Equation — Questions and Answers
Welcome to HSLC Guru. This guide covers ASSEB Class 7 New Mathematics Chapter 15, Equation — the concept of equality, forming equations from daily-life situations, solving simple linear equations by different methods, and verifying the solution. It gives full worked answers to every “Work it Out” box, every “Try yourself” box, worked Examples 1–17, all of Exercise 15 and Puzzles 1–3.
Summary
When both pans of a weighing balance carry the same weight, the balance stays level — this is the idea of equality. In the same way, a statement of equality between two algebraic expressions, or between an algebraic expression and an arithmetic expression, is called an equation. For example $n + 5 = 12$, $10x = 100$ and $2n – 1 = 51$. An unknown quantity is denoted by a letter such as $x, n, y, z, l, m, p$.
The part on the left of the ‘=’ sign is the Left Hand Side (LHS) and the part on the right is the Right Hand Side (RHS). The two sides become equal only for one particular value of the unknown; that value is the solution of the equation. This chapter solves equations by the trial-and-error method, the symbolic method, and the addition-subtraction and multiplication-division methods.
Applying the same operation (adding, subtracting, multiplying, or dividing by a non-zero number) to both sides keeps the equality unchanged — these axioms are the basis of solving equations. Indian mathematicians Aryabhatta, Brahmagupta and Bhaskaracharya set out systematic ways to solve equations, and the modern word “algebra” comes from “al-jabr”.
Summary: This ASSEB Class 7 New Mathematics Chapter 15 (Equation) guide explains equality using a balance scale, defines an equation, and forms equations from daily-life situations. It solves every Work it Out box (15.1–15.4), all Try-yourself boxes, worked Examples 1–17 and the full Exercise 15 using the trial-and-error, symbolic, addition-subtraction and multiplication-division methods, verifying each solution, and covers the brief history of equations and Puzzles 1–3.
Equality, the Balance and an Equation
The balance above is level because both sides carry the same weight — just as the LHS and RHS of an equation are equal. Adding, subtracting, multiplying, or dividing (by a non-zero number) both sides by the same number keeps this equality; this is exactly how we isolate the unknown and solve an equation.
Textbook Questions and Answers
15.1 Concept of Equality — Think yourself
Some biscuits are removed from one side of the weighing scale. Will the scale still remain balanced? What is the minimum weight that must be added or removed to balance the scales in Figure 15.2?
Answer: No. Removing biscuits from one side makes that side lighter, so the balance no longer stays level — the lighter side rises. To rebalance, either add a weight equal to the difference to the lighter side or remove a weight equal to the difference from the heavier side. So the minimum weight to add or remove equals the difference between the two sides.
15.2 Unknown Value — Try it out (Figure 15.4)
The number written on top of a balanced scale gives the total weight on both sides. Find the unknown objects’ values in Figure 15.4 (a)–(f).
Answer: Since each scale is balanced, both sides are equal, and the number on top is the total of both sides. So half of the total is the weight of one side. Subtracting the known objects from this gives the unknown object. For example, in Figure 15.3 a total of 10 means one side is $5 + 5$, and a total of 11 means $6 + 5$. In the same way, for each scale in Figure 15.4 subtract the known weights on one side from half the given total (18, 24, 36, 50, …) to find the value of each unknown object.
15.3 Basic Idea of Equations — Marble pattern (Think yourself)
In the marble pattern the number of marbles in the $n$-th position is $(2n – 1)$. In which position will there be 51 marbles?
Answer: Let the $n$-th position have 51 marbles. Then $2n – 1 = 51$. Adding 1 to both sides, $2n = 52$, and dividing both sides by 2, $n = 26$. So the 26th position will have 51 marbles.
Forming equations (Example 1): (a) 7 subtracted from 5 times $x$ gives 8. (b) Adding 2 to one-fifth of a number $t$ gives 3. Form the equations.
Answer: (a) 5 times $x$ is $5x$; subtracting 7 gives $5x – 7$; this equals 8. So the equation is $5x – 7 = 8$. (b) One-fifth of $t$ is $\frac{t}{5}$; adding 2 gives $\frac{t}{5} + 2$; this equals 3. So the equation is $\frac{t}{5} + 2 = 3$.
Find the unknowns in the examples of 15.3: (a) $n + 5 = 12$, (b) $10x = 100$, (c) $2n – 1 = 51$.
Answer: (a) Writing 12 as $7 + 5$ gives $n + 5 = 7 + 5$; removing 5 from both sides gives $n = 7$. (b) Writing 100 as $10 \times 10$ gives $10x = 10 \times 10$; dividing both sides by 10 gives $x = 10$. (c) $2n – 1 = 51 = 52 – 1$; removing $-1$ from both sides gives $2n = 52$; since $52 = 2 \times 26$, dividing by 2 gives $n = 26$.
Work it Out 15.1
1. Express the following as equations: (a) 1 is subtracted from one-third of a number, the result is 2. (b) Four times of $p$ is 20. (c) To get 40, a number is divided by 10 and then 10 is subtracted. (d) 5 is added to five times of a number, giving 20.
Answer: Taking the number as $x$ — (a) $\frac{x}{3} – 1 = 2$; (b) $4p = 20$; (c) $\frac{x}{10} – 10 = 40$; (d) $5x + 5 = 20$.
