CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Worksheet Set 06

Question. Determine graphically the co-ordinates of the vertices of the triangle, the equations of whose sides are: \( y = x, 3y = x, x + y = 8 \)
(a) 13 sq. units
(b) 21 sq. units
(c) 11 sq. units
(d) 12 sq. units
Answer: (d) 12 sq. units
Explanation: \( y = x \)

\( 3y = x \Rightarrow y = \frac{x}{3} \)

\( x + y = 8 \Rightarrow y = (-x + 8) \)

\( \Delta POQ \) is formed by the given three lines.
\( \therefore \text{ar } \Delta POQ = \text{ar } \Delta ROQ - \text{ar } \Delta POR = \frac{1}{2} \times 8 \times 4 - \frac{1}{2} \times 4 \times 2 = 16 - 4 = 12 \text{ sq.units} \)
The lines intersect at point A(4,4), B(0,0), C(6,2)

Question. A system of two linear equations in two variables is consistent, if their graphs
(a) do not intersect at any point
(b) coincide
(c) cut the x – axis
(d) intersect only at a point or they coincide with each other
Answer: (d) intersect only at a point or they coincide with each other
Explanation: A system of two linear equations in two variables is consistent, if their graphs intersect only at a point, because it has a unique solution or they may coincide with each other giving infinite solutions.

Question. The solution of \( px + qy = p - q \) and \( qx - py = p + q \) is
(a) \( x = -1 \) and \( y = 1 \)
(b) \( x = 1 \) and \( y = 1 \)
(c) \( x = 0 \) and \( y = 0 \)
(d) \( x = 1 \) and \( y = -1 \)
Answer: (d) \( x = 1 \) and \( y = -1 \)
Explanation: Given: \( a_1 = p, a_2 = q, b_1 = q, b_2 = -p, c_1 = p - q \) and \( c_2 = p + q \)
Using the cross-multiplication method,
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{-1}{a_1b_2 - a_2b_1} \)
\( \Rightarrow \frac{x}{q(p+q) - (-p)(p-q)} = \frac{y}{(p-q)q - (p+q)p} = \frac{-1}{p \times (-p) - q \times q} \)
\( \Rightarrow \frac{x}{pq + q^2 + p^2 - pq} = \frac{y}{pq - q^2 - p^2 - pq} = \frac{-1}{-p^2 - q^2} \)
\( \Rightarrow \frac{x}{q^2 + p^2} = \frac{y}{-q^2 - p^2} = \frac{-1}{-(p^2 + q^2)} \)
\( \Rightarrow \frac{x}{p^2 + q^2} = \frac{y}{-(p^2 + q^2)} = \frac{1}{p^2 + q^2} \)
\( \Rightarrow \frac{x}{p^2 + q^2} = \frac{1}{p^2 + q^2} \) and \( \frac{y}{-(p^2 + q^2)} = \frac{1}{p^2 + q^2} \)
\( \Rightarrow x = 1 \) and \( \Rightarrow y = -1 \)

Question. The value of ‘k’ so that the system of equations \( 3x - y - 5 = 0 \) and \( 6x - 2y - k = 0 \) have infinitely many solutions is
(a) \( k = -10 \)
(b) \( k = 10 \)
(c) \( k = -8 \)
(d) \( k = 8 \)
Answer: (b) \( k = 10 \)
Explanation: Given: \( a_1 = 3, a_2 = 6, b_1 = -1, b_2 = -2, c_1 = -5 \) and \( c_2 = -k \)
If there is infinitely many solutions, then \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{3}{6} = \frac{-1}{-2} = \frac{-5}{-k} \)
Taking \( \frac{-1}{-2} = \frac{-5}{-k} \Rightarrow k = 5 \times 2 \Rightarrow k = 10 \)

