CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Worksheet Set 08

Question. The area of the triangle formed by \( 2x - y + 6 = 0 \), \( 4x + 5y - 16 = 0 \) and the x – axis is 
(a) 15 sq. units
(b) 16 sq. units
(c) 14 sq. units
(d) 12 sq. units
Answer: (c) 14 sq. units

Question. The system of equations \( 2x + 3y - 7 = 0 \) and \( 6x + 5y - 11 = 0 \) has  
(a) unique solution
(b) infinite many solutions
(c) no solution
(d) non zero solution
Answer: (a) unique solution

Question. The pair of linear equations \( 5x - 3y = 11 \) and \( -10x + 6y = -22 \) are 
(a) dependent(consistent)
(b) None of the options
(c) consistent
(d) inconsistent
Answer: (a) dependent(consistent)

Question. The system of linear equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) has no solution if (1)
(a) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
(b) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
(c) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
(d) None of the options
Answer: (a) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)

Question. Ten students of class X took part in Mathematics quiz. The number of girls is 4 more than that of the boys. The algebraic representation of the above situation is 
(a) None of the options
(b) \( x - y = 10 \) and \( x + y = 4 \)
(c) \( x = y - 12 \) and \( y = 6 + x \)
(d) \( y = x + 4 \) and \( x = 10 - y \)
Answer: (d) \( y = x + 4 \) and \( x = 10 - y \)

Question. Given the linear equation \( 3x + 4y - 8 = 0 \), write another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines. 
Answer: Given equation \( 3x + 4y - 8 = 0 \).
Lines are parallel when \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
One of the linear equation in two variables can be \( 6x + 8y + k = 0 \) where \( k \) is constant not equal to -16.

Question. Write whether the following pair of linear equations is consistent or not.  
\( x + y = 14 \)
\( x - y = 4 \)
Answer: Given \( x + y = 14 \) and \( x - y = 4 \) such that \( a_1 = 1, b_1 = 1, c_1 = -14 \) and \( a_2 = 1, b_2 = -1, c_2 = -4 \).
Now, \( \frac{a_1}{a_2} = 1 \) and \( \frac{b_1}{b_2} = -1 \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the equations have unique solution. Therefore pair of linear equations is consistent.

Question. In a deer park, the number of heads and the number of legs of deer and human visitors were counted and it was found that there were 39 heads and 132 legs. Find the number of deer and human visitors in the park.  
Answer: Let a number of humans be \( x \) and deer be \( y \).
According to the condition, \( x + y = 39 ....(i) \)
and \( 2x + 4y = 132 \implies x + 2y = 66 .....(ii) \)
On solving (i) and (ii), we get
\( (x + 2y) - (x + y) = 66 - 39 \)
\( \implies y = 27 \)
Putting \( y = 27 \) in (i), we get \( x + 27 = 39 \implies x = 12 \).
So, \( y = 27, x = 12 \).

Question. Write the value of k for which the system of equations \( x + y - 4 = 0 \) and \( 2x + ky - 3 = 0 \) has no solution. 
Answer: Given system of equations \( x + y - 4 = 0 \) and \( 2x + ky - 3 = 0 \). With \( a_1=1, b_1=1, c_1=-4 \) and \( a_2=2, b_2=k, c_2=-3 \).
For no solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \implies \frac{1}{2} = \frac{1}{k} \neq \frac{-4}{-3} ...... (1) \)
Then \( \frac{1}{2} = \frac{1}{k} \implies k = 2 \).
Substitute \( k = 2 \) in (1) we get \( \frac{1}{2} = \frac{1}{2} \neq \frac{4}{3} \).
Hence, \( k = 2 \) for no solution.

Question. Determine k for which the system of equations has infinite solutions: (1) \( 4x + y = 3 \) and \( 8x + 2y = 5k \)
Answer: For infinite many solutions \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \frac{4}{8} = \frac{1}{2} = \frac{-3}{-5k} \)
\( \implies \frac{1}{2} = \frac{3}{5k} \)
\( \implies 5k = 6 \implies k = \frac{6}{5} \)

Question. Solve the following system of equations: (2) \( 2x - \frac{3}{y} = 9, 3x + \frac{7}{y} = 2, y \neq 0 \)
Answer: The given system of equations is
\( 2x - \frac{3}{y} = 9 .......... (i) \)
\( 3x + \frac{7}{y} = 2, y \neq 0 ............ (ii) \)
Taking \( \frac{1}{y} = u \), the given equations become,
\( 2x - 3u = 9 ........... (iii) \)
\( 3x + 7u = 2 ........... (iv) \)
From (iii), we get \( 2x = 9 + 3u \implies x = \frac{9+3u}{2} \)
Substituting \( x = \frac{9+3u}{2} \) in (iv), we get \( 3 \left( \frac{9+3u}{2} \right) + 7u = 2 \)
\( \implies \frac{27 + 9u + 14u}{2} = 2 \)
\( \implies 27 + 23u = 4 \)
\( \implies 23u = 4 - 27 \implies 23u = -23 \implies u = -1 \).
Hence, \( y = \frac{1}{u} = \frac{1}{-1} = -1 \).
Putting \( u = -1 \) in \( x = \frac{9+3u}{2} \), we get \( x = \frac{9+3(-1)}{2} = \frac{9-3}{2} = \frac{6}{2} = 3 \).
Hence, Solution of the given system of equation is \( x = 3, y = -1 \).

