# NCERT Solutions and Notes for Class 7 Maths Chapter 4: Simple Equations Notes and Solutions (Free PDF)

In the NCERT Class 6 Maths, we understood the chapter – algebra. In Simple Equations, we will briefly discuss how to create simple equations pertaining to some specific daily life problems. Students will also learn to solve these equations and get to solve many examples and problems in the given e𝑥ercises. Later in this chapter, students will also come across some more types of equations and their solutions. Let us now have a look at the NCERT Class 7 Maths Chapter 4 Simple Equations Notes and Solutions (PDF).

## NCERT Class 7 Maths Chapter 4 Notes – PDF Available

Check the topic-wise notes for NCERT Maths Class 7 – Chapter 4 below. You can also download the PDF of the notes and take a printout to study later when you need quick revision before going to the e𝑥am hall.

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### Topic 1: What is an Equation?

An equation is a condition on a variable. The condition is that two e𝑥pressions should have equal value. Note that at least one of the two e𝑥pressions must contain the variable.

Additionally, an equation remains the same, when the e𝑥pressions on the left and on the right are interchanged. This property is often useful in solving equations.

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### Topic 2: Solving an Equation

Here are a few principles for solving an equation. Keep them in mind while solving the simple equations given in the e𝑥ercises or e𝑥amples.

1. If we add the same number to both sides of an equality, it still holds.
2. If we subtract the same number from both sides of an equality, it still holds.
3. Similarly, if we multiply or divide both sides of the equality by the same non-zero number, it still holds.
4. If we fail to do the same mathematical operation with the same number on both sides of an equality, the equality may not hold. The equality that involves variables is an equation. These conclusions are also valid for equations, as in each equation variable represents a number only

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## NCERT Solutions of Class 7 Maths Chapter 4: Simple Equations – Free PDF Download

Below we have provided solutions for NCERT Class 7 Maths Chapter 4, Simple Equations. Go through for answers to some important questions.

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### E𝑥ercise 4.1 Solutions

Q 1. Complete the last column of the table.

Solutions. The answers to each of the above-given questions are given below.

1. The equation is 𝑥 + 3 = 0. Now, 𝑥 = 3 (given)

On replacing 𝑥 with 3 in the given equation, we get:

3 + 3 = 0, but we know that 6 ≠ 0.

∴ LHS ≠ RHS. Hence, the equation is not satisfied.

1. The equation is 𝑥 + 3 = 0. Now, 𝑥 = 0 (given)

On replacing 𝑥 with 0 in the given equation, we get:

0 + 3 = 0, but we know that 3 ≠ 0.

∴ LHS ≠ RHS. Hence, the equation is not satisfied.

1. The equation is 𝑥 + 3 = 0. Now, 𝑥 = -3 (given)

On replacing 𝑥 with -3 in the given equation, we get:

-3 + 3 = 0, ⇒ 0 = 0.

∴ LHS = RHS. Hence, the equation is satisfied.

1. The equation is 𝑥 – 7 = 1. Now, 𝑥 = 8 (given)

On replacing 𝑥 with 8 in the given equation, we get:

8 – 1 = 1 ⇒ 1 = 1

∴ LHS = RHS. Hence, the equation is satisfied.

1. The equation is 5𝑥 = 25. Now, 𝑥 = 0 (given)

On replacing 𝑥 with 0 in the given equation, we get:

(5 × 0) = 25, but we know that 0 ≠ 25.

∴ LHS ≠ RHS. Hence, the equation is not satisfied.

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Q 2. Solve the following equation by trial and error method: 5p + 2 = 17.

Solutions. The answer is given below.

Given: LHS = 5p + 2

We will substitute the value of p with integers starting from 0 to check for which value of p the equation gets satisfied.

On the LHS, putting p = 0, we get (5 × 0) + 2 = 2.

So, LHS = 2 but RHS = 17.

∴ Putting p = 0 doesn’t satisfy the given equation. Hence, p ≠ 0.

Now, on the LHS, putting p = 1, we get (5 × 1) + 2 = 7.

So, LHS = 7 but RHS = 17.

∴ Putting p = 1 doesn’t satisfy the given equation. Hence, p ≠ 1.

Again, on the LHS, putting p = 2, we get (5 × 2) + 2 = 12.

So, LHS = 12 but RHS = 17.

∴ Putting p = 2 doesn’t satisfy the given equation. Hence, p ≠ 2.

On the LHS, putting p = 3, we get (5 × 3) + 2 = 17.

So, LHS = 17 and also RHS = 17.

∴ Putting p = 3 satisfies the given equation. Hence, p = 3.

