In earlier lessons, you learned about algebraic expressions like 5x, 2x – 3, or more complex ones like x² + y². These expressions turn into equations when we add an equality sign (=). Equations usually contain expressions with just one variable and are linear, which means the variable’s highest power is 1.
An algebraic equation is an equality involving variables, split into two parts: the Left Hand Side (LHS) and the Right Hand Side (RHS). An equation is only true for certain values of the variable, called the solutions.
To find a solution, we keep the equation balanced. We do the same math operations on both sides of the equation. This way, we find the solution without upsetting the balance.
For example, in the equation 3x + 2 = 11, 3x + 2 and 11 are the expressions. While we often see numbers on the RHS, it’s not always the case. Sometimes both sides have variable expressions, like in the equation x + 5 = 2x – 1, where x + 5 is on the LHS and 2x – 1 is on the RHS.
NCERT Solutions for Class 8 Maths Exercise 2.1 Chapter 1 Linear Equations in One Variable
Solve the following equations and check your results.
1. 3x = 2x + 18
Step 1: Subtract 2x from both sides of the equation.
3x – 2x = 2x + 18 – 2x
Step 2: Simplify both sides of the equation.
x = 18
Step 3: Check the solution by substituting x with 18 in the original equation.
3(18) = 2(18) + 18
54 = 36 + 18
54 = 54 (This confirms the solution is correct.)
2. 5t – 3 = 3t – 5
Add 3 to both sides of the equation.
5t – 3 + 3 = 3t – 5 + 3
5t = 3t – 2
Subtract 3t from both sides of the equation.
5t – 3t = 3t – 3t – 2
2t = -2
Divide both sides by 2 to solve for t.
2t / 2 = -2 / 2
t = -1
Check the solution by substituting t with -1 in the original equation.
5(-1) – 3 = 3(-1) – 5
-5 – 3 = -3 – 5
-8 = -8 (This confirms the solution is correct.)
3. 5x + 9 = 5 + 3x
5x + 9 = 5 + 3x
5x – 3x + 9 = 5 (Subtract 3x from both sides)
2x + 9 = 5
2x = 5 – 9 (Subtract 9 from both sides)
2x = -4
x = -4 / 2 (Divide both sides by 2)
x = -2
5(-2) + 9 = 5 + 3(-2) (Check by substituting x = -2)
-10 + 9 = 5 – 6
-1 = -1 (True)
4. 4z + 3 = 6 + 2z
4z – 2z + 3 = 6 (Subtract 2z from both sides)
2z + 3 = 6
2z = 6 – 3 (Subtract 3 from both sides)
2z = 3
z = 3 / 2 (Divide both sides by 2)
z = 3/2
4(3/2) + 3 = 6 + 2(3/2) (Check by substituting z = 3/2)
6 + 3 = 6 + 3
9 = 9 (True)
5. 2x – 1 = 14 – x
2x + x – 1 = 14 (Add x to both sides)
3x – 1 = 14
3x = 14 + 1 (Add 1 to both sides)
3x = 15
x = 15 / 3 (Divide both sides by 3)
x = 5
2(5) – 1 = 14 – 5 (Check by substituting x = 5)
10 – 1 = 9
9 = 9 (True)
6. 8x + 4 = 3 (x – 1) + 7
8x + 4 = 3x – 3 + 7 (Expand the right side)
8x + 4 = 3x + 4 (Combine like terms on the right side)
8x – 3x = 4 – 4 (Subtract 3x from both sides and subtract 4 from both sides)
5x = 0
x = 0 / 5 (Divide both sides by 5)
x = 0
8(0) + 4 = 3(0 – 1) + 7 (Check by substituting x = 0)
4 = 3(-1) + 7
4 = 4 (True)
7. x = 4/5 (x + 10)
x = 4/5 * x + 8 (Expand the right side)
x – 4/5 * x = 8 (Subtract 4/5 * x from both sides)
1/5 * x = 8 (Combine like terms)
x = 8 / (1/5) (Divide both sides by 1/5)
x = 40
Check by substituting x = 40
40 = 4/5 (40 + 10)
40 = 40 (True)
8. (2x/3) + 1 = (7x/15) + 3
(2x/3) + 1 = (7x/15) + 3
2x/3 – 7x/15 = 3 – 1 (Subtract 7x/15 from both sides and subtract 1 from both sides)
10x/15 – 7x/15 = 2 (Combine like terms)
3x/15 = 2
x = 2 / (3/15) (Divide both sides by 3/15)
x = 10
Check by substituting x = 10
(210/3) + 1 = (710/15) + 3
20/3 + 1 = 14/3 + 3 (True)
9. 2y + 5/3 = 26/3 – y
2y + 5/3 = 26/3 – y
2y + y = 26/3 – 5/3 (Add y to both sides and subtract 5/3 from both sides)
3y = 21/3 (Combine like terms)
y = 21/3 / 3 (Divide both sides by 3)
y = 7
Check by substituting y = 7
2*7 + 5/3 = 26/3 – 7
14 + 5/3 = 26/3 – 7 (True)
10. 3m = 5 m – 8/5
3m = 5m – 8/5
3m – 5m = – 8/5 (Subtract 5m from both sides)
-2m = – 8/5 (Combine like terms)
m = – 8/5 / (-2) (Divide both sides by -2)
m = 4/5
Check by substituting m = 4/5
3*(4/5) = 5*(4/5) – 8/5
12/5 = 20/5 – 8/5 (True)
