NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

1. Give first the step you will use to separate the variable and then solve the equation:

a. $x - 1 = 0$

Ans: The first step is to add 1 to both sides.

$x - 1 + 1 = 0 + 1$

$x = 1$

Thus, the value of $x$ is 1.

b. $x + 1 = 0$

Ans: The first step is to subtract 1 from both sides.

$x + 1 - 1 = 0 - 1$

$x =  - 1$

Thus, the value of $x$ is $ - 1$.

c. $x - 1 = 5$

Ans: The first step is to add 1 to both sides.

$x - 1 + 1 = 5 + 1$

$x = 6$

Thus, the value of $x$ is 6.

d. $x + 6 = 2$

Ans: The first step is to subtract 6 from both sides.

$x + 6 - 6 = 2 - 6$

$x =  - 4$

Thus, the value of $x$ is $ - 4$.

e. $y - 4 =  - 7$

Ans: The first step is to add 4 to both sides.

$y - 4 + 4 =  - 7 + 4$

$y =  - 3$

Thus, the value of $y$ is $ - 3$.

f. $y - 4 = 4$

Ans: The first step is to add 4 to both sides.

$y - 4 + 4 = 4 + 4$

$y = 8$

Thus, the value of $y$ is 8.

g. $y + 4 = 4$

Ans: The first step is to subtract 4 from both sides.

$y + 4 - 4 = 4 - 4$

$y = 0$

Thus, the value of $y$ is 0.

h. $y + 4 =  - 4$

Ans: The first step is to subtract 4 from both sides.

$y + 4 - 4 =  - 4 - 4$

$y =  - 8$

Thus, the value of $y$ is $ - 8$.

2. Give first the step you will use to separate the variable and then solve the equations:

a. $3l = 42$

Ans: The first step is to divide both sides by 2.

$l = \dfrac{{42}}{3}$

$l = 14$

Hence, the value of \[l\] is 14.

b. $\dfrac{b}{2} = 6$

Ans: The first step is to multiply both sides by 2.

$\dfrac{b}{2} \times 2 = 2 \times 6$

$b = 2 \times 6$

$b = 12$

Hence, the value of $b$ is 12.

c. $\dfrac{p}{7} = 4$

Ans: The first step is to multiply both sides by 7.

$\dfrac{p}{7} \times 7 = 4 \times 7$

$p = 28$

Hence, the value of $p$ is 28.

d. $4x = 25$

Ans: The first step is to divide both sides by 4.

$x = \dfrac{{25}}{4}$

The value of $x$ is $\dfrac{{25}}{4}$.

e. $8y = 36$

Ans: The first step is to divide both sides by 8.

$\dfrac{{8y}}{8} = \dfrac{{36}}{8}$$$

$y = \dfrac{{36}}{8}$

$y = \dfrac{9}{2}$

The value of $y$ is $\dfrac{9}{2}$.

f. $\dfrac{z}{3} = \dfrac{5}{4}$

Ans: The first step is to multiply both sides by 3.

$\dfrac{z}{3} \times 3 = \dfrac{5}{4} \times 3$

$z = \dfrac{{15}}{4}$

The value of $z$ is $\dfrac{{15}}{4}$.

g. $\dfrac{a}{3} = \dfrac{7}{{15}}$

Ans: The first step is to multiply both sides by 3.

$\dfrac{a}{3} \times 3 = \dfrac{7}{{15}} \times 3$

$a = \dfrac{7}{5}$

The value of $a$ is $\dfrac{7}{5}$.

h. $20t =  - 10$

Ans: The first step is to divide both sides by 20.

$\dfrac{{20t}}{{20}} = \dfrac{{ - 10}}{{20}}$

$t =  - \dfrac{1}{2}$

The value of $t$ is $ - \dfrac{1}{2}$.

3. Give first the step you will use to separate the variable and then solve the equations:

a. $3n - 2 = 46$

Ans: The first step is to add 2 to both sides of the equation.

$3n - 2 + 2 = 46 + 2$

$3n = 48$

Now, we will divide both sides by 3 and simplify.

$n = \dfrac{{48}}{3}$

$n = 16$

Thus, the value of $n$ is 16.

b. $5m + 7 = 17$

Ans: The first step is to subtract 7 from both sides.

$5m + 7 - 7 = 17 - 7$

$5m = 10$

Divide both sides by 5.

$m = \dfrac{{10}}{5}$

$m = 2$

Thus, the value of $m$ is 2.

c. $\dfrac{{20p}}{3} = 40$

Ans: The first step is to multiply both sides by 3.

