NCERT Solutions for Class 7 Maths Chapter 4: Simple Equations — Complete Guide with Step-by-Step Solutions

Think of an equation as a balance scale (weighing scale). The left-hand side (LHS) and right-hand side (RHS) are the two pans.

When the equation is true, both pans are balanced. To keep the balance:
- Whatever you add to one side, you must add to the other.
- Whatever you subtract from one side, you must subtract from the other.
- Whatever you multiply one side by, you must multiply the other by the same number.
- Whatever you divide one side by, you must divide the other by the same number.

This is the essence of the balancing method — and it guarantees that the equation remains true at every step.

To check if a given value is a solution, substitute it into the equation and see if LHS $=$ RHS.

Example: Is $x = 4$ a solution of $3x - 5 = 7$?

Since LHS $=$ RHS, yes, $x = 4$ is a solution.

Example: Is $x = 3$ a solution of $2x + 1 = 9$?

Since $7 \neq 9$, no, $x = 3$ is NOT a solution.

Problem: Check whether $x = 5$ is a solution of $2x - 3 = 7$.

Solution:
Substitute $x = 5$:

Yes, $x = 5$ is a solution. $\square$

Problem: Write the equation for: "$7$ added to twice a number gives $19$."

Solution:
Let the number be $x$.
"Twice a number" $= 2x$.
"$7$ added to twice a number" $= 2x + 7$.
"Gives $19$" $\to$ equals $19$.

Problem: Write the equation for: "If you subtract $3$ from one-third of a number, you get $4$."

Solution:
Let the number be $x$.

Problem: Ravi's age after $15$ years will be four times his present age. Write the equation. If his present age is $5$, verify.

Solution:
Let present age $= x$.
Age after $15$ years $= x + 15$.
Four times present age $= 4x$.

**Verification for $x = 5$:**
LHS $= 5 + 15 = 20$.
RHS $= 4 \times 5 = 20$.
LHS $=$ RHS. Verified.

Problem: A piggy bank contains Rs. $5$ and Rs. $2$ coins. There are twice as many Rs. $2$ coins as Rs. $5$ coins. The total amount is Rs. $72$. Set up the equation.

Solution:
Let the number of Rs. $5$ coins $= x$.
Number of Rs. $2$ coins $= 2x$.

Total amount: $5x + 2(2x) = 72$

Problem: The length of a rectangle is $3$ cm more than its breadth. The perimeter is $30$ cm. Write the equation.

Solution:
Let breadth $= x$ cm. Then length $= x + 3$ cm.
Perimeter $= 2(l + b) = 2(x + 3 + x) = 2(2x + 3)$.

Problem: Check which value is the solution of $5x + 2 = 17$: (a) $x = 2$, (b) $x = 3$, (c) $x = 4$.

Solution:
(a) $5(2) + 2 = 12 \neq 17$. Not a solution.
(b) $5(3) + 2 = 17 = 17$. This IS the solution.
(c) $5(4) + 2 = 22 \neq 17$. Not a solution.

Answer: $x = 3$.

Problem: Three consecutive integers add up to $42$. Write the equation.

Solution:
Let the integers be $x$, $x + 1$, $x + 2$.

Problem: Solve $x + 5 = 8$.

Solution:
Subtract $5$ from both sides:

Verification: $3 + 5 = 8$. Correct.

Problem: Solve $3x = 18$.

Solution:
Divide both sides by $3$:

Verification: $3 \times 6 = 18$. Correct.

Problem: Solve $2x + 4 = 12$.

Solution:
Step 1: Subtract $4$ from both sides:

Step 2: Divide both sides by $2$:

Verification: $2(4) + 4 = 8 + 4 = 12$. Correct.

Problem: Solve $\frac{x}{2} = 7$.

Solution:
Multiply both sides by $2$:

Verification: $\frac{14}{2} = 7$. Correct.

Problem: Solve $3x - 5 = 16$.

Solution:
Step 1: Add $5$ to both sides: $3x = 21$.
Step 2: Divide by $3$: $x = 7$.

Verification: $3(7) - 5 = 21 - 5 = 16$. Correct.

Problem: Solve $4x + 7 = 3$.

Solution:
Step 1: Subtract $7$: $4x = 3 - 7 = -4$.
Step 2: Divide by $4$: $x = -1$.

Verification: $4(-1) + 7 = -4 + 7 = 3$. Correct.

Note: Equations can have negative solutions. This is perfectly valid.

Problem: Solve $\frac{2x}{3} = 8$.

