Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
A linear equation in two variables x and y has the form
Every solution (x, y) is a point on a straight line, and every point on that line is a solution. So one equation has infinitely many solutions.
Two such equations taken together form a pair:
A solution of the pair is a pair of values (x, y) that satisfies both equations, that is, a point lying on both lines.
Name the two unknowns, then turn each condition into an equation. Akhila's fair: if x = rides and y = Hoopla games, then
This gives x = 4 rides and y = 2 games.
| Compare the ratios | Lines | Solutions | Pair is |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | intersecting | exactly one | consistent |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | coincident | infinitely many | dependent (consistent) |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | parallel | none | inconsistent |
Write both equations in the form ax + by + c = 0 before taking the ratios.
If the solution has non-integer coordinates like (4/13, 1/19), it is hard to read from a graph. Algebraic methods give exact answers.
Solve 7x − 15y = 2 and x + 2y = 3.
Incomes in ratio 9 : 7, expenditures in ratio 4 : 3, each saves ₹2000. With incomes 9x, 7x and expenditures 4y, 3y:
Incomes are ₹18000 and ₹14000.
Use substitution when one equation already gives a variable with coefficient 1 or −1. Use elimination when coefficients are larger or can be matched easily. Both always give the same answer.
| Problem type | Let | Typical equations |
|---|---|---|
| Two numbers | x, y | x − y = 26, x = 3y |
| Ages | present ages x, y | n years ago: x − n, y − n; n years hence: x + n, y + n |
| Two-digit number | tens digit x, units digit y | number = 10x + y; reversed = 10y + x |
| Fraction | numerator x, denominator y | x + 2y + 2 = 911 ⇒ 11x − 9y = −4 |
| Fixed + variable charge | fixed x, rate y | x + 10y = 105 |
| Costs | price of each item x, y | 5x + 7y = 50 |
| Angles | x, y | supplementary: x + y = 180° |
Always put the values back into the words of the problem, not only into your equations. A slip while forming the equations will not show up otherwise.
| Situation | Graph | Algebra gives |
|---|---|---|
| Unique solution | lines cross | one value of x and of y |
| Infinitely many | same line | a true statement, e.g. 0 = 0 |
| No solution | parallel lines | a false statement, e.g. 0 = 9 |
Try each problem first, then open the solution to check your working.
Solve x + y = 6 and x − y = 2 graphically.
Use: Plot two points for each line
x + y = 6: (0, 6), (6, 0). x − y = 2: (2, 0), (0, −2).
The lines meet at (4, 2). Check: 4 + 2 = 6, 4 − 2 = 2.
Answer: x = 4, y = 2
Solve 2x − y = 4 and x + y = 5 graphically. Also find the vertices and area of the triangle formed by the two lines and the y-axis.
Use: Area = ½ × base × height
The lines meet at (3, 2).
They meet the y-axis at (0, −4) and (0, 5).
Base on the y-axis = 5 − (−4) = 9; height = distance of (3, 2) from the y-axis = 3.
Area = ½ × 9 × 3 = 13.5 square units
Answer: x = 3, y = 2; vertices (0, −4), (0, 5), (3, 2); area 13.5 square units
Say whether each pair has a unique solution, no solution or infinitely many: (a) 3x + 2y = 7, 6x + 4y = 14 (b) x − 2y = 3, 3x − 6y = 5 (c) 4x + y = 9, x − 3y = 1
Use: Compare a₁/a₂, b₁/b₂, c₁/c₂
(a) 3/6 = 2/4 = 7/14 = 1/2: coincident
(b) 1/3 = −2/−6 = 1/3, but 3/5 ≠ 1/3: parallel
(c) 4/1 ≠ 1/(−3): intersecting
Answer: (a) infinitely many (b) no solution (c) unique solution
Find k for which kx + 3y = k − 3 and 12x + ky = k have no solution.
Use: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
k/12 = 3/k gives k² = 36, so k = 6 or k = −6.
k = 6: c₁/c₂ = (k − 3)/k = 3/6 = 1/2 = b₁/b₂, so the lines coincide. Reject.
k = −6: b₁/b₂ = 3/(−6) = −1/2 and c₁/c₂ = (−9)/(−6) = 3/2. Different, so the lines are parallel.
