CBSE Class 8 Maths Linear Equations in One Variable Question Bank - Free Printable

Problem: Solve the following equations (1 to 7)

#### Equation 1:
$$
\frac{5(1-x) + 3(1+x)}{1-2x} = 8
$$

Solution:

1. Simplify the numerator:
$$
5(1-x) + 3(1+x) = 5 - 5x + 3 + 3x = 8 - 2x
$$

2. Substitute back into the equation:
$$
\frac{8 - 2x}{1 - 2x} = 8
$$

3. Cross-multiply to eliminate the fraction:
$$
8 - 2x = 8(1 - 2x)
$$

4. Expand the right-hand side:
$$
8 - 2x = 8 - 16x
$$

5. Rearrange terms to isolate \( x \):
$$
8 - 2x - 8 + 16x = 0
$$
$$
14x = 0
$$

6. Solve for \( x \):
$$
x = 0
$$

Answer:
$$
\boxed{x = 0}
$$

---

#### Equation 2:
$$
\frac{y - 4 - 3y}{2y - (3 + 4y)} = \frac{1}{5}
$$

Solution:

1. Simplify the numerator and denominator:
- Numerator: \( y - 4 - 3y = -2y - 4 \)
- Denominator: \( 2y - (3 + 4y) = 2y - 3 - 4y = -2y - 3 \)

So the equation becomes:
$$
\frac{-2y - 4}{-2y - 3} = \frac{1}{5}
$$

2. Cross-multiply to eliminate the fraction:
$$
5(-2y - 4) = 1(-2y - 3)
$$

3. Expand both sides:
$$
-10y - 20 = -2y - 3
$$

4. Rearrange terms to isolate \( y \):
$$
-10y + 2y = -3 + 20
$$
$$
-8y = 17
$$

5. Solve for \( y \):
$$
y = -\frac{17}{8}
$$

Answer:
$$
\boxed{y = -\frac{17}{8}}
$$

---

#### Equation 3:
$$
4t - 3(3t + 1) = 5t - 4
$$

Solution:

1. Distribute the \(-3\) in the left-hand side:
$$
4t - 9t - 3 = 5t - 4
$$

2. Combine like terms on the left-hand side:
$$
-5t - 3 = 5t - 4
$$

3. Add \( 5t \) to both sides:
$$
-3 = 10t - 4
$$

4. Add 4 to both sides:
$$
1 = 10t
$$

5. Solve for \( t \):
$$
t = \frac{1}{10}
$$

Answer:
$$
\boxed{t = \frac{1}{10}}
$$

---

#### Equation 4:
$$
\frac{5x + 1}{2x} = \frac{1}{3}
$$

Solution:

1. Cross-multiply to eliminate the fraction:
$$
3(5x + 1) = 2x
$$

2. Expand the left-hand side:
$$
15x + 3 = 2x
$$

3. Subtract \( 2x \) from both sides:
$$
15x - 2x + 3 = 0
$$
$$
13x + 3 = 0
$$

4. Subtract 3 from both sides:
$$
13x = -3
$$

5. Solve for \( x \):
$$
x = -\frac{3}{13}
$$

Answer:
$$
\boxed{x = -\frac{3}{13}}
$$

---

#### Equation 5:
$$
m - \frac{m-1}{2} = 1 - \frac{m-2}{3}
$$

Solution:

1. Eliminate the fractions by finding a common denominator (LCM of 2 and 3 is 6):
$$
m - \frac{m-1}{2} = 1 - \frac{m-2}{3}
$$

2. Multiply every term by 6:
$$
6m - 3(m-1) = 6 - 2(m-2)
$$

3. Distribute the constants:
$$
6m - 3m + 3 = 6 - 2m + 4
$$

4. Combine like terms:
$$
3m + 3 = 10 - 2m
$$

5. Add \( 2m \) to both sides:
$$
3m + 2m + 3 = 10
$$
$$
5m + 3 = 10
$$

6. Subtract 3 from both sides:
$$
5m = 7
$$

7. Solve for \( m \):
$$
m = \frac{7}{5}
$$

Answer:
$$
\boxed{m = \frac{7}{5}}
$$

---

#### Equation 6:
$$
0.16(5x - 2) = 0.4x + 7
$$

Solution:

1. Distribute \( 0.16 \) on the left-hand side:
$$
0.16 \cdot 5x - 0.16 \cdot 2 = 0.4x + 7
$$
$$
0.8x - 0.32 = 0.4x + 7
$$

2. Subtract \( 0.4x \) from both sides:
$$
0.8x - 0.4x - 0.32 = 7
$$
$$
0.4x - 0.32 = 7
$$

3. Add \( 0.32 \) to both sides:
$$
0.4x = 7.32
$$

4. Solve for \( x \):
$$
x = \frac{7.32}{0.4} = 18.3
$$

Answer:
$$
\boxed{x = 18.3}
$$

---

#### Equation 7:
The sum of three consecutive natural numbers is 48. Find the greatest of these numbers.

Solution:

1. Let the three consecutive natural numbers be \( n \), \( n+1 \), and \( n+2 \).

2. Their sum is given as 48:
$$
n + (n+1) + (n+2) = 48
$$

3. Simplify the equation:
$$
3n + 3 = 48
$$

4. Subtract 3 from both sides:
$$
3n = 45
$$

5. Solve for \( n \):
$$
n = 15
$$

6. The three consecutive numbers are \( 15 \), \( 16 \), and \( 17 \). The greatest number is \( 17 \).

Answer:
$$
\boxed{17}
$$

---

Final Answers:

1. \( \boxed{x = 0} \)
2. \( \boxed{y = -\frac{17}{8}} \)
3. \( \boxed{t = \frac{1}{10}} \)
4. \( \boxed{x = -\frac{3}{13}} \)
5. \( \boxed{m = \frac{7}{5}} \)
6. \( \boxed{x = 18.3} \)
7. \( \boxed{17} \)