Solution:
Multiply both sides by $3$: $2x = 24$.
Divide by $2$: $x = 12$.

Verification: $\frac{2 \times 12}{3} = \frac{24}{3} = 8$. Correct.

Problem: Solve $5(x - 3) = 20$.

Solution:
Divide both sides by $5$: $x - 3 = 4$.
Add $3$: $x = 7$.

Verification: $5(7 - 3) = 5 \times 4 = 20$. Correct.

Problem: Solve $5x = 3x + 12$.

Solution:
Subtract $3x$ from both sides: $5x - 3x = 12$.

Verification: LHS $= 5(6) = 30$. RHS $= 3(6) + 12 = 18 + 12 = 30$. Correct.

Problem: Solve $\frac{x}{4} + 3 = 10$.

Solution:
Subtract $3$: $\frac{x}{4} = 7$.
Multiply by $4$: $x = 28$.

Verification: $\frac{28}{4} + 3 = 7 + 3 = 10$. Correct.

Problem: Solve $5x + 2 = 27$.

Solution:
Transpose $+2$: $5x = 27 - 2 = 25$.
Transpose $\times 5$: $x = \frac{25}{5} = 5$.

Verification: $5(5) + 2 = 25 + 2 = 27$. Correct.

Problem: Solve $\frac{3x}{2} = 9$.

Solution:
Multiply both sides by $2$: $3x = 18$.
Divide by $3$: $x = 6$.

Verification: $\frac{3 \times 6}{2} = \frac{18}{2} = 9$. Correct.

Problem: Solve $4(x + 2) = 20$.

Solution:
Divide by $4$: $x + 2 = 5$.
Transpose $+2$: $x = 3$.

Verification: $4(3 + 2) = 4 \times 5 = 20$. Correct.

Problem: Solve $7x - 9 = 16$.

Solution:
Transpose $-9$: $7x = 16 + 9 = 25$.

Verification: $7 \times \frac{25}{7} - 9 = 25 - 9 = 16$. Correct.

Problem: Solve $\frac{2x}{3} + 1 = 7$.

Solution:
Transpose $+1$: $\frac{2x}{3} = 6$.
Multiply by $3$: $2x = 18$.
Divide by $2$: $x = 9$.

Verification: $\frac{2 \times 9}{3} + 1 = \frac{18}{3} + 1 = 6 + 1 = 7$. Correct.

Problem: Solve $8x - 3 = 5x + 12$.

Solution:
Transpose $5x$ to the left: $8x - 5x - 3 = 12$.
Simplify: $3x - 3 = 12$.
Transpose $-3$: $3x = 15$.
Divide: $x = 5$.

Verification: LHS $= 8(5) - 3 = 37$. RHS $= 5(5) + 12 = 37$. Correct.

Problem: Solve $-3x + 7 = -8$.

Solution:
Transpose $+7$: $-3x = -8 - 7 = -15$.
Divide by $-3$: $x = \frac{-15}{-3} = 5$.

Verification: $-3(5) + 7 = -15 + 7 = -8$. Correct.

Problem: Solve $2.5x = 10$.

Solution:

Verification: $2.5 \times 4 = 10$. Correct.

Problem: Solve $3(2x - 1) = 15$.

Solution:
Expand: $6x - 3 = 15$.
Transpose $-3$: $6x = 18$.
Divide: $x = 3$.

Verification: $3(2 \times 3 - 1) = 3(6 - 1) = 3 \times 5 = 15$. Correct.

Problem: Solve $\frac{x + 3}{2} = \frac{x - 1}{3}$.

Solution:
Cross-multiply: $3(x + 3) = 2(x - 1)$.
Expand: $3x + 9 = 2x - 2$.
Transpose $2x$: $3x - 2x + 9 = -2$.
Simplify: $x + 9 = -2$.
Transpose $+9$: $x = -11$.

Verification: LHS $= \frac{-11 + 3}{2} = \frac{-8}{2} = -4$. RHS $= \frac{-11 - 1}{3} = \frac{-12}{3} = -4$. Correct.

Problem: The sum of two numbers is $74$. One number is $10$ more than the other. Find the numbers.

Solution:
Let the smaller number $= x$. Then the larger $= x + 10$.

The numbers are $32$ and $42$.

Verification: $32 + 42 = 74$ and $42 - 32 = 10$. Both conditions satisfied.

Problem: The perimeter of a rectangle is $40$ cm. Its length is $4$ cm more than its breadth. Find the dimensions.