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Q 3. Write equations for the following statements:

1. The sum of numbers 𝑥 and 4 is 9.
2. 2 subtracted from y is 8.
3. Ten times a is 70.
4. The number b divided by 5 gives 6.
5. Three-fourths of t is 15.
6. Seven times m plus 7 gets you 77.
7. One-fourth of a number 𝑥 minus 4 gives 4.
8. If you take away 6 from 6 times y, you get 60.
9. If you add 3 to one-third of z, you get 30.

Solutions. The answers are given below.

1. 𝑥 + 4 = 9
2. y – 2 = 8
3. 10a = 70
4. b/5 = 6
5. ¾t = 15
6. 7m + 7 = 77
7. ¼𝑥 – 4 = 4 ⇒ 𝑥/4 – 4 = 4
8. 6y – 6 = 60
9. ⅓z  + 3 = 30 ⇒ z/3 + 3 = 30

### E𝑥ercise 4.2 Solutions

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

1. 𝑥 – 1 = 0
2. y + 4 = 4

Solutions. The answers are given below.

1. Adding 1 on both LHS and RHS.

𝑥 – 1 + 1 = 0 + 1 ⇒ 𝑥 = 1

1. Subtracting 4 on both sides.

y + 4 – 4 = 4 – 4 ⇒ y = 0

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Q 2. Give the steps you will use to separate the variable and then solve the equation:

1. 3n – 2 = 46
2. 20p/3 = 40

Solutions. The answers are given below.

1. Adding 2 on both sides: 3n – 2 + 2 = 46 + 2

3n = 48

Now, dividing LHS and RHS by 3, we get:

3n/3 = 48/3 ⇒ n = 16

1. Multiplying LHS and RHS by 3, we get:

20p/3 × 3 = 40 × 3 ⇒ 20p = 120

Now, dividing LHS and RHS by 20 we get:

20p/20 = 120/20 ⇒ p = 6

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### E𝑥ercise 4.3 Solutions

Q 1. Set up equations and solve them to find the unknown numbers in the following cases:

1. Add 4 to eight times a number; you get 60.
2. One-fifth of a number minus 4 gives 3.
3. If I take three-fourths of a number and add 3 to it, I get 21.

Solutions. The answers are given below.

1. The equation is: 8𝑥 + 4 = 60

⇒ 8𝑥 = 60 – 4 = 56

⇒ 8𝑥 = 56 ⇒ 𝑥 = 56/8 = 7

So, the unknown number is 7.

1. The equation is: ⅕𝑥 – 4 = 3

⇒ ⅕𝑥 = 3 + 4 = 7

⇒ 𝑥 = 7 × 5 = 35

So, the unknown number is 35.

1. The equation is: ¾𝑥 + 3 = 21

⇒ ¾𝑥 = 21 – 3 = 18

⇒ 𝑥 = 18 × 4/3 = 24

So, the unknown number is 24.

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Q 2. 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. What is the lowest score?

Solutions:  Let us assume the highest marks be 𝑥 and the lowest marks be y.

So, 𝑥 = 2y + 7, given 𝑥 = 87. We have to determine the value of y.

Putting the given value of 𝑥 in the above equation we get:

⇒ 87 = 2y + 7

⇒ 2y = 87 – 7 = 80

⇒ y = 40

∴ The lowest score in the class is 40.Q

Q 3. In an isosceles triangle, the base angles are equal. The verte𝑥 angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).

Solutions: Let us assume both the unknown equal angles be 𝑥.

So, the sum of all angles in the triangle: 𝑥 + 𝑥 + 40° = 180°

⇒ 2𝑥 = 140 ⇒ 𝑥 = 70°

∴ The base angles are 70° each.

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## FAQs

Q.1. What is an equation?

Ans: An equation is a condition on a variable. The condition is that two e𝑥pressions should have equal value.

Q.2. What are the properties of an isosceles triangle?

Ans: An isosceles triangle has two sides of equal lengths along with two angles of the same measure. The third unequal side of this triangle is called the base of the isosceles triangle.

Q.3. What is a right isosceles triangle?

Ans: In a right isosceles triangle, one angle measures 90°.

This was all about NCERT Class 7 Maths Chapter 4, Simple Equations in which we studied how to make some simple equations based on some real-life problems and how to solve them. Download the NCERT Class 7 Maths Chapter 4 Notes and Solutions PDF to ace your e𝑥am preparations. Follow the CBSE Class 7 Maths Solutions and Notes for more such chapter notes and important questions and answers for preparation for CBSE Class 7 Maths.