Question. 5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. The cost of 1 pen is
(a) Rs. 5
(b) Rs. 6
(c) Rs. 3
(d) Rs. 4
Answer: (a) Rs. 5
Explanation: Let, cost (in Rs.) of one pencil = x
and cost (in Rs.) of one pen = y
Therefore, according to question
\( 5x + 7y = 50 \) ........ (1)
\( 7x + 5y = 46 \) ......... (2)
Multiply equation (1) by 7 and equation (2) by 5 we get
\( 7(5x + 7y) = 7 \times 50 \)
\( 35x + 49y = 350 \) ....... (3)
and \( 5(7x + 5y) = 5 \times 46 \)
\( 35x + 25y = 230 \) ....... (4)
Subtract equation (4) from equation (3), we get
\( 35x + 49y - 35x - 25y = 350 - 230 \)
\( 24y = 120 \)
\( y = \frac{120}{24} = 5 \)
Substitute \( y = 5 \) in equation 1, we get
\( 5x + 7 \times 5 = 50 \)
\( 5x + 35 = 50 \)
\( 5x = 50 - 35 = 15 \)
\( x = \frac{15}{5} = 3 \)
Hence, Cost of One Pen = \( y = 5 \)

Question. Solve for x and y: \( \frac{2x+5y}{xy} = 6 \); \( \frac{4x-5y}{xy} = -3 \).
Answer: \( \frac{2x+5y}{xy} = 6 \) ..........(i)
\( \frac{4x-5y}{xy} = -3 \) .......(ii)
Adding (i) and (ii),
\( \frac{6x}{xy} = 3 \Rightarrow \frac{6}{y} = 3 \Rightarrow y = 2 \)
Substituting \( y = 2 \) in (i), we get \( x = 1 \)
\( \therefore x = 1 \) and \( y = 2 \)

Question. Determine the values of m and n so that the following system of linear equations have infinite number of solutions: \( (2m - 1)x + 3y - 5 = 0: 3x + (n - 1)y - 2 = 0 \)
Answer: We have to determine the values of m and n so that the following system of linear equations have infinite number of solutions:
\( (2m - 1)x + 3y - 5 = 0 \)
\( 3x + (n - 1)y - 2 = 0 \)
It is given that \( (2m - 1)x + 3y - 5 = 0 \) ...(i)
\( a_1 = 2m - 1, b_1 = 3, c_1 = -5 \)
\( 3x + (n - 1)y - 2 = 0 \) ...(ii)
\( a_2 = 3, b_2 = n - 1, c_2 = -2 \)
On comparing with the general form of eqn.
For a pair of linear equations to have infinite number of solutions
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
or \( \frac{2m - 1}{3} = \frac{3}{n - 1} = \frac{5}{2} \)
\( \frac{2m - 1}{3} = \frac{5}{2} \)
or \( 2(2m - 1) = 15 \)
or, \( 4m - 2 = 15 \)
or, \( 4m = 17 \)
\( m = \frac{17}{4} \)
and \( \frac{3}{n - 1} = \frac{5}{2} \)
or, \( 5(n - 1) = 6 \)
or, \( 5n - 5 = 6 \)
or, \( 5n = 11 \)
or, \( n = \frac{11}{5} \)
Hence, \( m = \frac{17}{4}, n = \frac{11}{5} \)

Question. Find whether the following pair of equations has no solution, unique solution or infinitely many solutions.
\( 5x - 8y + 1 = 0 \);
\( 3x - \frac{24}{5}y + \frac{3}{5} = 0 \)
Answer: \( a_1 = 5, b_1 = -8, c_1 = 1 \) and \( a_2 = 3, b_2 = \frac{-24}{5}, c_2 = \frac{3}{5} \)
\( \frac{a_1}{a_2} = \frac{5}{3} \) ...(i)
\( \frac{b_1}{b_2} = \frac{-8}{-24/5} = \frac{5}{3} \) ...(ii)
and \( \frac{c_1}{c_2} = \frac{1}{3/5} = \frac{5}{3} \) ...(iii)
Form (i), (ii) and (iii)
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \therefore \) The pair of equations has infinitely many solutions.

Question. For what value of k the following pair of linear equation has unique solution?
\( kx + 3y = 3 \)
\( 12x + ky = 6 \)
Answer: Given pair of equations
\( kx + 3y = 3, 12x + ky = 6 \)
For unique solutions \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
\( \frac{k}{12} \neq \frac{3}{k} \)
\( \Rightarrow k^2 \neq 36 \)
\( \Rightarrow k \neq \pm 6 \).

Question. If am = bl, then find whether the pair of linear equations \( ax + by = c \) and \( lx + my = n \) has no solution, unique solution or infinitely many solutions.
Answer: Since, \( am = bl \)
\( \therefore \frac{a}{l} = \frac{b}{m} \neq \frac{c}{n} \)
So, \( ax + by = c \) and \( lx + my = n \) has no solution.

