This MCQ module is based on: Solving Equations Systematically
TOPIC 21 OF 23
Solving Equations Systematically
🎓 Class 7
Mathematics
CBSE
Theory
Ch 7 — Simple Equations
⏱ ~35 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Solving Equations Systematically
Targeting Class 7 level in General Mathematics, with Basic difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
7.2 Solving Equations Systematically
In Part 1 we framed equations. Now, how do we actually find the value of the unknown? One quick approach is trial and error?: keep guessing values of the letter-number until both sides become equal. Let us revisit the equation \(2n + 1 = 99\).
If we try \(n = 49\): LHS \(= 2 \times 49 + 1 = 99\). Equal to RHS. Done.
But what about \(5x - 4 = 7\)? Let us try a few values:
| Try \(x=\) | LHS = 5x − 4 | Matches RHS (=7)? |
|---|---|---|
| 1 | \(5-4=1\) | No, too small |
| 2 | \(10-4=6\) | Close, still small |
| 3 | \(15-4=11\) | Too big |
| 2.2 | \(11-4=7\) ✓ | Yes! |
It took several guesses. Trial and error works but is inefficient, especially for fractional answers. Let us learn a systematic method.
The golden rule — do the same to both sides
The Balance Principle for Equations
When an equation is solved, the value of the unknown does not change if we perform the same operation on both sides of the '=' sign. This is exactly like adding, removing or multiplying equal weights on both pans of a balance — the equality stays intact.
We can check whether adding, subtracting, multiplying or dividing by the same number on both sides keeps the equality. Let's verify with a numerical example.
Example 1 — Subtraction on both sides
It is known that \(14593 - 1459 + 145 - 14 = 13353\). What is the value of \(14593 - 1459 + 145 - 14^{\square}\)?
If we subtract 8 from both sides: \(14593 - 1459 + 145 - 14 - 8 = 13353 - 8 = 13345\). So \(14593 - 1459 + 145 - 22 = 13345\).
Example 2 — Multiplication on both sides
It is known that \(23 \div 41 \times 11 \times 8 \times 7 = 5{,}80{,}888\). We can multiply both sides by 9 to get \(23 \div 41 \times 11 \times 8 \times 7 \times 9 = 5{,}80{,}888 \times 9 = 52{,}27{,}992\).
🔵 Is the same as dividing both sides by 7? No — multiplying by 9 is not the same as dividing by 7. But dividing a product by 7 leaves the expression evaluable if 7 is a factor. This is how inverse operations help us isolate the unknown.
Example 3 — Fraction on both sides
It is known that \(12345 − 5432 × 133 − 24 = 6471 = 7091\). What is the value of the expression \(\bigl(\tfrac{35}{113}\bigr) \times 24 \times 14^{?}\)?
(Worked example in NCERT p. 170 illustrates the same principle — multiply both sides by the reciprocal to retain the factor you need.)
Solving equations — Example 4: \(5x - 4 = 7\)
Our aim: isolate \(x\) on one side.
| Step | Reason | Result |
|---|---|---|
| \(5x - 4 = 7\) | Original | — |
| \(5x - 4 + 4 = 7 + 4\) | Add 4 to both sides (inverse of \(-4\)) | \(5x = 11\) |
| \(\dfrac{5x}{5} = \dfrac{11}{5}\) | Divide by 5 on both sides (inverse of \(\times 5\)) | \(x = \dfrac{11}{5}\) |
Check: Substitute \(x = \tfrac{11}{5}\) into LHS: \(5 \times \tfrac{11}{5} - 4 = 11 - 4 = 7\) = RHS ✓. So \(x = \tfrac{11}{5}\) is the correct solution.
Example 5 — The variable carries a coefficient: \(11y - 5 = 61\)
To retain only the unknown term \(11y\) on LHS, remove the \(-5\) by adding 5 to both sides:
\(11y - 5 + 5 = 61 + 5 \Rightarrow 11y = 66 \Rightarrow y = \dfrac{66}{11} = 6\).
Check: \(11 \times 6 - 5 = 66 - 5 = 61\) ✓.
Example 6 — Variable on both sides: \(6y + 7 = 4y + 21\)
Strategy: collect variable terms on one side, constants on the other.
- Subtract \(4y\) from both sides: \(6y - 4y + 7 = 4y - 4y + 21\), i.e. \(2y + 7 = 21\).
- Subtract 7 from both sides: \(2y = 14\).
- Divide by 2: \(y = 7\).
