MathematicsClass 8Some examples of expressions we have so far worked with are

Some examples of expressions we have so far worked with are | Class 8 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 5 min read

Some examples of expressions we have so far worked with are | Class 8 Mathematics Notes

Some examples of expressions we have so far worked with are – this guide gives you a concise, exam-ready overview of Some examples of expressions we have so far worked with are from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

2.2 Solving Equations having the Variable on both Sides

An equation is an equality between two expressions. In many cases, the right-hand side (RHS) of an equation is a number, but sometimes both sides contain expressions with variables. For example, the equation 2x - 3 = x + 2 has variables on both sides. Here, the LHS is 2x - 3 and the RHS is x + 2.

To solve such equations, we perform algebraic operations to isolate the variable on one side. The key principle is to maintain equality by performing the same operation on both sides.

Example 1: Solve 2x - 3 = x + 2.

Step 1: Add 3 to both sides to eliminate the constant term on the LHS: 2x - 3 + 3 = x + 2 + 3 which simplifies to: 2x = x + 5

Step 2: Subtract x from both sides to bring variable terms to one side: 2x - x = x + 5 - x which simplifies to: x = 5

Thus, the solution is x = 5.

Example 2: Solve 5x + 7/2 = (3/2)x - 14.

Step 1: Multiply both sides by 2 (the LCM of denominators) to clear fractions: 2 × (5x + 7/2) = 2 × ((3/2)x - 14) which simplifies to: 10x + 7 = 3x - 28

Step 2: Transpose 3x to the LHS by subtracting 3x from both sides: 10x - 3x + 7 = -28 which simplifies to: 7x + 7 = -28

Step 3: Subtract 7 from both sides: 7x = -28 - 7 which simplifies to: 7x = -35

Step 4: Divide both sides by 7: x = -5

Thus, the solution is x = -5.

This section emphasizes that variables can be transposed from one side to the other by performing inverse operations, and fractions can be cleared by multiplying both sides by the LCM of denominators. These techniques simplify the solving process.

📊 Diagram: Figure on page 2 illustrating balance and operations on equations.

🧪 Activity: No specific activity; includes worked examples demonstrating solving equations with variables on both sides.

🔗 Connection: Prepares for the next section on reducing equations to simpler forms by opening brackets and combining like terms.

Frequently asked questions

Solve the following equations and check your results. 1. $3x = 2x + 18$ 2. $5x - 3 = 3x - 5$ 3. $5x + 9 = 5 + 3x$ 4. $4x + 3 = 6 + 2x$ 5. $2x - 1 = 14 - x$ 6. $8x + 4 = 3(x - 1) + 7$ 7. $x = \frac{4}{5}(x + 10)$ 8. $\frac{2x}{3} + 1 = \frac{7x}{15} + 3$ 9. $2x + \frac{5}{3} = \frac{26}{3} - x$ 10. $3x = 5x - \frac{8}{5}$

1. Solve $3x = 2x + 18$: Subtract $2x$ from both sides: $3x - 2x = 18$ $x = 18$ Check: LHS = $3 \times 18 = 54$ RHS = $2 \times 18 + 18 = 36 + 18 = 54$ LHS = RHS

2. Solve $5x - 3 = 3x - 5$: Bring variables to one side and constants to other: $5x - 3x = -5 + 3$ $2x = -2$ $x = -1$ Check: LHS = $5(-1) - 3 = -5 - 3 = -8$ RHS = $3(-1) - 5 = -3 - 5 = -8$ LHS = RHS

3. Solve $5x + 9 = 5 + 3x$: $5x - 3x = 5 - 9$ $2x = -4$ $x = -2$ Check: LHS = $5(-2) + 9 = -10 + 9 = -1$ RHS = $5 + 3(-2) = 5 - 6 = -1$ L

Solve the following linear equations. 1. \(\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\) 2. \(\frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21\) 3. \(x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{2}\) 4. \(\frac{x - 5}{3} = \frac{x - 3}{5}\) 5. \(\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t\) 6. \(m - \frac{m - 1}{2} = 1 - \frac{m - 2}{3}\) Simplify and solve the following linear equations. 7. \(3(t - 3) = 5(2t + 1)\) 8. \(15(y - 4) - 2(y - 9) + 5(y + 6) = 0\) 9. \(3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17\) 10. \(0.25(4f - 3) = 0.05(10f - 9)\)

1. Solve \(\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\) Multiply both sides by 60 (LCM of 2,5,3,4): \(30x - 12 = 20x + 15\) \(30x - 20x = 15 + 12\) \(10x = 27\) \(x = \frac{27}{10} = 2.7\)

2. Solve \(\frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21\) LCM of 2,4,6 is 12: \(6n - 9n + 10n = 252\) \(7n = 252\) \(n = 36\)

3. Solve \(x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{2}\) Multiply both sides by 6: \(6x + 42 - 16x = 17 - 15x\) \(6x - 16x + 15x = 17 - 42\) \(5x = -25\) \(x = -5\

Which of the following is a linear expression in one variable?

2x + 3

Identify the parts labeled in the algebraic expression $3x + 5$ where '3' is the coefficient, 'x' is the variable, and '5' is the constant.

The number 3 is the coefficient, 'x' is the variable, and 5 is the constant in the expression $3x + 5$.

Ready to ace this chapter?

Get the full Some examples of expressions we have so far worked with are chapter — interactive notes, diagrams, worked solutions, polls and a free practice quiz — in the ConceptScroll app.

Open in ConceptScroll →

Study smarter with ConceptScroll

Daily NCERT-aligned reels, AI doubt solving and chapter quizzes — all free.

Start learning free
#cbse notes#class 8#mathematics#ncert

Continue reading