2. The total cost of a book and a geometry box is ₹350. If the book costs ₹$x$ and the geometry box ₹50, which equation gives $x$? (A) $50x = 350$ (B) $x + 50 = 350$ (C) $x – 50 = 350$ (D) $\frac{x}{50} = 350$
Answer: (B) $x + 50 = 350$. The sum of the two costs is the total, so $x + 50 = 350$ (giving $x = 300$).
3. Anjan’s grandfather is 72 years old, which is 2 years more than seven times Anjan’s age. Which equation is suitable? (A) $7x = 72$ (B) $7x – 2 = 72$ (C) $\frac{x}{7} = 72$ (D) $7x + 2 = 72$
Answer: (D) $7x + 2 = 72$. If Anjan’s age is $x$, seven times is $7x$, and 2 more is $7x + 2$, which equals 72.
4. Create a daily-life problem statement for each equation: (a) $5m = 10$ (b) $\frac{2}{3}t = 8$ (c) $8x – 3 = 5$ (d) $p – 5 = 2$.
Answer: (sample) (a) The total cost of $m$ pens at ₹5 each is ₹10, so $5m = 10$. (b) Two-thirds of a bottle holds 8 litres, so $\frac{2}{3}t = 8$. (c) After eating 3 of eight times $x$ laddus, 5 remain, so $8x – 3 = 5$. (d) After giving away 5 of $p$ mangoes, 2 are left, so $p – 5 = 2$.
5. Each Class VII student decided to give ₹10 for flood relief, and the teacher gave ₹50. Altogether ₹350 was collected. Form an equation for the number of students.
Answer: With $x$ students, the collection from them is $10x$, plus ₹50 gives $10x + 50$. So the equation is $10x + 50 = 350$ (giving $10x = 300$, $x = 30$ students).
15.4 Solving Equations — Flour packet (Figure 15.5)
In Figure 15.5 a flour packet with a 1 kg weight on one side, and a 1 kg weight with a 5 kg weight on the other, keep the scale balanced. Find the weight of the flour packet.
Answer: Let the flour weigh $x$ kg, so $x + 1 = 5 + 1$. Removing an equal 1 kg from both sides does not change the balance, so $x = 5$. The flour packet weighs 5 kg.
15.4.1 Trial and Error Method
Solve $x – 5 = 1$ by the trial and error method.
Answer: Substitute different values in LHS $= x – 5$ — $x = 1$: $-4$; $x = 2$: $-3$; $x = 3$: $-2$; $x = 4$: $-1$; $x = 5$: $0$; $x = 6$: $6 – 5 = 1 = $ RHS; $x = 7$: $2$ (greater than RHS). Only $x = 6$ makes LHS = RHS, so the solution is $x = 6$. (For a value less than 6 the LHS decreases; for a value greater than 6 it increases.)
Try yourself: Complete the following table.
Answer:
- $x – 3 = 4$, $x = 2$ → $2 – 3 = -1 \ne 4$, so not a solution (correct solution $x = 7$).
- $-2 = m + 4$, $m = -6$ → $-6 + 4 = -2$, so yes, a solution.
- $\frac{y}{5} = 4$ → for it to be a solution $y = 20$ (since $\frac{20}{5} = 4$).
- $10 = 2t$ → for it to be a solution $t = 5$ (since $2 \times 5 = 10$).
- $p + 1 = 2$ has the correct solution $p = 1$; any other value (say $p = 0$) gives $0 + 1 = 1 \ne 2$, so for that value it is not a solution.
- An equation whose solution is $t = 4$ — for example $2t = 8$ (since $2 \times 4 = 8$).
Solve $10x + 50 = 500$.
Answer: Trying values — $x = 1$: $60 \ne 500$; $x = 10$: $150$; $x = 50$: $550$ (greater than 500); $x = 45$: $10 \times 45 + 50 = 500 = $ RHS. So the solution is $x = 45$. (The symbol ‘$\ne$’ means “not equal”.)
Work it Out 15.2
1. Solve by trial and error: (a) $6p + 3 = 27$ (b) $\frac{t}{2} + 1 = 5$ (c) $3x – 4 = 5$ (d) $\frac{t}{2} – 2 = 0$.
Answer: (a) $p = 4$ ($6 \times 4 + 3 = 27$). (b) $t = 8$ ($\frac{8}{2} + 1 = 5$). (c) $x = 3$ ($3 \times 3 – 4 = 5$). (d) $t = 4$ ($\frac{4}{2} – 2 = 0$).
2. Does the value in brackets satisfy the equation? (a) $x + 6 = 0$, $(x = -6)$ (b) $\frac{z}{5} = 20$, $(z = 100)$ (c) $7 = 2x – 8$, $(x = 4)$ (d) $\frac{n}{3} + 6 = 8$, $(n = 3)$ (e) $4p + 8 = 4$, $(p = -1)$.
Answer: (a) $-6 + 6 = 0$ — Yes. (b) $\frac{100}{5} = 20$ — Yes. (c) $2 \times 4 – 8 = 0 \ne 7$ — No. (d) $\frac{3}{3} + 6 = 7 \ne 8$ — No. (e) $4 \times (-1) + 8 = 4$ — Yes.
15.4.2 Solving Using Symbols
Solve $x – 5 = 6$ using symbols. (‘+’ stands for $+1$, ‘−’ for $-1$, and a pair of ‘+’ and ‘−’ represents a zero.)
Answer: To remove the five ‘−’ from $x$, add five ‘+’ to both sides. On the LHS the five ‘−’ and five ‘+’ cancel (become zero), leaving only $x$; on the RHS six ‘+’ plus five ‘+’ make eleven ‘+’. So $x = 11$.