$20p = 40 \times 3$

$20p = 120$

Divide both sides by 20.

$p = \dfrac{{120}}{{20}}$

$p = 6$

Hence, the value of $p$ is 6.

d. $\dfrac{{3p}}{{10}} = 6$

Ans: The first step is to multiply both sides by 10.

$3p = 6 \times 10$

$3p = 60$

Divide both sides by 3.

$p = \dfrac{{60}}{3}$

$p = 30$

Hence, the value of $p$ is 30.

4. Solve the following equation.

a. $10p = 100$

Ans: We will divide both sides by 10, to get the value of $p$.

$p = \dfrac{{100}}{{10}}$

$p = 10$

Thus, the value of $p$ is 10.

b. $10p + 10 = 100$

Ans: To solve this, subtract 10 from both sides and then, divide both sides by 10.

$10p = 100 - 10$

$10p = 90$

$p = \dfrac{{90}}{{10}}$

$p = 9$

Hence, the value of $p$ is 9.

c. $\dfrac{p}{4} = 5$

Ans: To solve this, we will multiply both sides by 4.

$\dfrac{p}{4} \times 4 = 5 \times 4$

$p = 20$

Hence, the value of $p$ is 20.

d. $ - \dfrac{p}{3} = 5$

Ans: First, we will multiply both sides by 3.

$ - \dfrac{p}{3} \times 3 = 5 \times 3$

$ - p = 15$

Now, multiply both sides by $ - 1$.

$p =  - 15$

Hence, the value of $p$ is $ - 15$.

e. $\dfrac{{3p}}{4} = 6$

Ans: First, we will multiply both sides by 4.

$\dfrac{{3p}}{4} \times 4 = 6 \times 4$

$3p = 24$

Now, divide both sides by 3.

$p = \dfrac{{24}}{3}$

$p = 8$

Hence, the value of $p$ is 8.

f. $3s =  - 9$

Ans: Multiply both sides by 3 to get the value of $s$.

$s =  - \dfrac{9}{3}$

$s =  - 3$

Hence, the value of $p$ is $ - 3$.

g. $3s + 12 = 0$

Ans: To solve the given equation, we will subtract 12 from both sides.

$3s + 12 - 12 =  - 12$

$3s =  - 12$

$s = \dfrac{{ - 12}}{3}$

$s =  - 4$

Hence, the value of $s$ is $ - 4$.

h. $3s = 0$

Ans: to solve the given equation, divide both sides by 3.

$3s = 0$

$s = \dfrac{0}{3}$

$s = 0$

Hence, the value of $s$ is 0.

i. $2q = 6$

Ans: To solve the given equation, divide both sides by 2.

$q = \dfrac{6}{2}$

$q = 3$

Hence, the value of $q$ is 3.

j. $2q - 6 = 0$

Ans: To solve the given equation, add 6 to both sides.

$2q - 6 + 6 = 0 + 6$

$2q = 6$

Divide both sides by 2.

$q = \dfrac{6}{2}$

$q = 3$

Hence, the value of $q$ is 3.

k. $2q + 6 = 0$

Ans: To solve the given equation, subtract 6 from both sides.

$2q + 6 - 6 = 0 - 6$

$2q =  - 6$

Divide both sides by 2.

$q =  - \dfrac{6}{2}$

$q =  - 3$

Hence, the value of $q$ is $ - 3$.

l. $2q + 6 = 12$

Ans: To solve the given equation, subtract 6 from both sides.

$2q + 6 - 6 = 12 - 6$

$2q = 6$

Divide both sides by 2.

$q = \dfrac{6}{2}$

$q = 3$

Hence, the value of $q$ is 3.

Conclusion

NCERT Solutions for Class 7 Maths 4.2 Chapter 4 - Simple Equations is essential for learning algebra. This exercise focuses on solving simple equations, an important skill for understanding advanced mathematical concepts. It is crucial to follow the step-by-step process of isolating the variable correctly. Vedantu's solutions provide detailed explanations for each problem, ensuring a solid grasp of the concepts. Students should focus on balancing equations and applying addition, subtraction, multiplication, and division accurately. Mastering these techniques will build a strong foundation for future algebraic studies and help tackle complex problems with confidence in class 7 maths 4.2.

Class 7 Maths Chapter 4: Exercises Breakdown

CBSE Class 7 Maths Chapter 4 Other Study Materials

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.