NCERT Mathematics · Class 10 · English · 2026-27
Pair of Linear Equations in Two Variables — Exercise 3.1
Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.
Choose a question
Question 1
Model and solve graphically: (i) 10 quiz participants, girls exceed boys by 4; (ii) pencil/pen purchases costing ₹50 and ₹46 as specified in the textbook.
Step-by-step solution
(i) Quiz participants
Let x be the number of boys and y the number of girls. The total and difference give:
\[x+y=10,\qquad y-x=4\]
Plot x+y=10 through (0,10),(10,0), and y=x+4 through (0,4),(6,10). Their intersection is (3,7).
(ii) Pencil and pen prices
Let x rupees be the price of a pencil and y rupees the price of a pen. The purchases give:
\[5x+7y=50,\qquad7x+5y=46\]
For the first line use (3,5),(10,0). For the second use (3,5),(0,46/5). Plotting these lines gives the common point (3,5).
Resource ID: CB-10-M-CH3-EX3.1-Q1
Question 2
Classify the three equation pairs in Question 2 by comparing coefficient ratios.
Step-by-step solution
Write both equations in ax+by+c=0 form before comparing ratios. Different a-ratios and b-ratios mean intersection. Equal a/b/c ratios mean coincidence; equal a/b ratios with a different c-ratio mean parallel distinct lines.
(i) 5x−4y+8=0; 7x+6y−9=0
\[\frac{a_1}{a_2}=\frac57,\quad\frac{b_1}{b_2}=-\frac23\]
(ii) 9x+3y+12=0; 18x+6y+24=0
\[\frac9{18}=\frac36=\frac{12}{24}=\frac12\]
(iii) 6x−3y+10=0; 2x−y+9=0
\[\frac62=\frac{-3}{-1}=3\ne\frac{10}{9}\]
Resource ID: CB-10-M-CH3-EX3.1-Q2
Question 3
Decide consistency for each of the five equation pairs in Question 3.
Step-by-step solution
(i) 3x+2y=5; 2x−3y=7
\[\frac32\ne\frac2{-3}\]
(ii) 2x−3y=8; 4x−6y=9
\[\frac24=\frac{-3}{-6}=\frac12\ne\frac{-8}{-9}\]
(iii) (3/2)x+(5/3)y=7; 9x−10y=14
\[\frac{3/2}{9}=\frac16,\qquad\frac{5/3}{-10}=-\frac16\]
(iv) 5x−3y=11; −10x+6y=−22
\[\frac5{-10}=\frac{-3}6=\frac{-11}{22}=-\frac12\]
(v) (4/3)x+2y=8; 2x+3y=12
\[\frac{4/3}{2}=\frac23,\quad\frac23=\frac23,\quad\frac{-8}{-12}=\frac23\]
Resource ID: CB-10-M-CH3-EX3.1-Q3
Question 4
Classify each pair in Question 4; for consistent pairs, describe and draw their graphical solutions.
Step-by-step solution
(i) x+y=5; 2x+2y=10
The second equation is twice the first. Both pass through (0,5) and (5,0), so their graphs coincide.
(ii) x−y=8; 3x−3y=16
The first gives x−y=8; the second gives x−y=16/3. The left-side coefficients agree after scaling but the constants differ.
(iii) 2x+y−6=0; 4x−2y−4=0
Plot y=6−2x through (0,6),(3,0), and y=2x−2 through (0,−2),(1,0).
(iv) 2x−2y−2=0; 4x−4y−5=0
The equations reduce to x−y=1 and x−y=5/4. Their distinct parallel lines cannot meet.
Resource ID: CB-10-M-CH3-EX3.1-Q4
Question 5
A rectangle has half-perimeter 36 m, with length exceeding width by 4 m. Find its dimensions.
Step-by-step solution
Let length be l metres and width w metres. Half-perimeter is l+w, not 2(l+w).
\[l+w=36,\qquad l-w=4\]
Add the equations: 2l=40, giving l=20. Then w=36−20=16.
Resource ID: CB-10-M-CH3-EX3.1-Q5
Question 6
Choose a partner for 2x+3y−8=0 that gives (i) intersecting, (ii) parallel, (iii) coincident lines.
Step-by-step solution
Many choices are valid. The examples below use the coefficient test.
(i) Intersecting
\[x+y-1=0\]
The coefficient ratios are 2/1=2 and 3/1=3, which differ. Therefore the lines intersect.
(ii) Parallel distinct
\[2x+3y-9=0\]
The x- and y-coefficients match the original, but −8 and −9 differ. The two lines are parallel and distinct.
(iii) Coincident
\[4x+6y-16=0\]
Every coefficient is twice that in the original equation, so this is the same line.
Resource ID: CB-10-M-CH3-EX3.1-Q6
Question 7
Graph x−y+1=0 and 3x+2y−12=0. Find the triangle’s vertices with the x-axis and show its shaded region.
Step-by-step solution
For the first line y=x+1, plot (−1,0) and (0,1). For the second line y=6−3x/2, plot (0,6) and (4,0).
\[x+1=6-\frac32x\]
\[\frac52x=5,\qquad x=2,\quad y=3\]
Thus the two lines meet at C(2,3). Their x-intercepts are A(−1,0) and B(4,0).
Resource ID: CB-10-M-CH3-EX3.1-Q7