Try yourself: Solve using symbols — (a) $x + 3 = 12$ (b) $x – 2 = -3$ (c) $5 + x = -6$ (d) $8 = x – 2$.
Answer: (a) Remove 3 from both sides: $x = 9$. (b) Add 2 to both sides: $x = -1$. (c) Remove 5 from both sides: $x = -11$. (d) $8 + 2 = x$, so $x = 10$.
15.4.3 Using Addition and Subtraction
Example 2: Given $4123 – 512 – 659 + 36 – 94 = 2894$, find $4123 – 512 – 659 + 36 = ?$ without simplifying.
Answer: To remove $-94$ from the LHS, add 94 to it; to keep equality, add 94 to both sides (addition and subtraction are inverse operations). $4123 – 512 – 659 + 36 – 94 + 94 = 2894 + 94$, so $4123 – 512 – 659 + 36 = \textbf{2988}$.
Try yourself: Given $10298 – 6502 – 521 + 14 + 89 = 3378$, find $10298 – 6502 – 521 + 14 = ?$
Answer: To remove $+89$ from the LHS, subtract 89 from both sides. So $10298 – 6502 – 521 + 14 = 3378 – 89 = \textbf{3289}$.
Example 3: (a) Solve $x – 3 = 4$ and verify. (b) Solve $t + \frac{3}{4} = \frac{1}{4}$.
Answer: (a) Add 3 to both sides: $x – 3 + 3 = 4 + 3$, so $x = 7$. Check: LHS $= 7 – 3 = 4 = $ RHS — correct. (b) Subtract $\frac{3}{4}$ from both sides: $t = \frac{1}{4} – \frac{3}{4} = \frac{-2}{4} = -\frac{1}{2}$. Check: $-\frac{1}{2} + \frac{3}{4} = \frac{1}{4} = $ RHS — correct.
Try yourself: Solve — (a) $x + 5 = -6$ (b) $y – 8 = -1$ (c) $l + 16 = 5$ (d) $m + 9 = 13$.
Answer: (a) $x = -11$. (b) $y = 7$. (c) $l = -11$. (d) $m = 4$.
Try yourself: Solve — (a) $x + \frac{5}{7} = \frac{1}{14}$ (b) $y – \frac{1}{6} = \frac{5}{6}$ (c) $l – \frac{1}{4} = \frac{1}{4}$ (d) $m – 6 = -\frac{8}{9}$.
Answer: (a) $x = \frac{1}{14} – \frac{5}{7} = \frac{1 – 10}{14} = -\frac{9}{14}$. (b) $y = \frac{5}{6} + \frac{1}{6} = 1$. (c) $l = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$. (d) $m = 6 – \frac{8}{9} = \frac{46}{9}$.
15.4.4 Using Multiplication and Division
Example 4: Given $\frac{24}{315} \times 15 \times 32 \times \frac{7}{11} = \frac{80640}{3465}$, find $\frac{24}{315} \times 15 \times 32 = ?$
Answer: To remove $\frac{7}{11}$ (which is multiplied), divide both sides by $\frac{7}{11}$ (multiplication and division are inverse). So $\frac{24}{315} \times 15 \times 32 = \frac{80640}{3465} \times \frac{11}{7} = \frac{80640 \times 11}{3465 \times 7} = \frac{2304}{63}$.
Try yourself: Given $\frac{53}{87} \times \frac{18}{7} \times (-16) \times \frac{3}{5} = \frac{45792}{3045}$, find $\frac{53}{87} \times \frac{18}{7} \times (-16) = ?$
Answer: Divide both sides by $\frac{3}{5}$, i.e. multiply by $\frac{5}{3}$: $\frac{45792}{3045} \times \frac{5}{3} = \frac{45792 \times 5}{3045 \times 3} = \frac{45792}{1827} = \frac{15264}{609}$.
Example 5: Solve and verify (a) $3x = 6$ (b) $\frac{x}{4} = \frac{1}{2}$.
Answer: (a) Divide both sides by 3: $\frac{3x}{3} = \frac{6}{3}$, so $x = 2$. Check: $3 \times 2 = 6 = $ RHS — correct. (b) Multiply both sides by 4: $\frac{x}{4} \times 4 = \frac{1}{2} \times 4$, so $x = 2$. Check: $\frac{2}{4} = \frac{1}{2} = $ RHS — correct.
Find the error: $3x = 6$, or $x = 6 – 3$, or $x = 3$.
Answer: Since 3 and $x$ are multiplied, to isolate $x$ we must divide both sides by 3 ($x = \frac{6}{3} = 2$), but the working wrongly subtracts 3. The correct solution is $x = 2$.
Let us know — Axioms of equations
- Adding/subtracting the same number on both sides keeps the equality: if $a = b$, then $a + c = b + c$ and $a – c = b – c$.
- Multiplying both sides by the same number keeps the equality: if $a = b$, then $a \times c = b \times c$.
- Dividing both sides by the same non-zero number keeps the equality: if $a = b$, then $a \div c = b \div c$ ($c \ne 0$).
- [These postulates appear in Bhaskaracharya’s chapter ‘Ekavarnasamikaranam’.]
Work it Out 15.3
1. Solve and verify: (a) $x – 7 = 0$ (b) $z + 6 = -5$ (c) $7x = 35$ (d) $\frac{x}{7} = 5$ (e) $12p = -36$ (f) $\frac{t}{3} = 9$.