Answer: k = −6
Find k for which 2x + 3y = 7 and (k − 1)x + (k + 2)y = 3k have infinitely many solutions.
Use: a₁/a₂ = b₁/b₂ = c₁/c₂
2/(k − 1) = 3/(k + 2) gives 2k + 4 = 3k − 3, so k = 7.
Check: 3/(k + 2) = 3/9 = 1/3 and 7/(3k) = 7/21 = 1/3. ✓
Answer: k = 7
Solve 2x + y = 7 and 3x − 2y = 7 by substitution.
Use: y = 7 − 2x from the first equation
3x − 2(7 − 2x) = 7 ⇒ 7x − 14 = 7 ⇒ x = 3
y = 7 − 6 = 1
Answer: x = 3, y = 1
Solve x2 + y3 = 5 and x3 − y4 = 12.
Use: Multiply by the LCM of the denominators first
× 6: 3x + 2y = 30. × 12: 4x − 3y = 6.
From the first, y = (30 − 3x)/2.
4x − 3(30 − 3x)/2 = 6 ⇒ 8x − 90 + 9x = 12 ⇒ 17x = 102 ⇒ x = 6
y = (30 − 18)/2 = 6
Answer: x = 6, y = 6
Use substitution on (a) x − 2y = 3, 2x − 4y = 6 and (b) x − 2y = 3, 2x − 4y = 7.
Use: x = 3 + 2y
(a) 2(3 + 2y) − 4y = 6 ⇒ 6 = 6, true for every y.
(b) 2(3 + 2y) − 4y = 7 ⇒ 6 = 7, false.
Answer: (a) infinitely many solutions (b) no solution
A father is three times as old as his son. In 12 years he will be twice as old as his son. Find their present ages.
Given: father f, son s
Use: f = 3s; f + 12 = 2(s + 12)
3s + 12 = 2s + 24 ⇒ s = 12
f = 36
Check: in 12 years, 48 = 2 × 24 ✓
Answer: Father 36 years, son 12 years
Solve 3x + 4y = 18 and 5x − 4y = 14.
Use: The y-coefficients are already equal and opposite
Add: 8x = 32, so x = 4
3(4) + 4y = 18 ⇒ y = 3/2
Answer: x = 4, y = 3/2
Solve 4x − 3y = 1 and 3x − 2y = 2.
Use: Make the y-coefficients 6
× 2: 8x − 6y = 2. × 3: 9x − 6y = 6.
Subtract: x = 4
4(4) − 3y = 1 ⇒ y = 5
Answer: x = 4, y = 5
Solve 0.3x + 0.4y = 2.6 and 0.5x − 0.2y = 0.
Use: Multiply by 10 to clear decimals
3x + 4y = 26 and 5x − 2y = 0.
Multiply the second by 2: 10x − 4y = 0. Add: 13x = 26, so x = 2.
5(2) − 2y = 0 ⇒ y = 5
Answer: x = 2, y = 5
Solve 3x − 5y = 7 and 6x − 10y = 9 by elimination.
Use: Make the x-coefficients equal
× 2: 6x − 10y = 14
Subtract the second equation: 0 = 5, which is false.
Answer: No solution; the lines are parallel
The digits of a two-digit number add up to 12. The number obtained by reversing the digits is 18 more than the original. Find the number.
Given: tens digit x, units digit y
Use: number = 10x + y
x + y = 12
(10y + x) − (10x + y) = 18 ⇒ 9y − 9x = 18 ⇒ y − x = 2
Add: 2y = 14, y = 7, x = 5
Answer: 57
The numerator and denominator of a fraction add up to 18. If the denominator is increased by 2, the fraction becomes 1/3. Find the fraction.
Given: numerator x, denominator y
x + y = 18
x/(y + 2) = 1/3 ⇒ 3x = y + 2
Substitute y = 18 − x: 3x = 20 − x, so x = 5 and y = 13
Answer: 5/13
A cinema sold 50 tickets for ₹5200. Adult tickets cost ₹120 and child tickets ₹80. How many of each were sold?