Check: LHS \(= 6(7) + 7 = 49\). RHS \(= 4(7) + 21 = 49\) ✓.
Visualising Example 6 on a balance: keep both pans equal while peeling away common parts.
Example — Dividing both sides: \(\dfrac{u}{15} = 67\)
Multiplying both sides by 15 leaves only \(u\) on LHS: \(u = 67 \times 15 = 1005\).
Three Golden Observations
(a) When a term that is added or subtracted on one side of an equation is removed from that side, the inverse operation must be performed on the other side. e.g., \(2y + 7 = 21 \Rightarrow 2y = 21 - 7\).
(b) If the side of an equation is the product of two or more factors, then the side divided by one or more of the factors will equal the other side divided by the same factors, e.g., \(14 = 2y \Rightarrow y = 14 \div 2 = 7\).
(c) If one side of an equation is the quotient of two expressions, and we reverse the divisor, the other side must be multiplied by the divisor, e.g., \(\tfrac{u}{15} = 6 \Rightarrow u = 6 \times 15\).
Example 7 — Matchstick pattern (Ranjana's puzzle)
Ranjana arranges square tiles in a T-shape. In Step 1 she uses 4 tiles, Step 2 uses 7, Step 3 uses 10, and so on.
Ranjana's T-pattern: Step \(k\) uses \(3k+1\) tiles.
To check that Step \(k\) uses \(3k+1\) tiles: Step 1 \(\to 3(1)+1 = 4\) ✓, Step 2 \(\to 3(2)+1 = 7\) ✓, Step 3 \(\to 3(3)+1 = 10\) ✓.
Which step of the sequence will use 100 tiles? Solve \(3k + 1 = 100\): subtract 1, get \(3k = 99\); divide by 3, get \(k = 33\). So Step 33 uses exactly 100 tiles.
Example 8 — Madhubanti's Snack Party
Madhubanti wants to organise a party. She will deliver snacks by buying plates of snacks costing ₹25 each and paying a fixed delivery charge of ₹50 for the sacks. Her budget is ₹500.
Let \(p\) = number of plates. The equation is:
\(25p + 50 = 500\)
Step 1: Subtract 50 from both sides: \(25p = 450\). Step 2: Divide by 25: \(p = 18\).
So Madhubanti can afford 18 plates. If each family member has one plate, she can invite \(18 - 5 = 13\) friends (removing herself and 5 family members).
Example 9 — Srikanth's Variant
Srikanth represents the unknown quantity of friends as \(f\). The equation becomes \(25(f+5) = 450\). Dividing both sides by 25: \(f + 5 = 18\). Subtracting 5: \(f = 13\). Same answer — the algebra is blind to how you choose the variable!
Example 10 — Savings equality
Two friends want to save money. Jahnavi starts with ₹4000 and adds ₹650 every month. Sunita starts with ₹5050 and adds ₹500 every month. After how many months \(m\) will their savings be equal?
Equation: \(4000 + 650m = 5050 + 500m\). Subtract \(500m\) from both sides: \(4000 + 150m = 5050\). Subtract 4000: \(150m = 1050\). Divide by 150: \(m = 7\). So after 7 months their savings match (₹8550 each).
🧪 Activity: The Number-Cruncher Pair Game
Materials: Paper, pencil, a pair of partners
Predict: Can you solve your partner's equation in under 60 seconds using only "do the same to both sides"?
- Each partner writes one equation involving one variable (e.g., \(3x + 5 = 26\), \(7 - 2y = -1\), \(\tfrac{u}{4} - 3 = 8\)).
- Exchange equations.
- Solve the received equation in a timed 60 seconds. You must write every step and the reason (add, subtract, multiply, divide) beside it.
- Swap back, check each other's solutions by substitution.
- Score: 2 points for correct answer with correct reasoning; 1 point for correct answer only; 0 for wrong.
\(3x+5=26 \Rightarrow 3x=21 \Rightarrow x=7\). \(7-2y=-1 \Rightarrow -2y=-8 \Rightarrow y=4\). \(\tfrac{u}{4}-3=8 \Rightarrow \tfrac{u}{4}=11 \Rightarrow u=44\).
Figure it Out (Section 7.2)
Q1. Solve and check: (a) \(3x - 10 = 35\) (b) \(5x = 3s\) (where \(s = 10\)) (c) \(3a = 7 \times 2a + 3\) (d) \(4(m+6) - 8 = 2m - 4\) (e) \(\tfrac{x}{15} = 6\).