Solution:
Let breadth $= x$ cm. Length $= (x + 4)$ cm.

Breadth $= 8$ cm, Length $= 12$ cm.

Verification: Perimeter $= 2(8 + 12) = 2 \times 20 = 40$ cm. Correct.

Problem: Anu is $4$ years older than her brother. The sum of their ages is $28$. Find their ages.

Solution:
Let brother's age $= x$. Anu's age $= x + 4$.

Brother is $12$, Anu is $16$.

Verification: $12 + 16 = 28$ and $16 - 12 = 4$. Correct.

Problem: After $20$ years, Arun will be $5$ times as old as he is now. Find his present age.

Solution:
Let present age $= x$.

Arun is $5$ years old now.

Verification: After $20$ years: $5 + 20 = 25 = 5 \times 5$. Correct.

Problem: The sum of three consecutive odd numbers is $51$. Find them.

Solution:
Let the three consecutive odd numbers be $x$, $x + 2$, $x + 4$.

The numbers are $15, 17, 19$.

Verification: $15 + 17 + 19 = 51$. Correct.

Problem: A bag contains Rs. $5$ coins and Rs. $2$ coins. The number of Rs. $2$ coins is $3$ times the number of Rs. $5$ coins. If the total value is Rs. $55$, find the number of each type.

Solution:
Let the number of Rs. $5$ coins $= x$.
Number of Rs. $2$ coins $= 3x$.

Total value: $5x + 2(3x) = 55$

Rs. $5$ coins $= 5$, Rs. $2$ coins $= 15$.

Verification: $5 \times 5 + 15 \times 2 = 25 + 30 = 55$. Correct.

Problem: A number is $\frac{2}{3}$ of another number. If their sum is $50$, find the numbers.

Solution:
Let the larger number $= x$. The smaller $= \frac{2}{3}x$.

The numbers are $30$ and $\frac{2}{3} \times 30 = 20$.

Verification: $30 + 20 = 50$ and $20 = \frac{2}{3} \times 30$. Correct.

Problem: A teacher distributes $42$ chocolates among $3$ students such that the second gets twice as many as the first, and the third gets $6$ more than the first. How many does each get?

Solution:
Let the first student get $x$ chocolates.
Second student $= 2x$.
Third student $= x + 6$.

First $= 9$, Second $= 18$, Third $= 15$.

Verification: $9 + 18 + 15 = 42$. Correct.

Problem: A car and a bus start from the same point. The car travels at $60$ km/hr and the bus at $40$ km/hr. After how many hours will they be $100$ km apart (travelling in opposite directions)?

Solution:
In $t$ hours:
- Car covers $= 60t$ km
- Bus covers $= 40t$ km
- Total distance apart $= 60t + 40t = 100t$

Answer: They will be $100$ km apart after $1$ hour.

Problem: Meena has Rs. $200$ more than Reena. Together they have Rs. $800$. Find how much each has.

Solution:
Let Reena have Rs. $x$. Meena has Rs. $(x + 200)$.

Reena has Rs. $300$, Meena has Rs. $500$.

Verification: $300 + 500 = 800$ and $500 - 300 = 200$. Correct.

Answer 1: $4x = 25 + 7 = 32$. $x = 8$. Verify: $4(8) - 7 = 25$. Correct.

Answer 2: $\frac{3x}{4} = 11 - 2 = 9$. $3x = 36$. $x = 12$. Verify: $\frac{36}{4} + 2 = 9 + 2 = 11$. Correct.

Answer 3: Let the numbers be $x$ and $x + 2$. $x + x + 2 = 54 \Rightarrow 2x = 52 \Rightarrow x = 26$. The numbers are $26$ and $28$.

Answer 4: $5x - 7 = 28 \Rightarrow 5x = 35 \Rightarrow x = 7$.

Answer 5: Father $= 7x$, Son $= 2x$. After $5$ years: $\frac{7x + 5}{2x + 5} = \frac{8}{3}$. Cross-multiply: $3(7x + 5) = 8(2x + 5)$. $21x + 15 = 16x + 40$. $5x = 25$. $x = 5$. Father $= 35$, Son $= 10$.

Answer 6: Expand: $3x - 12 + 2 = 5x - 16$. $3x - 10 = 5x - 16$. Transpose: $-10 + 16 = 5x - 3x$. $6 = 2x$. $x = 3$. Verify: LHS $= 3(3-4) + 2 = -3 + 2 = -1$. RHS $= 5(3) - 16 = 15 - 16 = -1$. Correct.