Given: adults x, children y
x + y = 50
120x + 80y = 5200 ⇒ 3x + 2y = 130
Subtract 2 × first: x = 30, so y = 20
Answer: 30 adult and 20 child tickets
If the length of a rectangle is reduced by 5 m and the breadth increased by 3 m, the area falls by 9 m². If the length is increased by 3 m and the breadth by 2 m, the area rises by 67 m². Find the dimensions.
Given: length l, breadth b
Use: Expand and cancel lb
(l − 5)(b + 3) = lb − 9 ⇒ 3l − 5b = 6
(l + 3)(b + 2) = lb + 67 ⇒ 2l + 3b = 61
× 3 and × 5: 9l − 15b = 18, 10l + 15b = 305. Add: 19l = 323, l = 17
b = (61 − 34)/3 = 9
Answer: Length 17 m, breadth 9 m
A gym charges a joining fee plus a fixed monthly fee. The total for 4 months is ₹3700 and for 9 months is ₹6200. Find both fees.
Given: joining fee x, monthly fee y
x + 4y = 3700
x + 9y = 6200
Subtract: 5y = 2500, y = 500; x = 1700
Answer: Joining fee ₹1700; monthly fee ₹500
A bag has 20 coins of ₹2 and ₹5, worth ₹70 in all. How many of each are there?
Given: ₹2 coins x, ₹5 coins y
x + y = 20
2x + 5y = 70
Subtract 2 × first: 3y = 30, y = 10, x = 10
Answer: 10 coins of ₹2 and 10 coins of ₹5
The ages of A and B add up to 55 years. Five years ago, A was twice as old as B. Find their ages.
Given: A's age a, B's age b
a + b = 55
a − 5 = 2(b − 5) ⇒ a − 2b = −5
Subtract: 3b = 60, b = 20, a = 35
Answer: A is 35 years, B is 20 years
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(i) Let the number of girls be x and of boys be y. Then x + y = 10 and x − y = 4.
Points on x + y = 10: (10, 0), (0, 10), (5, 5). Points on x − y = 4: (4, 0), (0, −4), (5, 1).
The lines meet at (7, 3). So there are 7 girls and 3 boys.
(ii) Let a pencil cost ₹x and a pen ₹y. Then 5x + 7y = 50 and 7x + 5y = 46.
Points on 5x + 7y = 50: (10, 0), (3, 5), (−4, 10). Points on 7x + 5y = 46: (3, 5), (8, −2), (−2, 12).
The lines meet at (3, 5).
Answer: (i) 7 girls, 3 boys (ii) pencil ₹3, pen ₹5
(i) a₁/a₂ = 5/7, b₁/b₂ = −4/6 = −2/3. Since 5/7 ≠ −2/3, the lines intersect at a point.
(ii) a₁/a₂ = 9/18 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2. All equal, so the lines are coincident.
(iii) a₁/a₂ = 6/2 = 3, b₁/b₂ = −3/−1 = 3, c₁/c₂ = 10/9. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
Answer: (i) intersecting (ii) coincident (iii) parallel
Write each pair as ax + by + c = 0 before comparing.
(i) 3/2 ≠ 2/(−3). Lines intersect: consistent (unique solution).
(ii) 2/4 = 1/2, (−3)/(−6) = 1/2, (−8)/(−9) = 8/9. a₁/a₂ = b₁/b₂ ≠ c₁/c₂: parallel, inconsistent.
(iii) a₁/a₂ = 3/29 = 16, b₁/b₂ = 5/3−10 = −16. Not equal: lines intersect, consistent.
(iv) 5/(−10) = −1/2, (−3)/6 = −1/2, (−11)/22 = −1/2. All equal: coincident, consistent (dependent).
(v) a₁/a₂ = 4/32 = 23, b₁/b₂ = 23, c₁/c₂ = 812 = 23. All equal: coincident, consistent (dependent).