Answer: (a) $x = 7$ (check $7 – 7 = 0$). (b) $z = -11$ (check $-11 + 6 = -5$). (c) $x = \frac{35}{7} = 5$ (check $7 \times 5 = 35$). (d) $x = 5 \times 7 = 35$ (check $\frac{35}{7} = 5$). (e) $p = \frac{-36}{12} = -3$ (check $12 \times (-3) = -36$). (f) $t = 9 \times 3 = 27$ (check $\frac{27}{3} = 9$).
2. Bapukan’s mother gave him ₹100 on Children’s Day. He bought a park entry ticket at a discount of ₹10 on its actual price, and a sapling at ₹30. He then had ₹20 left. If the actual price of the ticket is $P$, form the equation and find $P$.
Answer: The discounted ticket costs $(P – 10)$; the sapling ₹30; ₹20 remains. So $(P – 10) + 30 + 20 = 100$, i.e. $P + 40 = 100$, giving $P = 60$. (Check: ticket $60 – 10 = 50$, sapling 30, spent 80, left $100 – 80 = 20$.)
3. Two-thirds of a beaker is filled with water. If the water-filled part is 4 cm high, form and solve an equation to find the height of the beaker.
Answer: Let the height be $h$ cm, so $\frac{2}{3}h = 4$. Multiplying both sides by $\frac{3}{2}$, $h = 4 \times \frac{3}{2} = 6$. The beaker is 6 cm tall.
A few more equations (Example 6)
Solve [a] $3x + 7 = 11$ [b] $-2 = 4 + 3t$ [c] $\frac{3}{4}x = 12$.
Answer: [a] Subtract 7: $3x = 4$; divide by 3: $x = \frac{4}{3}$. Check: $3 \times \frac{4}{3} + 7 = 11$ — correct. (Trial and error is not convenient here as the solution is a fraction.) [b] Subtract 4: $-6 = 3t$; divide by 3: $t = -2$. (Writing it as $4 + 3t = -2$ gives the same solution.) [c] Divide by $\frac{3}{4}$, i.e. multiply by $\frac{4}{3}$: $x = 12 \times \frac{4}{3} = 16$.
15.4.5 Unknowns on Both Sides (Example 7)
Solve [a] $2 + 3x = 4x + 5$ [b] $\frac{t}{2} + 5 = 1 – \frac{t}{2}$ [c] $4(z – 3) = 8z + 9$.
Answer: [a] Subtract $4x$: $2 – x = 5$; subtract 2: $-x = 3$; multiply by $(-1)$: $x = -3$. Check: LHS $= 2 + 3(-3) = -7$, RHS $= 4(-3) + 5 = -7$ — correct. [b] Add $\frac{t}{2}$: $t + 5 = 1$; subtract 5: $t = -4$. [c] $4(z – 3) = 4z – 12$, so $4z – 12 = 8z + 9$; subtract $8z$: $-4z – 12 = 9$; add 12: $-4z = 21$; multiply by $(-1)$: $4z = -21$; divide by 4: $z = -\frac{21}{4}$.
Work it Out 15.4
1. Solve: (a) $6x + 5 = 2x + 17$ (b) $\frac{l}{2} = \frac{l}{3} – 1$ (c) $13 – m = 3m + 5$ (d) $p + 5 = 2(1 – 2p)$.
Answer: (a) $6x – 2x = 17 – 5$ → $4x = 12$ → $x = 3$. (b) $\frac{l}{2} – \frac{l}{3} = -1$ → $\frac{l}{6} = -1$ → $l = -6$. (c) $13 – 5 = 3m + m$ → $8 = 4m$ → $m = 2$. (d) $p + 5 = 2 – 4p$ → $5p = -3$ → $p = -\frac{3}{5}$.
Example 8 — Two columns of one solution
$5x + 12 = 22$ is solved in Column A and Column B. Observe the difference.
| Column A (detailed) | Column B (short) |
|---|---|
| $5x + 12 – 12 = 22 – 12$ | $5x = 22 – 12$ |
| $5x = 10$ | $5x = 10$ |
| $\frac{5x}{5} = \frac{10}{5}$ | $x = \frac{10}{5}$ |
| $x = 2$ | $x = 2$ |
Answer: In Column A every step is written out. Since 12 and $-12$ are additive inverses ($12 + (-12) = 0$), we can write directly $5x = 22 – 12$ as in Column B. Both give the same solution, $x = 2$. (Once the method is clear, follow the shorter Column B procedure.)
Application examples (Example 9–17)
Example 9: Solve $5x – 8 = 2x + 7$.
Answer: $5x – 8 – 2x = 7$ → $3x – 8 = 7$ → $3x = 15$ → $x = 5$.
Example 10: The sum of three consecutive odd numbers is 75. Find the numbers.
Answer: Consecutive odd numbers increase by 2. If the first is $x$, the others are $x + 2$ and $x + 4$. $x + (x + 2) + (x + 4) = 75$ → $3x + 6 = 75$ → $3x = 69$ → $x = 23$. The numbers are 23, 25, 27. (Consecutive natural numbers increase by 1, consecutive even numbers by 2.)
Example 11: On World Environment Day, some students and three teachers of Bangaon Higher Secondary School went to Manas Sanctuary. The school gave ₹2000. They spent ₹200 on entry fees and ₹60 each on food. How many students went?
