Exercise 3.1 — Pair of Linear Equations in Two Variables

1. 14 marksNCERT Cl-10 Maths, Ex 3.1, Q1

   Form a pair of linear equations for each situation and find the solution graphically. (i) A class of 10 students took part in a quiz; there are 4 more girls than boys. (ii) Five pencils and seven pens cost ₹50, while seven pencils and five pens cost ₹46.

   Hint. Name the two unknowns first, write one equation per sentence of information, then plot two points per line.

   (i) The quiz.

   Step 1 — Let the number of girls be x and boys be y.

   Step 2 — Ten students in total: x + y = 10. Four more girls than boys: x − y = 4.

   Step 3 — To plot each line, pick two convenient points. For x + y = 10 take (0, 10) and (10, 0). For x − y = 4 take (4, 0) and (10, 6).

   Step 4 — The two lines cross at (7, 3), which you can confirm algebraically: adding the equations gives 2x = 14.

   ✦ 7 girls and 3 boys.

   (ii) Pencils and pens.

   Step 1 — Let one pencil cost ₹x and one pen ₹y.

   Step 2 — 5x + 7y = 50 and 7x + 5y = 46.

   Step 3 — For plotting, find two points on each. On 5x + 7y = 50: (10, 0) and (3, 5). On 7x + 5y = 46: (3, 5) and (8, −2).

   Step 4 — Both lines pass through (3, 5), so that is the intersection.

   ✦ A pencil costs ₹3 and a pen costs ₹5.

   Where students slip. Mixing up which unknown is which halfway through — writing x − y = 4 after having defined x as boys. Write down 'x = girls, y = boys' before anything else and keep referring back to it.

   Another way. Part (ii) falls out quickly by adding and subtracting: adding gives 12x + 12y = 96, so x + y = 8; subtracting gives 2x − 2y = −4, so x − y = −2.

2. 23 marksNCERT Cl-10 Maths, Ex 3.1, Q2

   By comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, decide whether the lines in each pair intersect, are parallel, or are coincident: (i) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0; (ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0; (iii) 6x − 3y + 10 = 0 and 2x − y + 9 = 0.

   Hint. Put every equation in the form ax + by + c = 0 first, keeping the signs, then compare the three ratios in order.

   The test itself is short, so make sure you know what each outcome means geometrically:

   Ratios	Lines	Solutions
   a₁/a₂ ≠ b₁/b₂	intersect at one point	exactly one
   a₁/a₂ = b₁/b₂ ≠ c₁/c₂	parallel	none
   a₁/a₂ = b₁/b₂ = c₁/c₂	coincident (same line)	infinitely many

   (i) 5x − 4y + 8 = 0, 7x + 6y − 9 = 0.

   a₁/a₂ = 5/7 and b₁/b₂ = −4/6 = −2/3. These are different, so the very first comparison settles it.

   ✦ The lines intersect at exactly one point.

   (ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0.

   a₁/a₂ = 9/18 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2. All three agree.

   ✦ The lines are coincident — the second equation is just the first doubled.

   (iii) 6x − 3y + 10 = 0, 2x − y + 9 = 0.

   a₁/a₂ = 6/2 = 3 and b₁/b₂ = −3/−1 = 3, so the first two match. But c₁/c₂ = 10/9, which does not.

   ✦ The lines are parallel.

   Where students slip. Dropping a minus sign when forming b₁/b₂. In (iii) it is −3/−1 = 3, not −3. Both signs have to travel with their coefficients.

   Another way. You can also compare slopes after rewriting as y = mx + k: equal slopes with different intercepts means parallel, equal both means coincident.

3. 35 marksNCERT Cl-10 Maths, Ex 3.1, Q3

   Using the same ratio test, state whether each pair is consistent or inconsistent: (i) 3x + 2y = 5 and 2x − 3y = 7; (ii) 2x − 3y = 8 and 4x − 6y = 9; (iii) (3/2)x + (5/3)y = 7 and 9x − 10y = 14; (iv) 5x − 3y = 11 and −10x + 6y = −22; (v) (4/3)x + 2y = 8 and 2x + 3y = 12.

   Hint. Consistent simply means 'has at least one solution'. So intersecting and coincident are both consistent; only parallel is inconsistent.

   Rearrange each to ax + by + c = 0 before comparing — moving the constant across changes its sign, and that is where marks are lost.