Question. Determine the values of a and b for which the following system of linear equations has infinite solutions: (2) \( 2x - (a - 4)y = 2b + 1, 4x - (a - 1)y = 5b - 1 \)
Answer: \( 2x - (a - 4)y = 2b + 1 ........ (i) \)
\( 4x - (a - 1)y = 5b - 1 ....... (ii) \)
Compare the equations with form \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \).
We get, \( a_1 = 2, b_1 = -(a - 4), c_1 = 2b + 1 \) and \( a_2 = 4, b_2 = -(a - 1), c_2 = 5b - 1 \).
Equations has infinite number of solutions, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Therefore, \( \frac{2}{4} = \frac{-(a-4)}{-(a-1)} = \frac{2b+1}{5b-1} \)
\( \implies \frac{1}{2} = \frac{a-4}{a-1} \) and \( \frac{1}{2} = \frac{2b+1}{5b-1} \)
\( \implies a - 1 = 2a - 8 \implies a = 7 \)
And \( 5b - 1 = 4b + 2 \implies b = 3 \).
Hence \( a = 7 \) and \( b = 3 \).

Question. Solve the following system of linear equation by substitution method: (2) \( 2x - y = 2 ...(i), x + 3y = 15 ....(ii) \)
Answer: Given, \( 2x - y = 2 ..(i) \) and \( x + 3y = 15 ..(ii) \).
From eqn. (i), we get \( y = 2x - 2 ... (iii) \)
Substituting the value of \( y \) in eqn. (ii), we get \( x + 3(2x - 2) = 15 \)
\( \implies x + 6x - 6 = 15 \implies 7x = 21 \implies x = 3 \).
Substituting this value of \( x \) in (iii), we get \( y = 2(3) - 2 \implies y = 6 - 2 = 4 \).
Hence the value of \( x \) and \( y \) of given equations are 3 and 4 respectively.

Question. Solve for x and y: \( \frac{3}{x+y} + \frac{2}{x-y} = 2, \frac{9}{x+y} - \frac{4}{x-y} = 1 \).  
Answer: According to the question,
\( \frac{3}{x+y} + \frac{2}{x-y} = 2 \)
\( \frac{9}{x+y} - \frac{4}{x-y} = 1 \)
Putting \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \),
\( 3u + 2v = 2 .......... (i) \)
\( 9u - 4v = 1 .......... (ii) \)
Multiplying (i) by 2 and (ii) by 1, we get
\( 6u + 4v = 4 ........ (iii) \)
\( 9u - 4v = 1 ......... (iv) \)
Adding (iii) and (iv), \( 15u = 5 \implies u = \frac{1}{3} \).
Putting \( u = \frac{1}{3} \) in (i), \( 3 \times \frac{1}{3} + 2v = 2 \implies 1 + 2v = 2 \implies 2v = 1 \implies v = \frac{1}{2} \).
Now, \( u = \frac{1}{3} \implies x + y = 3 ..........(v) \)
And \( v = \frac{1}{2} \implies x - y = 2 ............(vi) \)
Adding (v) and (vi), \( 2x = 5 \implies x = \frac{5}{2} \).
Putting \( x = \frac{5}{2} \) in (v), \( \frac{5}{2} + y = 3 \implies y = 3 - \frac{5}{2} = \frac{1}{2} \).
The solution is \( x = \frac{5}{2}, y = \frac{1}{2} \).

Question. Draw the graphs of the equations \( 4x - y - 8 = 0 \) and \( 2x - 3y + 6 = 0 \). Also, determine the vertices of the triangle formed by the lines and x-axis. 
Answer: \( 4x - y - 8 = 0 \implies y = 4x - 8 \)
Solution table for \( 4x - y - 8 = 0 \) is:
x: 0, 1, 2
y: -8, -4, 0
\( 2x - 3y + 6 = 0 \implies 3y = 2x + 6 \)
Solution table for \( 2x - 3y + 6 = 0 \) is:
x: 0, 3, -3
y: 2, 4, 0
Vertices of the triangle formed by lines and x-axis are \( (2, 0), (3, 4) \) and \( (-3, 0) \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations: (3) \( \frac{1}{(3x+y)} + \frac{1}{(3x-y)} = \frac{3}{4} \) and \( \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8} \)
Answer: The given equations are:
\( \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} ........... (1) \)
\( \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8} .......... (2) \)
Put \( \frac{1}{3x+y} = u ......... (3) \) and \( \frac{1}{3x-y} = v .......... (4) \).
Then, \( u + v = \frac{3}{4} .......(5) \)
\( \frac{1}{2}u - \frac{1}{2}v = -\frac{1}{8} \implies u - v = -\frac{1}{4} .......(6) \)
Adding (5) and (6), \( 2u = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \implies u = \frac{1}{4} \).
Subtracting (6) from (5), \( 2v = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 \implies v = \frac{1}{2} \).
From (3), \( 3x + y = 4 ....(7) \) and from (4), \( 3x - y = 2 ....(8) \).
Adding (7) and (8), \( 6x = 6 \implies x = 1 \).
Substituting \( x = 1 \) in (7), \( 3(1) + y = 4 \implies y = 1 \).
Hence, the solution is \( x = 1, y = 1 \).