Answer: After entry fees, $2000 – 200 = 1800$ remains. At ₹60 per person, total persons $= 1800 \div 60 = 30$. Excluding 3 teachers, students $= 30 – 3 = \textbf{27}$. (As an equation, with $n$ students: $60(n + 3) + 200 = 2000$ → $60(n + 3) = 1800$ → $n + 3 = 30$ → $n = 27$.)
Example 12: Juri is 7 years older than Banajit. If the sum of their ages is 25 years, find each present age.
Answer: Let Banajit be $y$ years; Juri is $(y + 7)$ years. $(y + 7) + y = 25$ → $2y + 7 = 25$ → $2y = 18$ → $y = 9$. So Banajit is 9 years and Juri is $9 + 7 = \textbf{16 years}$.
Example 13: In a tile pattern the $k$-th step needs $4k + 1$ tiles. Can the pattern be completed with 121 tiles? At which step?
Answer: In both ways the $k$-th step needs $4k + 1$ tiles. $4k + 1 = 121$ → $4k = 120$ → $k = 30$. So the pattern is completed at the 30th step.
Example 14: Ajay’s father gave him ₹150 for pens and tiffin. If he spends ₹30 on tiffin, how many ₹15 pens can he buy with the rest?
Answer: With $n$ pens, $15n + 30 = 150$ → $15n = 120$ → $n = 8$. He can buy 8 pens.
Example 15: Solve $12(x – 12) + 450 = 846$ in different ways.
Answer: Way 1: $12(x – 12) = 396$ → $x – 12 = 33$ → $x = 45$. Way 2 (expand): $12x – 144 + 450 = 846$ → $12x = 540$ → $x = 45$. Way 3 (divide by 3 first): $4(x – 12) + 150 = 282$ → $4x – 48 = 132$ → $4x = 180$ → $x = 45$. All three give $x = 45$.
Try yourself: Find the error in each and solve correctly.
Answer:
- (1) $x – 2 = 19$ → $x = 21$ — correct (no error).
- (2) $10 – 7t = -4$ — wrong ($10 – 7$ was treated as $3t$). Correct: $-7t = -14$ → $t = 2$.
- (3) $4p – 3 = 13$ → $4p = 16$ — wrong ($16 – 4$ written). Correct: $p = \frac{16}{4} = \textbf{4}$.
- (4) $5p + 2 = -10$ — wrong (only $5p$ divided by 5). Correct: $5p = -12$ → $p = -\frac{12}{5}$.
- (5) $y – 4 = 2y – 10$ — wrong (sign error). Correct: $-y = -6$ → $y = 6$.
- (6) $3(n – 2) = 12$ — wrong ($3n – 2$ instead of $3n – 6$). Correct: $3n = 18$ → $n = 6$.
Example 16: “Where are you going, hundred brothers? — We are not one hundred; as many as we came, that many more shall come, half shall come, and half of that again; with you we shall become a hundred.” How many were there at the beginning?
Answer: Let the number at the beginning be $x$ — as many again $x$, half $\frac{1}{2}x$, half of that $\frac{1}{4}x$, and you (1). So $x + x + \frac{1}{2}x + \frac{1}{4}x + 1 = 100$ → $\frac{11x}{4} = 99$ → $x = \frac{99 \times 4}{11} = 36$. So there were 36 at the beginning.
Example 17: Richa and Manpreet save ₹300 and ₹450 each month. They also saved prize money of ₹3600 and ₹3000 respectively. After how many months will their savings be equal?
Answer: Let savings be equal after $n$ months — Richa: $3600 + 300n$, Manpreet: $3000 + 450n$. So $3600 + 300n = 3000 + 450n$ → $600 = 150n$ → $n = 4$. Using Brahmagupta’s formula $x = \frac{D – B}{A – C}$, $n = \frac{3000 – 3600}{300 – 450} = \frac{-600}{-150} = 4$. So after 4 months.
15.5 Forming an Equation from a Given Solution
If the solution of an equation is 8 (that is, $x = 8$), what equations can be formed?
Answer: Applying the same operation to both sides of $x = 8$ forms a new equation — e.g. subtract 1: $x – 1 = 7$; multiply by 2: $2x = 16$; divide by 2: $\frac{x}{2} = 4$; add 5: $x + 5 = 13$; multiply by 2 then add 3: $2x + 3 = 19$; multiply by $(-1)$ then add 5: $-x + 5 = -3$. Thus many equations can be formed from one solution.
Try yourself: Construct any 5 equations whose solution is $x = -3$.
Answer: (sample) (1) $x + 3 = 0$; (2) $2x = -6$; (3) $x – 2 = -5$; (4) $3x + 1 = -8$; (5) $5x + 20 = 5$. Each has solution $x = -3$.
15.6 A Brief History of Equations
What do we learn here? Solve Bhaskaracharya’s horse problem and note the historical facts briefly.
Answer: The Gupta-period mathematician Brahmagupta used ‘Varna’ (a letter) for unknowns and gave the formula $x = \frac{D – B}{A – C}$ for equations of the form $Ax + B = Cx + D$. The ‘Bakhshali’ manuscript contains the ‘Anuman Samadhan’ method for simple equations. The medieval mathematician Bhaskaracharya began his ‘Ekavarnasamikaranam’ chapter with four axioms and is famous for the ‘Chakrawal’ method. In the 8th century these Indian ideas reached Arabic; Al-Khwarizmi wrote “Hisab al-jabr wal-muqabala”, and from ‘al-jabr’ came the word ‘algebra’. Horse problem: one merchant has 7 horses and 100 silvers, another 9 horses and 80 silvers; both are equally rich. With a horse costing $x$, $7x + 100 = 9x + 80$ → $20 = 2x$ → $x = 10$. So a horse costs 10 silvers; each merchant’s total is $7 \times 10 + 100 = 170$ silvers, and both together 340 silvers.