   (i) 3x + 2y − 5 = 0 and 2x − 3y − 7 = 0. Here 3/2 ≠ 2/(−3), so the lines intersect.

   ✦ Consistent (unique solution).

   (ii) 2x − 3y − 8 = 0 and 4x − 6y − 9 = 0. Now 2/4 = 1/2 and −3/−6 = 1/2 agree, but −8/−9 = 8/9 does not.

   ✦ Inconsistent — parallel lines, no solution.

   (iii) (3/2)x + (5/3)y − 7 = 0 and 9x − 10y − 14 = 0. Compare a₁/a₂ = (3/2)/9 = 1/6 with b₁/b₂ = (5/3)/(−10) = −1/6. Different.

   ✦ Consistent (unique solution).

   (iv) 5x − 3y − 11 = 0 and −10x + 6y + 22 = 0. Then a₁/a₂ = 5/(−10) = −1/2, b₁/b₂ = −3/6 = −1/2, c₁/c₂ = −11/22 = −1/2. All equal.

   ✦ Consistent — coincident lines, infinitely many solutions.

   (v) (4/3)x + 2y − 8 = 0 and 2x + 3y − 12 = 0. Then a₁/a₂ = (4/3)/2 = 2/3, b₁/b₂ = 2/3, c₁/c₂ = −8/−12 = 2/3. All equal.

   ✦ Consistent — coincident lines, infinitely many solutions.

   Where students slip. Calling a coincident pair 'inconsistent' because it has no single answer. Infinitely many solutions is still solutions — inconsistent means none at all, which only happens for parallel lines.

   Another way. For (iv) the second equation is exactly −2 times the first, which you can spot without any ratio work. Scanning for that first often saves the arithmetic.

4. 44 marksNCERT Cl-10 Maths, Ex 3.1, Q4

   Which of the following pairs are consistent and which inconsistent? Where consistent, obtain the solution graphically: (i) x + y = 5 and 2x + 2y = 10; (ii) x − y = 8 and 3x − 3y = 16; (iii) 2x + y − 6 = 0 and 4x − 2y − 4 = 0; (iv) 2x − 2y − 2 = 0 and 4x − 4y − 5 = 0.

   Hint. Classify with the ratio test first. Only the intersecting case needs an actual graph and a single answer.

   (i) x + y = 5, 2x + 2y = 10. The second is exactly twice the first, so all three ratios equal 1/2.

   ✦ Consistent, coincident. Every point on x + y = 5 is a solution — for example (0, 5), (5, 0), (2, 3). Do not give a single point here.

   (ii) x − y = 8, 3x − 3y = 16. Ratios 1/3 = −1/−3 for the coefficients, but −8/−16 = 1/2 differs.

   ✦ Inconsistent. Plotted, the lines run parallel and never meet — no solution.

   (iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0. Here 2/4 = 1/2 but 1/(−2) = −1/2, so they intersect.

   For the graph: on 2x + y = 6 take (0, 6) and (3, 0); on 4x − 2y = 4 take (1, 0) and (0, −2). The lines cross at (2, 2).

   Check: 2(2) + 2 = 6 ✓ and 4(2) − 2(2) = 4 ✓

   ✦ Consistent, x = 2 and y = 2.

   (iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0. Coefficient ratios both 1/2, but −2/−5 = 2/5 does not match.

   ✦ Inconsistent — parallel, no solution.

   Where students slip. Giving one point as 'the' answer in part (i). When lines are coincident the honest answer is that there are infinitely many solutions; quoting a single pair suggests you misread the situation.

   Another way. Rewriting in the form y = mx + k makes (ii) and (iv) obvious at a glance — both pairs have the same slope but different intercepts, which is exactly what parallel means.

5. 53 marksNCERT Cl-10 Maths, Ex 3.1, Q5

   Half the perimeter of a rectangular garden is 36 m, and its length is 4 m more than its width. Find the dimensions of the garden.

   Hint. Half the perimeter of a rectangle is length + width — you do not need the full perimeter formula.

   Step 1 — Let the length be x metres and the width y metres.

   Step 2 — The perimeter is 2(x + y), so half of it is x + y. That gives x + y = 36. This is the step worth pausing on: reading 'half the perimeter' as x + y saves you carrying a factor of 2 through the whole problem.