Question. For what value of k will the equations \( x + 2y + 7 = 0, 2x + ky + 14 = 0 \) represent coincident lines?
Answer: The given equation are
\( x + 2y + 7 = 0 \)
\( 2x + ky + 14 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \),
After comparing the given equation with standard equation
We get, \( a_1 = 1, b_1 = 2, c_1 = 7 \) and \( a_2 = 2, b_2 = k, c_2 = 14 \)
The given equations will represent coincident lines if they have infinitely many solutions.
The condition for which is
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \Rightarrow \frac{1}{2} = \frac{2}{k} = \frac{7}{14} \)
\( \Rightarrow k = 4 \)
Hence, the given system of equations will represent coincident lines, if \( k = 4 \).

Question. The expenses of a lunch are partly constant and partly proportional to the number of guests. The expenses amount to Rs. 65 for 7 guests and Rs. 97 for 11 guests. How much the expenses for 18 guests will amount to?
Answer: Let the fixed expenses = Rs. x
and proportional charges = Rs. y
As per given condition
The expenses amount to Rs. 65 for 7 guests .
\( x + 7y = 65 \) ..(i)
And the expenses amount Rs. 97 for 11 guests.
So, \( x + 11y = 97 \) ..(ii)
Subtracting (i) from (ii), we get
\( 4y = 32 \)
\( \Rightarrow y = 8 \)
Put \( y = 8 \) eq. (i) , we get
\( x + 7(8) = 65 \)
\( \Rightarrow x = 9 \)
Now, expenses for 18 guests
\( = x + 18y \)
\( = 9 + 18(8) \)
\( = 9 + 144 = Rs. 153 \)

Question. Write the value of k for which the system of equations \( 3x + ky = 0, 2x - y = 0 \) has a unique solution.
Answer: The given equations are
\( 3x + ky = 0 \) ........ (i)
\( 2x - y = 0 \) ......... (ii)
We know that,
The system of linear equations is in the form of
\( a_1x + b_1y + c_1 = 0 \)
and \( a_2x + b_2y + c_2 = 0 \)
Compare (i) and (ii), we get
\( a_1 = 3, b_1 = k \) and \( c_1 = 0 \)
\( a_2 = 2, b_2 = -1 \) and \( c_2 = 0 \)
The equations has a unique solution if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
So, \( \frac{3}{2} \neq \frac{k}{-1} \)
\( \Rightarrow k \neq -\frac{3}{2} \)
Thus, k can take any real values except \( -\frac{3}{2} \).

Question. Solve the following pair of linear equations by the elimination method and the substitution method: \( 3x + 4y = 10 \) and \( 2x – 2y = 2 \).
Answer: 1. By Elimination method,
The given system of equation is :
\( 3x + 4y = 10 \) ...................(1)
\( 2x - 2y = 2 \) ...................(2)
Multiplying equation (2) by 2, we get
\( 4x - 4y = 4 \) ...................(3)
Adding equation (1) and equation (3), we get \( 7x = 14 \)
\( \therefore x = \frac{14}{7} = 2 \)
Substituting this value of x in equation (2), we get \( 2(2) - 2y = 2 \)
\( \Rightarrow 4 - 2y = 2 \)
\( \Rightarrow 2y = 4 - 2 \)
\( \Rightarrow 2y = 2 \)
\( \Rightarrow y = \frac{2}{2} = 1 \)
So, the solution of the given system of equation is \( x = 2, y = 1 \)
2. By Substitution method,
The given system of equation is:
\( 3x + 4y = 10 \) .................(1)
\( 2x - 2y = 2 \) ....................(2)
From equation (2),
\( 2y = 2x - 2 \Rightarrow y = \frac{2x-2}{2} = x - 1 \) .................(3)
Substituting this value of y in equation (1), we get
\( 3x + 4(x-1) = 10 \Rightarrow 3x + 4x - 4 = 10 \Rightarrow 7x = 14 \Rightarrow x = 2 \)
Substituting this value of x in equation (3), we get \( y = 2 - 1 = 1 \)
So, the solution of the given system of equation is \( x = 2, y = 1 \)
Verification: Substituting \( x = 2, y = 1 \), we find that both the equation (1) and (2) are satisfied shown below:
\( 3x + 4y = 3(2) + 4(1) = 6 + 4 = 10 \)
\( 2x - 2y = 2(2) - 2(1) = 4 - 2 = 2 \)
Hence, the solution is correct.