(a) \(3x=45,\ x=15\). (b) \(s=10 \Rightarrow 5x=30,\ x=6\). (c) \(3a-14a=3 \Rightarrow -11a=3,\ a=-\tfrac{3}{11}\). (d) \(4m+16=2m+4 \Rightarrow 2m=-12,\ m=-6\). (e) \(x = 6\times15 = 90\).
Q2. Frame an equation that has no solution.
Any equation of the form \(x + a = x + b\) with \(a \neq b\). For example, \(2x + 7 = 2x + 9\) leads to \(7 = 9\), which is false — no solution.
Competency-Based Questions
Scenario: A mobile recharge pack costs ₹149 per month for the first three months and ₹99 per month afterwards. Nisha wants to calculate how many months \(m\) she must stay with the company so that her total spend equals ₹1{,}000.
Q1. Write and solve the correct equation.
L3 ApplyAnswer: First three months cost \(3 \times 149 = ₹447\). Remaining months are \(m - 3\) at ₹99 each. So: \(447 + 99(m - 3) = 1000\). Simplify: \(99(m-3) = 553 \Rightarrow m - 3 = \tfrac{553}{99} \approx 5.59\). So \(m \approx 8.59\) — she must recharge at least 9 months to cross ₹1000.
Q2. Nisha's friend claims the equation \(149m + 99 \times 3 = 1000\) is equally valid because addition is commutative. Analyse whether this is correct.
L4 AnalyseAnswer: Incorrect. Commutativity of addition does not let you swap which months cost ₹149 vs ₹99. The friend's equation would mean all \(m\) months cost ₹149 and exactly 3 months cost ₹99 — a different real-world story. Modelling must match the physical situation.
Q3. Evaluate: If the company raises the promotional rate to ₹159 for the first three months, does Nisha need more or fewer months to reach ₹1000? By how many months?
L5 EvaluateAnswer: New initial cost = 3 × 159 = ₹477. Equation: \(477 + 99(m-3) = 1000 \Rightarrow 99(m-3) = 523 \Rightarrow m - 3 \approx 5.28\), so \(m \approx 8.28\). She now crosses ₹1000 slightly earlier (about a third of a month), but practically still in 9 months. Increase of ₹30 in fixed cost ≈ 0.3 months sooner.
Q4. Create a word problem that leads to the equation \(15m + 80 = 5m + 200\) and solve it.
L6 CreateSample: "Arjun saves ₹15 per month and already has ₹80. Bhavya saves ₹5 per month and already has ₹200. After how many months will their savings match?" Solve: \(15m + 80 = 5m + 200 \Rightarrow 10m = 120 \Rightarrow m = 12\) months.
Assertion–Reason Questions
A: To solve \(11y - 5 = 61\), one correct first step is to add 5 to both sides.
R: Adding the same number to both sides of an equation keeps it balanced.
R: Adding the same number to both sides of an equation keeps it balanced.
Answer: (a) — Both statements are true and R explains A.
A: The equation \(\tfrac{u}{15} = 67\) can be solved by multiplying both sides by 15.
R: Multiplication is the inverse of division, so multiplying by 15 undoes the \(\div 15\) on the left.
R: Multiplication is the inverse of division, so multiplying by 15 undoes the \(\div 15\) on the left.
Answer: (a) — R is the exact mathematical justification for A.
A: Trial and error is the most efficient way to solve linear equations.
R: Any equation can be solved by checking a few integer values of the variable.
R: Any equation can be solved by checking a few integer values of the variable.
Answer: Both are false. Trial and error is inefficient and fails when the solution is a fraction. The systematic balance method is efficient and always works.
Frequently Asked Questions
What is the balancing method?
The balancing method applies the same arithmetic operation to both sides of the equation so the equality stays true. It reflects the fact that a balanced scale remains balanced when equal quantities are added to or removed from both sides.
What is transposition?
Transposition is a shortcut in which a term moved from one side of the equation to the other has its sign reversed: addition becomes subtraction, and multiplication becomes division.
How do you solve 2x + 5 = 17?
Subtract 5 from both sides: 2x = 12. Divide by 2: x = 6. Verify: 2 times 6 plus 5 equals 17, which matches.
Why do we verify the solution?
Verification confirms the answer by substituting it into the original equation. If both sides are equal, the solution is correct; if not, there was an error in the steps.
Can an equation have no solution?
Yes. Some equations simplify to a false statement such as 0 = 5, which means no value of x makes the original equation true.
AI Tutor
Mathematics Class 7 — Ganita Prakash-II
Ready