   Step 3 — The length exceeds the width by 4: x − y = 4.

   Step 4 — Add the two equations: 2x = 40, so x = 20.

   Step 5 — Substitute back: 20 + y = 36, so y = 16.

   Step 6 — Check against the original wording: the full perimeter is 2(20 + 16) = 72, and half of that is 36 ✓. And 20 is indeed 4 more than 16 ✓

   ✦ The garden is 20 m long and 16 m wide.

   Where students slip. Writing 2(x + y) = 36 by treating 36 as the whole perimeter. Re-read the phrase — it says *half* the perimeter is 36, so the full perimeter is 72.

   Another way. Graphically, x + y = 36 and x − y = 4 meet at (20, 16), which is what the question asks you to produce if it specifies a graphical solution.

6. 63 marksNCERT Cl-10 Maths, Ex 3.1, Q6

   Given the equation 2x + 3y − 8 = 0, write a second linear equation in two variables so that the pair represents (i) intersecting lines, (ii) parallel lines, (iii) coincident lines.

   Hint. Work backwards from the ratio conditions — each case tells you exactly what to do with the coefficients.

   There are infinitely many right answers here. What is being tested is whether you can turn each ratio condition into a construction.

   (i) Intersecting needs a₁/a₂ ≠ b₁/b₂. So just choose coefficients that are not in the ratio 2 : 3.

   Take 3x + 2y − 7 = 0. Check: 2/3 ≠ 3/2 ✓

   (ii) Parallel needs a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Scale the coefficients of x and y together but give the constant a different scaling.

   Take 4x + 6y − 5 = 0. Check: 2/4 = 3/6 = 1/2, while −8/−5 = 8/5 ✗ — exactly the mismatch parallel requires.

   (iii) Coincident needs all three ratios equal, so multiply the entire equation by a constant.

   Take 4x + 6y − 16 = 0 (the original doubled). Check: 2/4 = 3/6 = −8/−16 = 1/2 ✓

   ✦ For example: (i) 3x + 2y − 7 = 0, (ii) 4x + 6y − 5 = 0, (iii) 4x + 6y − 16 = 0.

   Where students slip. For the coincident case, doubling only the x and y terms and leaving the constant alone. That produces parallel lines instead — the constant must be scaled by the same factor.

   Another way. A quick self-check for any answer you invent: substitute a point from the first line, such as (4, 0), and see whether it satisfies your new equation. It should for coincident, and should not for parallel.

7. 74 marksNCERT Cl-10 Maths, Ex 3.1, Q7

   Draw the graphs of x − y + 1 = 0 and 3x + 2y − 12 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the x-axis, and shade that triangle.

   Hint. Three lines make three vertices. Two of them are where each line crosses the x-axis; the third is where the two lines cross each other.

   Step 1 — Find where each line meets the x-axis by putting y = 0.

   For x − y + 1 = 0: x + 1 = 0, so x = −1. That vertex is (−1, 0).

   For 3x + 2y − 12 = 0: 3x = 12, so x = 4. That vertex is (4, 0).

   Step 2 — Find where the two lines meet each other. From the first, y = x + 1. Substituting into the second:

   3x + 2(x + 1) − 12 = 0 ⟹ 5x − 10 = 0 ⟹ x = 2, and then y = 3.

   That vertex is (2, 3).

   Step 3 — Check the point lies on both lines: 2 − 3 + 1 = 0 ✓ and 3(2) + 2(3) − 12 = 0 ✓

   Step 4 — For the sketch, plot each line through its two known points and shade the region enclosed by the two lines and the x-axis.

   ✦ The vertices are (−1, 0), (4, 0) and (2, 3).

   As a bonus, the base along the x-axis runs from −1 to 4, a length of 5, and the height is the y-coordinate of the apex, 3 — so the area is ½ × 5 × 3 = 7.5 square units.

   Where students slip. Giving only the intersection of the two lines. The question says the triangle is formed by the two lines *and the x-axis*, so the x-axis contributes two of the three vertices.

   Another way. You can also get the apex by elimination: multiply x − y = −1 by 2 to get 2x − 2y = −2, then add 3x + 2y = 12 to obtain 5x = 10.

Solutions written by the tuition.in editorial team and checked against the rationalised NCERT Class 10 Mathematics textbook and the CBSE 2026-27 syllabus (the cross-multiplication method is no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.