Answer: (i) consistent (ii) inconsistent (iii) consistent (iv) consistent (v) consistent
(i) 1/2 = 1/2 = 5/10: coincident, so consistent with infinitely many solutions. Every point on x + y = 5 is a solution, for example (0, 5), (2, 3), (5, 0).
(ii) 1/3 = (−1)/(−3) = 1/3, but 8/16 = 1/2: parallel, so inconsistent.
(iii) 2/4 ≠ 1/(−2): consistent. Points on 2x + y = 6: (0, 6), (3, 0), (2, 2). Points on 4x − 2y = 4: (0, −2), (1, 0), (2, 2).
The lines meet at (2, 2), so x = 2, y = 2.
(iv) 2/4 = (−2)/(−4) = 1/2, but (−2)/(−5) = 2/5: parallel, so inconsistent.
Answer: (i) consistent, infinitely many solutions (ii) inconsistent (iii) consistent, x = 2, y = 2 (iv) inconsistent
Let the length be x m and the width y m.
x + y = 36 (half the perimeter) and x − y = 4.
Adding: 2x = 40, so x = 20. Then y = 16.
Graphically, x + y = 36 and x − y = 4 meet at (20, 16). Check: 20 + 16 = 36 and 20 − 16 = 4.
Answer: Length 20 m, width 16 m
(i) Need a₁/a₂ ≠ b₁/b₂. Example: 3x + 2y − 7 = 0 (2/3 ≠ 3/2). These meet at (1, 2).
(ii) Need a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Example: 4x + 6y − 12 = 0 (2/4 = 3/6 = 1/2 but −8/−12 = 2/3).
(iii) Need all three ratios equal. Example: 4x + 6y − 16 = 0 (twice the given equation).
Many other answers are possible.
Answer: (i) 3x + 2y − 7 = 0 (ii) 4x + 6y − 12 = 0 (iii) 4x + 6y − 16 = 0
Points on x − y + 1 = 0: (−1, 0), (0, 1), (2, 3). Points on 3x + 2y − 12 = 0: (4, 0), (0, 6), (2, 3).
The first line meets the x-axis at (−1, 0), the second at (4, 0), and they meet each other at (2, 3).
Answer: Vertices (−1, 0), (4, 0) and (2, 3)
(i) From x − y = 4, x = y + 4. Then (y + 4) + y = 14, so y = 5 and x = 9.
(ii) s = t + 3. The second equation × 6: 2s + 3t = 36. So 2(t + 3) + 3t = 36, 5t = 30, t = 6 and s = 9.
(iii) y = 3x − 3. Then 9x − 3(3x − 3) = 9 gives 9 = 9, true for all x. Infinitely many solutions: every (x, 3x − 3).
(iv) × 10: 2x + 3y = 13 and 4x + 5y = 23. x = (13 − 3y)/2. Then 2(13 − 3y) + 5y = 23, so 26 − y = 23, y = 3 and x = 2.
(v) From the first, x = −√3y√2. Then √3(−√3y√2) − 2√2y = 0, i.e. −(3√2 + 2√2)y = 0, so y = 0 and x = 0.
(vi) × 6: 9x − 10y = −12 and 2x + 3y = 13. x = (13 − 3y)/2. Then 9(13 − 3y)/2 − 10y = −12, so 117 − 27y − 20y = −24, 47y = 141, y = 3 and x = 2.
Answer: (i) x = 9, y = 5 (ii) s = 9, t = 6 (iii) infinitely many (y = 3x − 3) (iv) x = 2, y = 3 (v) x = 0, y = 0 (vi) x = 2, y = 3
From the first, 2x = 11 − 3y. Substituting in the second: 11 − 3y − 4y = −24, so 7y = 35, y = 5.
2x = 11 − 15 = −4, so x = −2.
y = mx + 3 gives 5 = −2m + 3, so m = −1.
Answer: x = −2, y = 5; m = −1
(i) Let the numbers be x and y (x > y): x − y = 26, x = 3y. Then 3y − y = 26, y = 13, x = 39.
