 ⇒ y = 8
substitute the value of y = 8 in the given equation
y – 4 = 4
8 – 4 = 4
4 = 4
∴ LHS = RHS (verified)
g) y + 4 = 4
Solution:
The given equation is y + 4 = 4
subtract 4 from both sides
= y – 4 + 4 = 4 – 4 ⇒ y = 0
substitute the value of y = 0 in the given equation.
y + 4 = 4
0 + 4 = 4
4 = 4
∴ LHS = RHS (verified)
h) y + 4 = – 4
Solution:
The given equation is y + 4 = – 4
subtract 4 from both the sides
y – 4 + 4 = – 4 – 4
y = – 8 (verified)
substitute the value of y = – 8 in the given equation
y + 4 = – 4
– 8 + 4 = – 4
– 4 = – 4
∴ LHS = RHS (verified)
Question 2.
Give first the step you will use to separate the variable and then solve the equation:
a) 3l = 42
Solution:
The given equation is 3l = 42
Divide by 3 both the sides
\(\frac{31}{3}\) = \(\frac{42}{3}\)
∴ l = 14
Substitute the value of l = 14 in the given equation
3l = 42
3(14) = 42
42 = 42
∴ LHS = RHS (verified)
b) \(\frac{\mathbf{b}}{2}\) = 6
Solution:
The given equation is \(\frac{\mathbf{b}}{2}\) = 6
Multiplied by 2 both the sides b
\(\frac{b}{2}\) × 2 = 6 × 2
∴ b = 12
substitute the value of b = 12 in the given equation
∴ LHS = RHS (verified)
c) \(\frac{\mathbf{p}}{7}\) = 4
Solution:
The given equation is \(\frac{p}{7}\) = 4
Multiplied by 7 to both the sides
\(\frac{p}{7}\) ×7 = 4 × 7
∴ p = 28
Substitute the value of p = 28 in the given equation
∴ LHS = RHS (verified)
d) 4x = 25
Solution:
The given equation is 4 x = 25
Divide both sides by 4
substitue the value of x = \(\frac{25}{4}\) in the given equation
4x = 25
4 × \(\frac{25}{4}\) = 25
∴ LHS = RHS (verified)
e) 8y = 36
Solution:
The given equation is 8y = 36
Divide both sides by 8
Substitute the equation by y = \(\frac{9}{2}\) in the given equation
8y = 36
8 × \(\frac{9}{2}\) = 36
36 = 36
∴ LHS = RHS
f) \(\frac{z}{3}=\frac{5}{4}\)
Solution:
Multiply both sides by 3
g) \(\frac{a}{5}=\frac{7}{15}\)
Solution:
Multiply both sides by 5
Substitute the value of a = \(\frac{7}{3}\) in the given equation
h) 20t = – 10
Solution:
The given equation is 20 t = – 10
Divide by 20 to both sides
Question 3.
Given the steps you will use to separate the variable and then solve the equation:
a) 3n – 2 = 46
Solution:
Add 2 to both sides 3n – 2 + 2 = 46 + 2
3n = 48
Divide the equation by 3
\(\frac{3n}{3}\) = \(\frac{48}{3}\)
∴ n = 16 This is required solution.
substitute the value of n = 16 in the given equation
3n – 2 = 46
3(16) – 2 = 46
48 – 2 = 46
46 = 46
∴ LHS = RHS (verified)
b) 5m + 7 = 17
The given equation is 5m + 7 = 17
Subtract 7 from both sides
5m + 7 – 7 = 17 – 7
5m = 10
Divide the equation by 5 both the sides
\(\frac{5m}{5}\) = \(\frac{10}{5}\)
∴ m = 2 (solution)
Substitute the value of m = 2 in the given equation
5m + 7 = 17
5(2) + 7 =17
10 + 7 = 17
∴ LHS = RHS (verified)
c) \(\frac{20 p}{3}\) = 40
Solution:
Multiply both sides by 3
\(\frac{20}{3}\) × 3 × P = 40 × 3
20P = 120
Divide both sides by 20
\(\frac{20}{20}\) × P = \(\frac{120}{20}\) ⇒ P = 6
d) \(\frac{3 P}{10}=6\)
Solution:
The given equation is \(\frac{3 P}{10}\) = 6
Multiply by 10 to both the sides
6 = 6 ∴ LHS = RHS (verified)
Question 4.
Solve the following equations :
a) 10p = 100
Solution:
b) 10p + 10 = 100
Solution:
10 P + 10 – 10 = 100 – 10
10 P = 90
c) \(\frac{\mathbf{p}}{4}\) = 5
Solution:
\(\frac{p}{4}\) × 4 = 5 × 4 ⇒ P = 20
d) \(\frac{-p}{3}\) = 5
Solution:
– P = 15 ⇒ P = – 15
e) \(\frac{3 P}{4}=6\)
Solution:
f) 3s = – 9
Solution:
\(\frac{3s}{3}=\frac{-9}{3}\)
S = – 3
g) 3s + 12 = 0
Solution:
3S + 12 + 12 – 12 = 0 – 12
h) 3s = 0
Solution:
j) 2q – 6 = 0
Solution:
2q – 6 + 6 = 0 + 6
k) 2q + 6 = 0
Solution:
2q + 6 + 6 = 0 – 6
2q = – 6
l) 2q + 6 = 12
2q + 6 – 6 = 12 – 6
2q = 6
