UP Board Solutions for Class 10 Maths Chapter 3 Exercise 3.2 Pair Of Linear Equations In Two Variables
Exercise 3.2
1.Form the pair of linear equations in the following problems, and find •their solution graphically.
Hence, the cost of one pencil is Rs 3 and that of one pen is Rs 5. Verification: Put x = 3 and y = 5 in (1) and (2), we find that both the equations are satisfied.
2.On comparing the ratios
find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide. (i) 5x- 4y + 8= 0; 7x +6y – 9= 0 (ii) 9x + 3y + 12 = 0; 18x+ 6y +24 =0 (iii)6x -3y+10 =0;2x- y+9=0
Sol. (i) The given pair of linear equations are 5x- 4y +8= 0 and, …(1)
and, 7x +6y – 9 = 0…(2)
Here,
=> (1) and (2) are intersecting lines. (ii) The given pair of linear equations are 9x +3y +12 = 0
and, 18x +6y +24 = 0 Here,
=> (1) and (2) are coincident lines. (iii) The given pair of linear equations are
6r – 3y + 10 = 0 …(1) and, x2 – y + 9 = 0 ….(2)
=> (1) and (2) are parallel lines.
3.Oncomparing the ratios
find out whether the following pairs of linear equations are consistent, or inconsistent. (i) 3x+2y=5;2x-3y=7
4. Which of the following pairs of linear equations are consistent ? Obtain the solution in such cases graphically. (i) x+y=5, 2x+2y=10 (ii) x – y = 8, 3x-3y = 16 (iii) 2x+3-6=0, 47-23-4=0 (iv) 2x-2y-2=0, 4x-4y-5=0
Sol. (i) Graph of x +y= 5 : Wehave,x+y=5 = > y=5-x
When x=0, y=5; when x=5, y=0
Thus, we have the following table:
Plotting the points A(0, 5) and B(5, 0) on the graph paper. Join A and B and extend it on both sides to obtain the graph of x +y=5.
Graph of 2x + 2y = 10 :
We have 2x+2y=10 => 2y=10-2x
=> y=5-x
When x=1,y=5- 1=4; when x=2,y=5- 2=3 Thus, we have the following table:
Plotting the points C(1, 4) and D(2, 3) on the graph paper and drawing a line passing through these points on the same graph paper, we obtain the graph of 2x +2y = 10. We find that C and db othlie on the graph of x + y = 5.
Thus, the graphs of the two equations are coincident. Consequently, every solution of one equation is a solution of the other. Hence, the system of equations has infinitely many solutions, i.e., consistent. (ii) Graph of x – y = 8 : We have,x-y=8 = > y=x-8 When x = 0, y = – 8; when x = 8, y= 0 Thus, we have the following table
Plotting the points A(0, – 8) and B(8, 0) on a graph paper. Join A and B and extend it on both sides to obtain the graph of x- y = 8 as shown. Graph of 3x – 3y = 16 : Wehave, 3x-3y=16 => 3y=3x-16 => y=
Thus, we have the following table:
graph paper. Join C and D and extend it on both sides to obtain the graph of 3x- 3y = 16 as shown.
We find the graphs of x- y=8and 3x – 3y= 16 are parallel. So, the two lines have no common point. Hence, the given equations has no solution, i.e., inconsistent.
(iii) Graph of 2x +y- 6 = 0 :
Wehave,2x+y-6=0 => y=6-2x When x = 0, y = 6- 0 = 6; when x = 3, y = 6 – 6 = 0 Thus, we have the following table:
Plot the points A(0, 6) and B(3, 0) on a graph paper. Join A and B and extend it on both sides to obtain the graph of 2x +y – 6 = 0 as shown. Graph of 4x- 2y – 4 =0 : We have, 4x- 2y 4-=0 => 2y=4x-4>= > y=2x- 2
When x = 0, y = – 2; when x = 1, y = 0 Thus, we have the following table :
Plotting the points C(0, -2) and D(2, 0) on the same graph and drawing a line joining them asP(2, )2 shown.
Clearly, the two lines intersect at point P(2, 2).
Hence, x= 2, y = 2 is the solution of the given equations, i.e., consistent.
Verification: Putting x = 2, y = 2 in the given equations, we find that both the equations are satisfied.
(iv) Graph of 2x – 2y – 2 =0 : We have, 2x- 2y-2=0 => 2y=2x-2 = > y=x-2
When x=2,y=0; whenx=0,y=- 2
Thus, we have the following table:
Plot the points A(2, 0) and B(0, – 2) on a graph paper. Join A and B and extend it on both sides to obtain thegraphof 2x-2y- 2=0. Graph of 4x – 4y – 5 =0 : We have, 4x-4y-5=0 => 4y=5 => 5-4x => y=
Thus, we have the following table:
graph paper. Join C and D and extend it on both sides to obtain the graph of 4x – 4y – 5 =0 as shown.
We find the graphs of these equations are parallel lines. So, the two lines have no common point. Hence, the given system of equations has no solution, i.e., 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. Sol.Let the length of the garden be x m and its width be y m.
Hence, Length = 11 m and width = 7 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 Sol. Given linear equation is 2x + 3y – 8 = 0
(i) For intersecting lines, we know that
Any line intersecting line may be taken as
5x+2y- 9=0
(ii) For parallel lines, . Any line parallel to (1) may be taken as
4x + 6y – 3 = 0
(iii) For coincident lines,
Any line coincident to (1) may be taken as
6x + 9х – 24 = 0
Note : Answers can differ from person to person.
7.Draw the graphs of the equation 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 Sol. For the graph of x- y+1=0:
We have, x-y + 1 =0 => y=x+1
when x=0,y=1; when x=-1, y=0 Thus, ew have the following table :
Plot the points A(0, 1) and B-( 1, 0) on a graph paper. Join A and B and extend it on both sides to obtain thegraphof x- y+1=0. For the graph of 3x +2y- 12=0: We have, 3x+2y-12=0 = > 2y=12-3x => y=
Clearly, we obtain a A PBC formed by the given lines and the x-axis. The co-ordinates of the vertices are P(2, 3), B(- 1, 0) and C(4, 0).
= 7;9x – 10y = 14
+2y=8;28+34=12.
b1= 2, c1, =8; a2= 2, b2 = 3, C2 =12