Question. A person rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.
Answer: Let the speed of the stream = x km/hr
Speed of the boat in upstream = (5 - x) km/hr
Speed of the boat in downstream = (5 + x) km/hr
Distance = 40 km
Time taken in upstream = \( \frac{40}{5-x} \) hr
Time taken in downstream = \( \frac{40}{5+x} \) hr
According to the question,
\( \frac{40}{5-x} = 3 \left( \frac{40}{5+x} \right) \)
\( \Rightarrow (5 + x) = 3(5 - x) \)
\( \Rightarrow 5 + x = 15 - 3x \)
\( \Rightarrow 4x = 10 \Rightarrow x = 2.5 \)
Therefore, speed of the stream = 2.5 km/hr.

Question. Solve the following system of equations in x and y \( ax + by = 1, bx + ay = \frac{(a+b)^2}{a^2+b^2} - 1 \) or \( bx + ay = \frac{2ab}{a^2+b^2} \)
Answer: The given system of equations are
\( ax + by - 1 = 0 \) .......... (i)
\( bx + ay - \frac{2ab}{a^2+b^2} = 0 \) .......... (ii)
By cross-multiplication, of equations (i) and (ii), we have
\( \frac{x}{b \times \left( \frac{-2ab}{a^2+b^2} \right) - a \times -1} = \frac{-y}{a \times \left( \frac{-2ab}{a^2+b^2} \right) - b \times -1} = \frac{1}{a \times a - b \times b} \)
\( \Rightarrow \frac{x}{\frac{-2ab^2}{a^2+b^2} + a} = \frac{-y}{\frac{-2a^2b}{a^2+b^2} + b} = \frac{1}{a^2-b^2} \)
\( \Rightarrow \frac{x}{\frac{-2ab^2 + a^3 + ab^2}{a^2+b^2}} = \frac{-y}{\frac{-2a^2b + a^2b + b^3}{a^2+b^2}} = \frac{1}{a^2-b^2} \)
\( \Rightarrow \frac{x}{\frac{a^3 - ab^2}{a^2+b^2}} = \frac{-y}{\frac{b^3 - a^2b}{a^2+b^2}} = \frac{1}{a^2-b^2} \)
\( \Rightarrow \frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}} = \frac{-y}{\frac{b(b^2-a^2)}{a^2+b^2}} = \frac{1}{a^2-b^2} \)
\( \Rightarrow x = \frac{a(a^2-b^2)}{a^2+b^2} \times \frac{1}{a^2-b^2} \) and \( y = \frac{b(a^2-b^2)}{a^2+b^2} \times \frac{1}{a^2-b^2} \)
\( \Rightarrow x = \frac{a}{a^2+b^2} \) and \( y = \frac{b}{a^2+b^2} \)
Hence, the solution of the given system of equations is \( x = \frac{a}{a^2+b^2}, y = \frac{b}{a^2+b^2} \).

Question. Examine whether the solution set of the system of equations \( 3x – 4y = – 7; 3x – 4y = –9 \). Is consistent or inconsistent.
Answer: The given system of equation is :
\( 3x - 4y = -7 \) .........(1)
\( 3x - 4y = -9 \) ..........(2)
Here, \( a_1 = 3, b_1 = -4, c_1 = 7 \)
\( a_2 = 3, b_2 = -4, c_2 = 9 \)
We see that \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Hence, the lines represented by the given pair of linear equations are parallel.
Therefore, equation (1) and (2) have no common solution, i.e., the solution set of the given system equations is inconsistent.