Try yourself: Solve $10x – 35 = -4x + 35$ and $4x – 9 = 5x – 10$ (using Brahmagupta’s formula).
Answer: $10x – 35 = -4x + 35$ → $14x = 70$ → $x = 5$ (formula: $\frac{35 – (-35)}{10 – (-4)} = \frac{70}{14} = 5$). $4x – 9 = 5x – 10$ → $x = 1$ (formula: $\frac{-10 – (-9)}{4 – 5} = \frac{-1}{-1} = 1$).
Exercise 15
1. Form equations: (a) 8 added to a number gives 15. (b) 5 subtracted from three times a number gives 16. (c) A number divided by 8 gives 7. (d) Two-thirds of $x$ is 6.
Answer: Taking the number as $x$ — (a) $x + 8 = 15$; (b) $3x – 5 = 16$; (c) $\frac{x}{8} = 7$; (d) $\frac{2}{3}x = 6$.
2. Write statements for: (a) $x – 7 = 14$ (b) $2y – 18 = 2$ (c) $\frac{3}{4}z = 9$ (d) $11 + 3t = 17$.
Answer: (a) 7 subtracted from a number gives 14. (b) 18 subtracted from twice a number gives 2. (c) Three-fourths of a number is 9. (d) 11 added to three times a number gives 17.
3. Solve by trial and error: (a) $3x – 5 = 13$ (b) $5p – 3 = p + 17$.
Answer: (a) $x = 6$ ($3 \times 6 – 5 = 13$). (b) $p = 5$ ($5 \times 5 – 3 = 22$ and $5 + 17 = 22$).
4. Match Column 1 with Column 2. Column 1: (P) $5x + 2 = 17$, (Q) $x – 2 = 17$, (R) $\frac{x}{2} + 5 = 20$; Column 2: (I) 30, (II) 3, (III) 19.
Answer: (P) $5x = 15$ → $x = 3$ = (II); (Q) $x = 19$ = (III); (R) $\frac{x}{2} = 15$ → $x = 30$ = (I). So P → II, Q → III, R → I — option (B).
5. Statement 1: the solution of $2x + 3 = 7$ is $x = 2$. Statement 2: $2(2x + 5) = 3 + 5x$ has no solution. Choose the correct option.
Answer: Statement 1 is true ($x = 2$). For Statement 2: $4x + 10 = 3 + 5x$ → $x = 7$, so a solution exists — Statement 2 is false. Correct option (C) (Statement 1 true but Statement 2 false).
6. The steps of solving $3(x + 2) – 2(x – 1) = 7$ are out of order: (I) $x + 8 = 7$ (II) $3x + 6 – 2x + 2 = 7$ (III) $x = -1$ (IV) $x = 7 – 8$. Find the correct order.
Answer: Expand first (II), simplify (I), then (IV), finally (III). Correct order II → I → IV → III — option (C).
7. The perimeter of a rectangular garden is 64 m. If its length is 6 m more than its breadth, find the length and breadth.
Answer: Let breadth $= b$, so length $= b + 6$. Perimeter $= 2(l + b)$: $2(2b + 6) = 64$ → $4b + 12 = 64$ → $4b = 52$ → $b = 13$. So breadth 13 m and length $13 + 6 = \textbf{19 m}$.
8. Solve: (a) $3(x + 6) = 24$ (b) $5(2 – 5t) = -20$ (c) $5p + 1 = 2(p + 17)$ (d) $\frac{n}{4} – 8 = 1$ (e) $4(r – 1) = r – 3$ (f) $3(x + 6) + 2(x + 3) = 64$ (g) $2 – 5z = 2(5 – z) – 6(2 – z)$ (h) $-4(m – 1) = 5(m + 2)$.
Answer: (a) $x + 6 = 8$ → $x = 2$. (b) $2 – 5t = -4$ → $t = \frac{6}{5}$. (c) $5p + 1 = 2p + 34$ → $3p = 33$ → $p = 11$. (d) $\frac{n}{4} = 9$ → $n = 36$. (e) $4r – 4 = r – 3$ → $3r = 1$ → $r = \frac{1}{3}$. (f) $5x + 24 = 64$ → $x = 8$. (g) $2 – 5z = -2 + 4z$ → $4 = 9z$ → $z = \frac{4}{9}$. (h) $-4m + 4 = 5m + 10$ → $-6 = 9m$ → $m = -\frac{2}{3}$.
9. Find $x$: (a) an isosceles triangle with two equal sides $x$ and base 6 cm, perimeter 20 cm. (b) a quadrilateral PQRS with sides $x$, $x$, $x + 5$, $x + 5$; perimeter 34 cm.
Answer: (a) Perimeter $= x + x + 6 = 20$ → $2x = 14$ → $x = 7$ cm. (b) Perimeter $= x + x + (x + 5) + (x + 5) = 4x + 10 = 34$ → $4x = 24$ → $x = 6$ cm.
10. Birdaw travels from home to school (home–park $= 2x + 3$, park–market $= 4$, market–school $= x$). If home–park $= 5$ km, find (a) the market–school distance and (b) the home–school distance.
