>Exercise 3.1 Solutions
Pair of Linear Equations in Two Variables
1. Form the pair of linear equations and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Step 1: Form the equations.
Let the number of boys be x and the number of girls be y.
Total students: x + y = 10 —(1)
Girls are 4 more than boys: y = x + 4 or y - x = 4 —(2)
Step 2: Find points for the graphical solution.
For equation (1), x + y = 10:
If x=5, y=5. Point: (5, 5)
If x=3, y=7. Point: (3, 7)
If x=7, y=3. Point: (7, 3)
For equation (2), y = x + 4:
If x=0, y=4. Point: (0, 4)
If x=1, y=5. Point: (1, 5)
If x=3, y=7. Point: (3, 7)
Step 3: Find the solution.
By plotting these points on a graph, the two lines will intersect at the point (3, 7).
Answer: Number of boys (x) = 3, and Number of girls (y) = 7.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Step 1: Form the equations.
Let the cost of one pencil be ₹ x and one pen be ₹ y.
First condition: 5x + 7y = 50 —(1)
Second condition: 7x + 5y = 46 —(2)
Step 2: Find points for the graphical solution.
For equation (1), 5x + 7y = 50:
If x=3, 15 + 7y = 50 => 7y=35 => y=5. Point: (3, 5)
If x=10, 50 + 7y = 50 => 7y=0 => y=0. Point: (10, 0)
For equation (2), 7x + 5y = 46:
If x=3, 21 + 5y = 46 => 5y=25 => y=5. Point: (3, 5)
If x=8, 56 + 5y = 46 => 5y=-10 => y=-2. Point: (8, -2)
Step 3: Find the solution.
By plotting these points, the two lines will intersect at the point (3, 5).
Answer: Cost of one pencil (x) = ₹ 3, and Cost of one pen (y) = ₹ 5.
2. On comparing the ratios a1a2, b1b2 and c1c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.
Conditions:
(a) Intersecting lines: a1a2 ≠b1b2
(b) Coincident lines: a1a2 = b1b2 = c1c2
(c) Parallel lines: a1a2 = b1b2 ≠c1c2
(i) 5x − 4y + 8 = 0; 7x + 6y − 9 = 0
a1a2 = 5/7, b1b2 = -4/6 = -2/3. Since 5/7 ≠-2/3, the lines are intersecting.
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
a1a2 = 9/18 = 1/2, b1b2 = 3/6 = 1/2, c1c2 = 12/24 = 1/2. Since all ratios are equal, the lines are coincident.
(iii) 6x − 3y + 10 = 0; 2x − y + 9 = 0
a1a2 = 6/2 = 3, b1b2 = -3/-1 = 3, c1c2 = 10/9. Since the first two ratios are equal but not equal to the third, the lines are parallel.
3. On comparing the ratios, find out whether the following pair of linear equations are consistent, or inconsistent.
Conditions:
(a) Consistent: Intersecting or Coincident lines.
(b) Inconsistent: Parallel lines.
(i) 3x + 2y = 5; 2x − 3y = 7
a1a2 = 3/2, b1b2 = 2/-3. Not equal. Intersecting lines. Consistent.
(ii) 2x − 3y = 8; 4x − 6y = 9
a1a2 = 2/4 = 1/2, b1b2 = -3/-6 = 1/2, c1c2 = -8/-9 = 8/9. Parallel lines. Inconsistent.
(iii) (3/2)x + (5/3)y = 7; 9x − 10y = 14
a1a2 = (3/2)/9 = 1/6, b1b2 = (5/3)/-10 = -1/6. Not equal. Intersecting lines. Consistent.
(iv) 5x − 3y = 11; −10x + 6y = −22
a1a2 = 5/-10 = -1/2, b1b2 = -3/6 = -1/2, c1c2 = -11/22 = -1/2. Coincident lines. Consistent.
(v) (4/3)x + 2y = 8; 2x + 3y = 12
a1a2 = (4/3)/2 = 2/3, b1b2 = 2/3, c1c2 = -8/-12 = 2/3. Coincident lines. Consistent.
4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5; 2x + 2y = 10
Ratios are 1/2, 1/2, -5/-10 = 1/2. The lines are coincident. Consistent.
There are infinitely many solutions. Any point on the line x + y = 5 is a solution (e.g., (1,4), (2,3), (5,0)).
(ii) x − y = 8; 3x − 3y = 16
Ratios are 1/3, -1/-3 = 1/3, -8/-16 = 1/2. The lines are parallel. Inconsistent.
(iii) 2x + y − 6 = 0; 4x − 2y − 4 = 0
Ratios are 2/4 = 1/2, 1/-2 = -1/2. The lines are intersecting. Consistent.
For 2x + y = 6: Points are (0,6), (3,0), (2,2).
For 4x – 2y = 4 (or 2x – y = 2): Points are (0,-2), (1,0), (2,2).
The graphical solution is the intersection point: x = 2, y = 2.
(iv) 2x − 2y − 2 = 0; 4x − 4y − 5 = 0
Ratios are 2/4 = 1/2, -2/-4 = 1/2, -2/-5 = 2/5. The lines are parallel. Inconsistent.
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Let the length be l and the width be w.
Equation 1 (from perimeter):
Half the perimeter = (1/2) * 2(l + w) = l + w.
So, l + w = 36.
Equation 2 (from length-width relation):
l = w + 4.
Step 3: Solve the equations.
Substitute equation (2) into (1):
(w + 4) + w = 36
2w + 4 = 36
2w = 32 => w = 16 m.
Now find the length: l = w + 4 = 16 + 4 = 20 m.
Answer: The dimensions are Length = 20 m and Width = 16 m.
6. Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines (ii) parallel lines (iii) coincident lines
The given equation is 2x + 3y – 8 = 0. (a1=2, b1=3, c1=-8)
(i) Intersecting lines: We need a1/a2 ≠b1/b2. We can choose almost any coefficients that don’t match the ratio 2:3.
Example: 3x + 5y – 7 = 0
(ii) Parallel lines: We need a1/a2 = b1/b2 ≠c1/c2. Multiply the x and y coefficients by a constant (e.g., 2) and change the constant term.
Example: 4x + 6y – 5 = 0
(iii) Coincident lines: We need a1/a2 = b1/b2 = c1/c2. Multiply the entire equation by a constant (e.g., 2).
Example: 4x + 6y – 16 = 0
7. Draw the graphs of the equations x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Step 1: Find points for each line.
Line 1: x - y + 1 = 0 (or y = x + 1)
Points: (-1, 0), (0, 1), (2, 3).
Line 2: 3x + 2y - 12 = 0 (or 2y = 12 – 3x)
Points: (4, 0), (0, 6), (2, 3).
Step 2: Find the vertices of the triangle.
Vertex 1 (Intersection of the two lines): From the points above, the common point is (2, 3).
Vertex 2 (Intersection of Line 1 with the x-axis, where y=0):
x – 0 + 1 = 0 => x = -1. The point is (-1, 0).
Vertex 3 (Intersection of Line 2 with the x-axis, where y=0):
3x + 2(0) – 12 = 0 => 3x = 12 => x = 4. The point is (4, 0).
Answer: The coordinates of the vertices of the triangle are (2, 3), (-1, 0), and (4, 0). After plotting, shade the region bounded by these three points.