Question. Solve for x and y: \( \frac{35}{x+y} + \frac{14}{x-y} = 19 \) and \( \frac{14}{x+y} + \frac{35}{x-y} = 37 \).
Answer: Put \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \)
So, we get
\( 35u + 14v = 19 \) ........(i)
and \( 14u + 35v = 37 \) ..........(ii)
Adding (i) and (ii), we get
\( 49u + 49v = 56 \)
\( \Rightarrow 7u + 7v = 8 \) ........(iii)
Subtract (i) from (ii), we get
\( 21u - 21v = -18 \)
\( \Rightarrow 7u - 7v = -6 \) ..........(iv)
Adding (iii) and (iv), we get
\( 14u = 2 \Rightarrow u = \frac{1}{7} \)
Substituting \( u = \frac{1}{7} \) in (iii), we get \( v = 1 \),
So, \( \frac{1}{x+y} = \frac{1}{7} \) and \( \frac{1}{x-y} = 1 \)
\( \Rightarrow x + y = 7 \) ........(iv)
and \( x - y = 1 \) .........(v)
Adding (iv) and (v), we get
\( 2x = 8 \Rightarrow x = 4 \)
Substituting \( x = 4 \) in (iv), we get \( y = 3 \).
\( x=4 \) and \( y = 3 \)

Question. One says, "Give me a hundred rupee, friend! I shall then become twice as rich as you are." The other replies, "If you give me ten rupees, I shall be six times as rich as you are." Tell me how much money both have initially?
Answer: Suppose initially, they had Rs x and Rs. y with them respectively.
as per condition given in the question, we obtain
\( x + 100 = 2(y - 100) \)
\( x + 100 = 2y - 200 \)
\( x - 2y = -300 \) ...(i)
and \( 6(x - 10) = (y + 10) \)
\( 6x - 60 = y + 10 \)
\( 6x - y = 70 \) .....(ii)
Multiplying equation (ii) by 2 & then subtracting equation (i) from it, we obtain:-
\( (12x - 2y) - (x - 2y) = 140 - (- 300 ) \)
\( 11x = 140 + 300 \)
\( 11x = 440 \)
\( x = 40 \)
Putting \( x = 40 \) in equation (i), we obtain
\( 40 - 2y = -300 \)
\( 40 + 300 = 2y \)
\( 2y = 340 \)
\( y = 170 \)
Therefore, initially they had Rs 40 and Rs 170 with them respectively.

Question. Half the perimeter of a rectangular garden, whose length is 4m more than its width is 36m. The area of the garden is (1)
(a) 320 \( m^2 \)
(b) 300 \( m^2 \)
(c) 400 \( m^2 \)
(d) 360 \( m^2 \)
Answer: (a) 320 \( m^2 \)

Question. A fraction becomes \( \frac{9}{11} \), if 2 is added to both the numerator and denominator. If 3 is added to both the numerator and denominator it becomes \( \frac{5}{6} \), then the fraction is (1)
(a) \( \frac{9}{7} \)
(b) \( \frac{-9}{7} \)
(c) \( \frac{7}{9} \)
(d) \( \frac{-7}{9} \)
Answer: (c) \( \frac{7}{9} \)

Question. If a pair of linear equation is consistent, then the lines will be (1)
(a) always intersecting
(b) intersecting or coincident
(c) always coincident
(d) parallel
Answer: (b) intersecting or coincident

Question. The system of equations \( 6x + 3y = 6xy \) and \( 2x + 4y = 5xy \) has (1)
(a) one solution
(b) one unique solution
(c) many solutions
(d) no solution
Answer: (b) one unique solution

Question. The sum of the numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to \( \frac{1}{3} \). The fraction is (1)
(a) \( \frac{-7}{11} \)
(b) \( \frac{5}{13} \)
(c) \( \frac{-5}{13} \)
(d) \( \frac{7}{11} \)
Answer: (b) \( \frac{5}{13} \)

Question. If \( ad \neq bc \), then find whether the pair of linear equations \( ax + by = p \) and \( cx + dy = q \) has no solution, unique solution or infinitely many. (1)
Answer: For the pair of linear equations \( ax + by = p \) and \( cx + dy = q \), we have \( \frac{a}{c} \) and \( \frac{b}{d} \). Given \( ad \neq bc \), which can be written as \( \frac{a}{c} \neq \frac{b}{d} \). Hence, the pair of given linear equations has a unique solution.