Answer: home–park $= 2x + 3 = 5$ → $x = 1$. (a) market–school $= x = \textbf{1 km}$. (b) home–school $= (2x + 3) + 4 + x = 5 + 4 + 1 = \textbf{10 km}$.
11. Write five equations whose solution is $x = -10$.
Answer: (sample) (1) $x + 10 = 0$; (2) $2x = -20$; (3) $x – 5 = -15$; (4) $3x + 10 = -20$; (5) $\frac{x}{2} = -5$. Each has solution $x = -10$.
12. An ant can carry 40 times its own weight. 25 such ants carry a raisin weighing 1500 mg. Find the weight of one ant.
Answer: Let one ant weigh $x$ mg; it can carry $40x$ mg. Together 25 ants carry $25 \times 40x = 1000x = 1500$ mg, so $x = 1.5$. One ant weighs 1.5 mg.
13. A farm has some goats and chickens: 50 heads and 132 legs in total. Find the number of each. [Let goats $= x$, chickens $= y$; $x + y = 50$, $4x + 2y = 132$.]
Answer: Substituting $x = 50 – y$: $4(50 – y) + 2y = 132$ → $200 – 2y = 132$ → $y = 34$, and $x = 16$. So 16 goats and 34 chickens.
14. If $15k – 33 = 6(k + 8)$, find (a) $19k + 4$ (b) $33k – 8$.
Answer: $15k – 33 = 6k + 48$ → $9k = 81$ → $k = 9$. (a) $19 \times 9 + 4 = \textbf{175}$. (b) $33 \times 9 – 8 = \textbf{289}$.
15. An autorickshaw charges ₹20 for the first km and ₹15 for each additional km. If the total fare is ₹125, find the total distance. (A) 6 km (B) 7 km (C) 8 km (D) 9 km
Answer: Let distance $= d$ km: $20 + 15(d – 1) = 125$ → $15(d – 1) = 105$ → $d – 1 = 7$ → $d = 8$. Correct option (C) 8 km.
16. A pattern of small squares made from equal line segments has $n$ squares in the $n$-th position (in a row). (a) How many squares in the 10th position? (b) In which position are there 65 squares? (c) How many line segments in the 15th position? (d) Is a position with 52 line segments possible?
Answer: For $n$ squares in a row, squares $= n$ and line segments $= 3n + 1$. (a) 10 squares. (b) $n = 65$, so the 65th position. (c) $3 \times 15 + 1 = \textbf{46}$ line segments. (d) $3n + 1 = 52$ → $3n = 51$ → $n = 17$, a whole number, so yes — the 17th position has 52 line segments.
17. In Nahar’s savings pot, ₹10 and ₹20 notes together amount to ₹800, and there are 50 notes in all. Find the number of each type of note.
Answer: Let ₹10 notes $= x$ and ₹20 notes $= y$: $x + y = 50$ and $10x + 20y = 800$ (i.e. $x + 2y = 80$). Subtracting, $y = 30$ and $x = 20$. So 20 ten-rupee notes and 30 twenty-rupee notes.
18. If 15 is added to six times a number, it equals seven times the number. Find the number.
Answer: With number $n$: $6n + 15 = 7n$ → $n = 15$.
19. One-third of a number is 10 less than two-thirds of the number. Find the number.
Answer: With number $n$: $\frac{1}{3}n = \frac{2}{3}n – 10$ → $10 = \frac{1}{3}n$ → $n = 30$.
20. The sum of three consecutive even numbers is 48. Find the numbers.
Answer: With first number $x$: $x + (x + 2) + (x + 4) = 48$ → $3x + 6 = 48$ → $x = 14$. The numbers are 14, 16, 18.
21. The difference between two complementary angles is 12°. Find the two angles.
Answer: Complementary angles add to 90°. Let them be $x$° and $y$°: $x + y = 90$ and $x – y = 12$. Adding, $2x = 102$ → $x = 51$, so $y = 39$. The angles are 51° and 39°.
22. Statement 1: $(x – 3) = 4(x – 6)$ has only one solution $x = -7$. Statement 2: many equations may exist whose solution is $x = 7$. Choose the correct option.
Answer: $(x – 3) = 4(x – 6)$ → $x – 3 = 4x – 24$ → $21 = 3x$ → $x = 7$ (not $-7$), so Statement 1 is false. Statement 2 is true. Correct option (D) (Statement 1 false but Statement 2 true).
23. Help the rabbit reach the basket of carrots by moving toward the correct solution of each equation.
Answer: $5 = 12 – x$ → $x = 7$; $3(2m + 1) = 27$ → $m = 4$; $3x + 12 = 42$ → $x = 10$; $7x – 9 = 3x + 11$ → $x = 5$; $6n = -54$ → $n = -9$; $2x + 1 = 5(x – 1)$ → $x = 2$; $y + 5 = 2y + 12$ → $y = -7$; $2x + 4 + x = -20$ → $x = -8$; $-18 = 2 + 4x$ → $x = -5$; $\frac{x}{5} + 3 = 10$ → $x = 35$; $5(m – 4) – 3 = 37$ → $m = 12$; the centre one $2x + 5 = x + 12$ → $x = 7$. Following each correct solution leads the rabbit to the carrot basket.
Puzzles 1, 2, 3
Puzzle 1: Find the missing numbers, applying $+5$, $\times 4$, $-10$, $\div 2$ in turn. (a) 9 → … (b) ? → … $= 15$ (c) ? → … $= 29$.