Question. 2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work. (3)
Answer: Suppose man's 1 day's work be \( \frac{1}{x} \) and boy's 1 day's work be \( \frac{1}{y} \).
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \).
By first condition, \( 2u + 5v = \frac{1}{4} ....... (i) \)
By second condition, \( 3u + 6v = \frac{1}{3} ........ (ii) \)
Multiplying (i) by 6 and (ii) by 5, we get
\( 12u + 30v = \frac{6}{4} ..........(iii) \)
\( 15u + 30v = \frac{5}{3} ...........(iv) \)
Subtracting (iii) from (iv), \( 3u = \frac{5}{3} - \frac{3}{2} = \frac{10-9}{6} = \frac{1}{6} \implies u = \frac{1}{18} \).
Putting \( u = \frac{1}{18} \) in (i), \( \frac{2}{18} + 5v = \frac{1}{4} \implies 5v = \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \implies v = \frac{1}{36} \).
Now, \( u = \frac{1}{18} \implies x = 18 \) and \( v = \frac{1}{36} \implies y = 36 \).
The man will complete the work in 18 days and the boy in 36 days when they work alone.

Question. A motor boat takes 6 hours to cover 100 km downstream and 30 km upstream. If the boat goes 75 km downstream and returns back to the starting point in 8 hours. Find the speed of the boat in still water and the speed of the stream.  
Answer: Let the speed of the motor boat in still water be \( x \) km / hour and the speed of the stream be \( y \) km/hour.
Speed downstream = \( (x + y) \) km / hour and Speed upstream = \( (x - y) \) km / hour.
In the first case, \( \frac{100}{x+y} + \frac{30}{x-y} = 6 ...(1) \)
In the second case, \( \frac{75}{x+y} + \frac{75}{x-y} = 8 ...(2) \)
Put \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \).
\( 100u + 30v = 6 \implies 50u + 15v = 3 ...(3) \)
\( 75u + 75v = 8 \implies 15u + 15v = \frac{8}{5} ...(4) \)
Subtracting (4) from (3), \( 35u = 3 - \frac{8}{5} = \frac{7}{5} \implies u = \frac{1}{25} \).
Putting \( u = \frac{1}{25} \) in (3), \( 50 \times \frac{1}{25} + 15v = 3 \implies 2 + 15v = 3 \implies v = \frac{1}{15} \).
Now, \( x + y = 25 \) and \( x - y = 15 \).
Adding these, \( 2x = 40 \implies x = 20 \). Subtracting them, \( 2y = 10 \implies y = 5 \).
Speed of boat in still water is 20 km/hour and speed of stream is 5 km/hour.

Question. Solve the following system of linear equation graphically \( 4x - 5y - 20 = 0 \) and \( 3x + 5y - 15 = 0 \). Also, find the coordinates of the vertices of the Triangle formed by these two lines and the Y-axis.  
Answer: Given equations are \( 4x - 5y - 20 = 0 \) and \( 3x + 5y - 15 = 0 \).
For \( 4x - 5y - 20 = 0 \): When \( y = 0, x = 5 \); when \( x = 0, y = -4 \).
For \( 3x + 5y - 15 = 0 \): When \( y = 0, x = 5 \); when \( x = 0, y = 3 \).
Graphing these, the two lines intersect at \( A(5, 0) \).
The lines meet y-axis at \( B(0, -4) \) and \( C(0, 3) \) respectively.
Therefore, the vertices of the triangle are \( (5, 0), (0, -4) \) and \( (0, 3) \).

Question. Solve the following system of linear equations graphically and shade the region between the two lines and x-axis: (4) \( 3x + 2y - 4 = 0, 2x - 3y - 7 = 0 \)
Answer: Given system of linear equations is \( 3x + 2y - 4 = 0 .....(1) \) and \( 2x - 3y - 7 = 0 .....(2) \).
For eqn (1), \( y = \frac{4-3x}{2} \). If \( x = 0, y = 2 \); if \( x = 2, y = -1 \).
For eqn (2), \( y = \frac{2x-7}{3} \). If \( x = 2, y = -1 \); if \( x = 5, y = 1 \).
Plotting these points on graph, we find the lines intersect at \( (2, -1) \). The region between the lines and x-axis is shaded.