Question. For what value of a the following pair of linear equation has infinitely many solutions? \( 2x + ay = 8 \); \( ax + 8y = a \) (1)
Answer: For infinite numbers of solution, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). Here \( \frac{2}{a} = \frac{a}{8} = \frac{8}{a} \). From \( \frac{2}{a} = \frac{a}{8} \Rightarrow a^2 = 16 \Rightarrow a = \pm 4 \). From \( \frac{a}{8} = \frac{8}{a} \Rightarrow a^2 = 64 \Rightarrow a = \pm 8 \). Since there is no single value of \( a \) that satisfies both conditions simultaneously, the system does not have infinite solutions for any value of \( a \).

Question. Find whether the following pair of linear equations is consistent or inconsistent: \( x + 3y = 5 \); \( 2x + 6y = 8 \) (1)
Answer: Given equations are \( x + 3y = 5 \) and \( 2x + 6y = 8 \). Here, \( \frac{a_1}{a_2} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \), and \( \frac{c_1}{c_2} = \frac{5}{8} \). Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel and the pair of linear equations is inconsistent.

Question. If \( 12x + 17y = 53 \) and \( 17x + 12y = 63 \) then find the value of (x + y). (1)
Answer: Adding the equations: \( (12x + 17y) + (17x + 12y) = 53 + 63 \Rightarrow 29x + 29y = 116 \Rightarrow 29(x + y) = 116 \Rightarrow x + y = \frac{116}{29} = 4 \).

Question. For what value of k, the following system of equations represent parallel lines? \( kx - 3y + 6 = 0 \), \( 4x - 6y + 15 = 0 \) (1)
Answer: For parallel lines (no solution), \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Here \( \frac{k}{4} = \frac{-3}{-6} \neq \frac{6}{15} \). Thus, \( \frac{k}{4} = \frac{1}{2} \Rightarrow k = \frac{4}{2} = 2 \).

Question. Is the system of linear equations \( 2x + 3y - 9 = 0 \) and \( 4x + 6y - 18 = 0 \) consistent? Justify your answer (2)
Answer: For \( 2x + 3y - 9 = 0 \), \( a_1 = 2, b_1 = 3, c_1 = -9 \). For \( 4x + 6y - 18 = 0 \), \( a_2 = 4, b_2 = 6, c_2 = -18 \). Comparing ratios: \( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \), and \( \frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2} \). Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the system is consistent and dependent (coincident lines).

Question. Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138. (2)
Answer: Let the numbers be \( x \) and \( y \). According to conditions: \( 3x + y = 142 \) ...(i) and \( 4x - y = 138 \) ...(ii). Adding (i) and (ii), we get \( 7x = 280 \Rightarrow x = 40 \). Putting \( x = 40 \) in (i), \( 3(40) + y = 142 \Rightarrow 120 + y = 142 \Rightarrow y = 22 \). The numbers are 40 and 22.

Question. Solve the following systems of equations by using the method of substitution: \( 3x - 5y = -1 \), \( x - y = -1 \) (2)
Answer: From \( x - y = -1 \), we get \( y = x + 1 \). Substituting this in \( 3x - 5y = -1 \): \( 3x - 5(x + 1) = -1 \Rightarrow 3x - 5x - 5 = -1 \Rightarrow -2x = 4 \Rightarrow x = -2 \). Putting \( x = -2 \) in \( y = x + 1 \), we get \( y = -2 + 1 = -1 \). Solution is \( x = -2, y = -1 \).

Question. Solve for x and y: \( \frac{6}{x+y} = \frac{7}{x-y} + 3 \), \( \frac{1}{2(x+y)} = \frac{1}{3(x-y)} \) where \( x + y \neq 0 \) and \( x - y \neq 0 \). (3)
Answer: Let \( \frac{1}{x+y} = A \) and \( \frac{1}{x-y} = B \). Equations become: \( 6A = 7B + 3 \Rightarrow 6A - 7B = 3 \) ...(i) and \( \frac{1}{2}A = \frac{1}{3}B \Rightarrow 3A = 2B \Rightarrow 3A - 2B = 0 \) ...(ii). Multiplying (ii) by 2 gives \( 6A - 4B = 0 \). Subtracting from (i): \( (6A - 7B) - (6A - 4B) = 3 - 0 \Rightarrow -3B = 3 \Rightarrow B = -1 \). Substituting \( B = -1 \) in (ii): \( 3A - 2(-1) = 0 \Rightarrow 3A = -2 \Rightarrow A = -\frac{2}{3} \). Thus, \( x + y = -\frac{3}{2} \) and \( x - y = -1 \). Solving these: \( 2x = -\frac{5}{2} \Rightarrow x = -\frac{5}{4} \) and \( 2y = -\frac{1}{2} \Rightarrow y = -\frac{1}{4} \).