Answer: (a) $9 \xrightarrow{+5} 14 \xrightarrow{\times 4} 56 \xrightarrow{-10} 46 \xrightarrow{\div 2} 23$. (b) Working backwards: $15 \times 2 = 30$, $30 + 10 = 40$, $40 \div 4 = 10$, $10 – 5 = 5$ — the start is 5 (row: 5, 10, 40, 30, 15). (c) $29 \times 2 = 58$, $58 + 10 = 68$, $68 \div 4 = 17$, $17 – 5 = 12$ — the start is 12 (row: 12, 17, 68, 58, 29).
Puzzle 2: Fill the boxes. A number $x$ goes $\times 5$ ($= 5x$) one way and $+5$ ($= x + 5$) the other, then the difference $5x – (x + 5) = 4x – 5$ is taken. (a) $x = 13$ (b) result 75 (c) result 595.
Answer: (a) $5 \times 13 = 65$, $13 + 5 = 18$, $65 – 18 = 47$. (b) $4x – 5 = 75$ → $x = 20$ ($5 \times 20 = 100$, $20 + 5 = 25$, $100 – 25 = 75$). (c) $4x – 5 = 595$ → $x = 150$ ($5 \times 150 = 750$, $150 + 5 = 155$, $750 – 155 = 595$).
Puzzle 3: Take any number from 1–100 → double it → add 4 → multiply by 5 → divide by 10 → subtract 2. What number do you get?
Answer: Let the number be $x$ — $2x \to 2x + 4 \to 5(2x + 4) = 10x + 20 \to \frac{10x + 20}{10} = x + 2 \to (x + 2) – 2 = x$. So you get back the original number; that is the similarity between the first and last numbers.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. Which of the following is an equation? (a) $2x + 3$ (b) $x + 5 = 9$ (c) $7 – 4$ (d) $3x$
Answer: (b) $x + 5 = 9$ (it has an equals sign).
2. The solution of $x + 5 = 12$ is — (a) 5 (b) 7 (c) 17 (d) 12
Answer: (b) 7.
3. The part on the right of the equals sign is called — (a) LHS (b) RHS (c) variable (d) constant
Answer: (b) RHS.
4. The solution of $3x = 15$ is — (a) 3 (b) 45 (c) 5 (d) 12
Answer: (c) 5.
5. In the trial and error method we — (a) substitute different values (b) draw a graph (c) only divide (d) do nothing
Answer: (a) substitute different values.
6. If $a = b$, then $a + c =$ ? (a) $b$ (b) $b – c$ (c) $b + c$ (d) $bc$
Answer: (c) $b + c$.
7. The solution of $\frac{x}{4} = 2$ is — (a) 2 (b) 6 (c) 8 (d) $\frac{1}{2}$
Answer: (c) 8.
8. The word “algebra” comes from — (a) al-jabr (b) varna (c) siddhanta (d) lilavati
Answer: (a) al-jabr.
9. If the sum of three consecutive odd numbers is 75, the middle number is — (a) 23 (b) 25 (c) 27 (d) 24
Answer: (b) 25.
10. If $2x + 3 = 7$, then $x =$ ? (a) 1 (b) 2 (c) 5 (d) 4
Answer: (b) 2.
Fill in the Blanks
1. A statement of equality between two expressions is called an ______. Answer: equation.
2. The value for which both sides of an equation are equal is called its ______. Answer: solution.
3. Both sides of an equation may be divided by the same ______ number ($c \ne 0$). Answer: non-zero.
4. Addition and subtraction are ______ operations. Answer: inverse.
5. ______ equations can be formed from one solution. Answer: Many.
True or False
1. The two sides of an equation are equal only for one particular value of the unknown. Answer: True.
2. If a number is removed from one side of an equation, the same need not be removed from the other side. Answer: False (to keep equality, remove the same number from both sides).
3. $x + 5 = 12$ and $x + 6 = 13$ have the same solution. Answer: True (both give $x = 7$).
4. The symbol ‘$\ne$’ means “not equal”. Answer: True.
5. Brahmagupta gave a formula to solve equations. Answer: True.
Short Answer Questions
1. What is an equation?
Answer: A statement of equality between two algebraic expressions, or between an algebraic expression and an arithmetic expression, is called an equation.
2. What do LHS and RHS mean?
Answer: The part left of the equals sign is the LHS (Left Hand Side) and the part on the right is the RHS (Right Hand Side).
3. Solve $2x + 3 = 11$ and verify.
Answer: $2x = 11 – 3 = 8$ → $x = 4$. Check: $2 \times 4 + 3 = 11 = $ RHS — correct.
4. Describe the trial and error method briefly.
Answer: Different values are substituted for the unknown until one particular value makes LHS = RHS; that value is the solution.
Key Terms
| Term | Meaning |
|---|---|
| Equation | A statement of equality between two expressions |
| Equality | The state of both sides having the same value |
| Unknown / Variable | A letter standing for an unknown quantity (e.g. $x, n$) |
| Left Hand Side (LHS) | The part left of the equals sign |
| Right Hand Side (RHS) | The part right of the equals sign |
| Solution / Root | The value for which LHS = RHS |
| Trial and Error method | Finding the solution by substituting values |
| Axiom | An accepted truth that needs no proof |
| Inverse operation | Opposite operations such as add–subtract or multiply–divide |
| Verification | Checking whether a solution is correct |
| Algebraic expression | An expression containing letters (variables) |
| Arithmetic expression | An expression containing only numbers |