Finding The | Class 7 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 3 min read
Finding The – this guide gives you a concise, exam-ready overview of Finding The from Class 7 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
More on Solving Equations
This section extends the concept of solving equations to more complex forms, including those with variables on both sides and involving brackets. The chapter explains how to simplify both sides of the equation first by removing brackets using distributive property, combining like terms, and then applying the properties of equality to solve. For example, in the equation 3(x + 2) = 2x + 9, first expand the bracket: 3x + 6 = 2x + 9. Then bring variables to one side and constants to the other: 3x - 2x = 9 - 6 ⇒ x = 3. The chapter also discusses equations where the variable cancels out, leading to either no solution or infinite solutions. For instance, 2x + 3 = 2x + 5 has no solution because 3 ≠ 5, while 3x + 4 = 3x + 4 has infinite solutions. The section provides detailed examples and encourages students to verify solutions by substitution. This deepens understanding of equation types and solution sets.
📊 Diagram: No specific diagrams; solution steps illustrated through examples.
🧪 Activity: No specific activity in this section.
🔗 Connection: Leads to the next section where students apply equation-solving skills to word problems.
Frequently asked questions
We have the expression 3k + 1 which gives the number of tiles needed to make an arrangement in Step k. To check whether an arrangement is possible using 100 tiles at some Step k, we can solve the equation: 3k + 1 = 100. Find the value of k.
Given the equation 3k + 1 = 100, Subtract 1 from both sides: 3k = 100 - 1 3k = 99 Divide both sides by 3: k = 99 ÷ 3 k = 33 Therefore, the arrangement using 100 tiles corresponds to Step 33.
Madhubanti wants to organise a party. She decides to buy snacks for the party from the chaat shop in town. Each plate of snacks costs ₹25. The shop charges an additional fixed amount of ₹50 to deliver the snacks to Madhubanti’s house. There are 5 members in Madhubanti’s family, including herself. Her parents tell her she can spend ₹500 on this party. How many friends can she invite to the party if she wants to give a plate of snacks to each person, including her family and friends?
Total money available = ₹500 Delivery charge = ₹50 Money left for snacks = 500 - 50 = ₹450 Cost per plate = ₹25 Number of plates that can be bought = 450 ÷ 25 = 18 Number of family members = 5 Number of friends invited = 18 - 5 = 13 Therefore, Madhubanti can invite 13 friends.
Two friends want to save money. Jahnavi starts with an initial amount of ₹4000, and in addition, saves ₹650 per month. Sunita starts with ₹5050 and saves ₹500 per month. After how many months will they have the same amount of money?
Let m be the number of months after which their savings are equal. Jahnavi's savings after m months = 4000 + 650m Sunita's savings after m months = 5050 + 500m Set equal: 4000 + 650m = 5050 + 500m Subtract 500m from both sides: 4000 + 150m = 5050 Subtract 4000 from both sides: 150m = 1050 Divide both sides by 150: m = 1050 ÷ 150 = 7 Therefore, after 7 months, both will have the same amount of money.
Solve 28 (x + 4) + 300 = 1000.
Given equation: 28(x + 4) + 300 = 1000 Step 1: Subtract 300 from both sides: 28(x + 4) = 1000 - 300 28(x + 4) = 700 Step 2: Divide both sides by 28: x + 4 = 700 ÷ 28 x + 4 = 25 Step 3: Subtract 4 from both sides: x = 25 - 4 x = 21 Therefore, the solution is x = 21.
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