Question. The cost of 2 kg of apples and 1 kg of grapes in a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes in Rs. 300. Represent the situation algebraically and geometrically. (3)
Answer: Let cost of 1 kg apples be Rs. \( x \) and grapes be Rs. \( y \). Algebraic representation: \( 2x + y = 160 \) and \( 4x + 2y = 300 \) (or \( 2x + y = 150 \)). Since \( \frac{2}{2} = \frac{1}{1} \neq \frac{160}{150} \), the lines are parallel. Geometrically, plotting points like (50, 60) and (40, 80) for the first line, and (50, 50) and (30, 90) for the second line results in two parallel lines that do not intersect.

Question. The sum of a two digit number and the number formed by interchanging the digit is 132. If 12 is added to the number, the new number becomes 5 times the sum of the digits. Find the number. (3)
Answer: Let digits be \( x \) (units) and \( y \) (tens). Number is \( 10y + x \), interchanged is \( 10x + y \). Condition 1: \( (10y + x) + (10x + y) = 132 \Rightarrow 11x + 11y = 132 \Rightarrow x + y = 12 \). Condition 2: \( (10y + x) + 12 = 5(x + y) \Rightarrow 10y + x + 12 = 5x + 5y \Rightarrow 4x - 5y = 12 \). Solving \( x + y = 12 \) and \( 4x - 5y = 12 \) using cross-multiplication or substitution gives \( x = 8 \) and \( y = 4 \). The number is \( 10(4) + 8 = 48 \).

Question. The coach of the cricket team buys 3 bat and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Representing this situation algebraically and geometrically. (3)
Answer: Let cost of one bat be Rs. \( x \) and one ball be Rs. \( y \). Equations: \( 3x + 6y = 3900 \Rightarrow x + 2y = 1300 \) and \( x + 3y = 1300 \). Geometrically, line 1 passes through (0, 650) and (1300, 0). Line 2 passes through (400, 300) and (1000, 100). The lines intersect at (1300, 0).

Question. On selling a T.V. at 5% gain and a fridge at 10% gain. A shopkeeper gains Rs. 2000. But if he sells the T.V. at 10% gain and the fridge at 5% loss. He gains Rs. 1500 on the transaction. Find the actual price of the T.V. and the fridge. (4)
Answer: Let T.V. price be Rs. \( x \) and Fridge price be Rs. \( y \). From first condition: \( 0.05x + 0.10y = 2000 \Rightarrow x + 2y = 40000 \). From second: \( 0.10x - 0.05y = 1500 \Rightarrow 2x - y = 30000 \). Solving these: Multiplying the second by 2 gives \( 4x - 2y = 60000 \). Adding to first: \( 5x = 100000 \Rightarrow x = 20000 \). Then \( 20000 + 2y = 40000 \Rightarrow 2y = 20000 \Rightarrow y = 10000 \). T.V. is Rs. 20000 and Fridge is Rs. 10000.

Question. Write the number of solutions of the following pair of linear equations: \( x + 2y - 8 = 0, 2x + 4y = 16 \). (4)
Answer: Rewrite equations: \( x + 2y - 8 = 0 \) and \( 2x + 4y - 16 = 0 \). Comparing ratios: \( \frac{a_1}{a_2} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2} \), and \( \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \). Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the system has an infinite number of solutions.

Question. Five years hence, father's age will be three times the age of his son. Five years ago, father was seven times as old as his son. Find their present ages. (4)
Answer: Let present ages be \( x \) (father) and \( y \) (son). Five years hence: \( x + 5 = 3(y + 5) \Rightarrow x - 3y = 10 \) ...(i). Five years ago: \( x - 5 = 7(y - 5) \Rightarrow x - 7y = -30 \) ...(ii). Subtracting (ii) from (i): \( 4y = 40 \Rightarrow y = 10 \). Putting \( y = 10 \) in (i): \( x - 30 = 10 \Rightarrow x = 40 \). Present age of father is 40 years and son is 10 years.