RD Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Solutions
Question - 31 : - Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and y-axis calculate the area of the triangle so formed.
Answer - 31 : -
5x – y = 5
=> y = 5x – 5
Substituting some different values of x, we get their corresponding values of y as shown below:
Plot these points on the graph and join them. Similarly in the equation
3x – y = 3
=> y = 3x – 3
Now plot these points and join them We see that these two lines intersect each other at (1, 0)
Question - 32 : - Form the pair of linear equations in the following problems, and find their solution graphically.
(i) 10 students of class X took part in 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.
(ii) 5 pencils and 7 pens together costs Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and a pen.
(iii) Champa went to a ‘sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased.” Help her friends to find how many pants and skirts Champa bought. [NCERT]
Answer - 32 : -
Let number of boys = x
and number of girls = y
According to the given conditions
x + y = 10
y – x = 4
Now, x + y = 10
=> x = 10 – y
Substituting some different values of y, we get their corresponding values of x as given below
Plot the points on the graph and join them Similarly in the equation
y – x = 4
=> y = 4 + x
Now plot the points and join them we see that these two lines intersect each other at (3, 7)
x = 3, y = 7
Number of boys = 3
and number of girls = 7
(ii) Let cost of 1 pencil = Rs. x
and cost of 1 pen = Rs. y
According to the given conditions,
5x + 7y = 50
2x + 5y = 46
5x + 7y = 50
5x = 50 – 7y
x =
Substituting some different values of y, we get them corresponding values of x as given below
Plot these points and join them Similarly in the equation
7x + 5y = 46
=> 7x = 46 – 5y
=> x =
Now plot the points on the graph and join them. We see that these two lines intersect each other at (3, 5)
x = 3, y = 5
or cost of pencil = Rs. 3
and cost of a pen = Rs. 5
(iii) Let number of skirts = x
and number of pants = y
According to the given condition,
x = 2y – 2 and x = 4y – 4
2y – 2 = 4y – 4
4y – 2y = -2 + 4
2y = 2
y = 1
and x = 2y – 2 = 2 x 1 – 2 = 2 – 2 = 0
Number of skirts = 0
and number of pants = 1
Question - 33 : - Solve the following system of equations graphically shade the region between the lines and the y -axis
(i) 3x – 4y = 7
5x + 2y = 3 (C.B.S.E. 2006C)
(ii) 4x – y = 4
3x + 2y = 14 (C.B.S.E. 2006C)
Answer - 33 : -
(i) 3x – 4y = 7
3x = 7 + 4y
x =
Substituting some different values of y, we get their corresponding values of x as shown below
Plot these points on the graph and join them. Similarly in the equation
5x + 2y = 3
=> 5x = 3 – 2y
x =
Plot these points and join them We see that the lines intersect each other at (1, -1)
x = 1, y = -1
Now the region between the these lines and y-axis has been shaded as shown
(ii) 4x – y = 4
3x + 2y = 14
4x – y = 4
y = 4x – 4
Substituting some different values of x, we get their corresponding values of y as given below
Plot these points and join them Similarly in equation
3x + 2y = 14
3x = 14 – 2y
x =
Now plot these points and join them
We see that these lines intersect each other at (2, 4)
x = 2, y = 4
The region between these two lines and y-axis has been shaded as shown
Represent the following pair of equations graphically and write the coordinates of points where the lines intersects y-axis
x + 3y = 6
2x – 3y = 12 (C.B.S.E. 2008)
Answer - 34 : -
x + 3y = 6
x = 6 – 3y
Substituting some different values of y, we get their corresponding values of x as shown below
Now plot these points on the graph and join them
Similarly in the equation
2x – 3y = 12 => 2x = 12 + 3y
x =
Now plot there points and join them We see that these two lines meet y-axis at (0, 2) and (0, -4)
Question - 35 : - 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 [NCERT]
Answer - 35 : -
Given a linear equation 2x + 3y – 8 = 0
(i) When the lines are intersecting, then
Question - 36 : - Determine graphically the co-ordinates of the vertices of a triangle, the equations of whose sides are :