(ii) Larger x, smaller y: x + y = 180, x − y = 18. x = y + 18, so 2y + 18 = 180, y = 81, x = 99.
(iii) Bat ₹x, ball ₹y: 7x + 6y = 3800, 3x + 5y = 1750. From the second, x = (1750 − 5y)/3. Then 7(1750 − 5y) + 18y = 11400, so 12250 − 17y = 11400, y = 50, x = (1750 − 250)/3 = 500.
(iv) Fixed ₹x, rate ₹y per km: x + 10y = 105, x + 15y = 155. x = 105 − 10y, so 105 + 5y = 155, y = 10, x = 5. For 25 km: 5 + 25 × 10 = ₹255.
(v) Fraction xy: x + 2y + 2 = 911 gives 11x − 9y = −4; x + 3y + 3 = 56 gives 6x − 5y = −3. From the second, x = (5y − 3)/6. Then 11(5y − 3) − 54y = −24, so y − 33 = −24, y = 9, x = 7.
(vi) Jacob x, son y: x + 5 = 3(y + 5) gives x − 3y = 10; x − 5 = 7(y − 5) gives x − 7y = −30. x = 3y + 10, so 3y + 10 − 7y = −30, y = 10, x = 40.
Answer: (i) 39 and 13 (ii) 99° and 81° (iii) bat ₹500, ball ₹50 (iv) fixed ₹5, ₹10 per km; ₹255 (v) 7/9 (vi) Jacob 40 years, son 10 years
(i) Elimination: 3 × first + second: 3x + 3y + 2x − 3y = 15 + 4, so 5x = 19, x = 19/5; y = 5 − 19/5 = 6/5.
Substitution: y = 5 − x, 2x − 15 + 3x = 4, x = 19/5, y = 6/5.
(ii) Elimination: first + 2 × second: 7x = 14, x = 2; then y = 1. Substitution: x = y + 1, 3y + 3 + 4y = 10, y = 1, x = 2.
(iii) Write as 3x − 5y = 4 and 9x − 2y = 7. Elimination: 3 × first − second: −13y = 5, y = −5/13; 3x = 4 + 5(−5/13) = 27/13, x = 9/13.
Substitution: x = (4 + 5y)/3; 9(4 + 5y)/3 − 2y = 7, so 12 + 13y = 7, y = −5/13, x = 9/13.
(iv) × 6: 3x + 4y = −6; × 3: 3x − y = 9. Elimination: subtract: 5y = −15, y = −3; then 3x = 9 + y = 6, x = 2.
Substitution: y = 3x − 9; 3x + 4(3x − 9) = −6, 15x = 30, x = 2, y = −3.
Answer: (i) x = 19/5, y = 6/5 (ii) x = 2, y = 1 (iii) x = 9/13, y = −5/13 (iv) x = 2, y = −3
(i) Fraction xy: x + 1y − 1 = 1 gives x − y = −2; xy + 1 = 12 gives 2x − y = 1. Subtract: x = 3, then y = 5. Fraction 3/5.
(ii) Nuri x, Sonu y: x − 5 = 3(y − 5) gives x − 3y = −10; x + 10 = 2(y + 10) gives x − 2y = 10. Subtract: y = 20, then x = 50.
(iii) Tens digit x, units digit y: x + y = 9, and 9(10x + y) = 2(10y + x) gives 88x − 11y = 0, i.e. 8x − y = 0. Add to x + y = 9: 9x = 9, x = 1, y = 8. Number 18.
(iv) x notes of ₹50 and y of ₹100: x + y = 25, 50x + 100y = 2000 i.e. x + 2y = 40. Subtract: y = 15, x = 10.
(v) Fixed charge ₹x, extra ₹y per day: 7 days = 3 + 4 extra, so x + 4y = 27; 5 days: x + 2y = 21. Subtract: 2y = 6, y = 3, x = 15.
Answer: (i) 3/5 (ii) Nuri 50 years, Sonu 20 years (iii) 18 (iv) ten ₹50 notes, fifteen ₹100 notes (v) fixed charge ₹15, ₹3 per extra day