(i) y = x, y = 2x and y + x = 6 (C.B.S.E. 2000)
(ii) y = x, 3y = x, x + y = 8 (C.B.S.E. 2000)
Answer - 36 : -
(i) y = x
Substituting some different values of x, we get their corresponding values of y as shown below
Now plot the points on the graph and join them. Similarly in the equation y = 2x
and y + x = 6 => x = 6 – y
Now plot the points on the graph and join them. We see that these lines intersect each other at (0, 0), (3, 3) and (2, 4)
Vertices of the triangle so formed by these lines are (0, 0), (3, 3) and (2, 4)
(ii) y = x, 3y = x, x + y = 8
y = x
Substituting some different values of x, we get corresponding values of y as shown below
Plot these points on the graph and join them Similarly in the equation 3y = x
and x + y = 8 => x = 8 – y
Now plot the points and join them. We see that these lines intersect each other at (0,0), (4, 4), (6, 2)
The vertices of the triangle so formed are (0, 0), (4, 4) and (6, 2)
Graphically, solve the following pair of equations:
2x + y = 6
2x – y + 2 = 0
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis. [NCERT Exemplar]
Answer - 37 : -
Given equations are 2x + y – 6 and 2x – y + 2 = 0
Table for equation 2x + y = 6
Hence, the pair of equations intersect graphically at point E (1, 4), i.e., x = 1 and y = 4
Question - 38 : - Determine, graphically, the vertices of the triangles formed by the lines y = x, 3y = x, x + y = 8. [NCERT Exemplar]
Answer - 38 : -
Given linear equations are y = x …….(i)
3y = x ………(ii)
and x + y = 8 …….(iii)
For equation y = x,
If x = 1, then y = 1
If x = 0, then y = 0
If x = 2, then y = 2
Table for line y = x,
For equation x = 3y
If x = 0, then y = 0,
if x = 3, then y = 1
and if x = 6, then y = 2
Table for line x = 3y,
For equation,
If x = 0, then y = 8
if x = 8, then y = 0
and if x = 4, then y = 4
Table for line x + y = 8,
Plotting the points A (1, 1) and B (2,2), we get the straight line AB. Plotting the points C (3, 1) and D (6, 2), we get the straight line CD. Plotting the points P (0, 8), Q (4, 4) and R (8, 0), we get the straight line PQR. We see that lines AB and CD intersecting the line PR on Q and D, respectively.
So, ∆OQD is formed by these lines. Hence, the vertices of the ∆OQD formed by the given lines are O (0, 0), Q (4, 4) and D (6, 2).
Question - 39 : - Draw the graph of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also, find the area of the quadrilateral formed by the lines and the x-axis. |NCERT Exemplar]
Answer - 39 : -
Given equation of lines 2x – y – 4 = 0, x = 3 and x = 5
Table for line 2x – y – 4 = 0,
Draw the points P (0, -4) and Q (2,0) and join these points and form a line PQ also draw the lines x = 3 and x = 5.
Area of quadrilateral ABCD =
x distance between parallel lines (AB) x (AD + BC) [since, quadrilateral ABCD is a trapezium]=
x 2 x (6 + 2) [∵ AB = OB – OA = 5 – 3 = 2, AD = 2 and BC = 6]= 8 sq. units
Hence, the required area of the quadrilateral formed by the lines and the x-axis is 8 sq. units.
Question - 40 : - Draw the graphs of the lines x = -2, and y = 3. Write the vertices of the figure formed by these lines, the x-axis and the y-axis. Also, find the area of the figure. [NCERT Exemplar]
Answer - 40 : -
We know that the graph of x = -2 is a line parallel to y-axis at a distance of 2 units to the left of it. So, the line l is the graph of x = -2
The graph of y = 3 is a line parallel to the x-axis at a distance of 3 units above it.
So, the line m is the graph of y = 3
The figure enclosed by the line x = -2, y = 3, the x-axis and the y-axis is OABC, which is a rectangle.
A is a point on the y-axis at a distance of 3 units above the x-axis. So, the coordinates of A are (0, 3).
C is a point on the x-axis at a distance of 2 units to the left of y-axis. So, the coordinates of C are (-2, 0).
B is the solution of the pair of equations x = -2 and y = 3. So, the coordinates of B are (-2, 3).
So, the vertices of the rectangle OABC are O (0, 0), A (0, 3), B (-2, 3), C (-2, 0).
The length and breadth of this rectangle are 2 units and 3 units, respectively.
As the area of a rectangle = length x breadth, the area of rectangle OABC = 2 x 3 = 6 sq. units.