CBSE Class 10 Mathematics Pair Of Linear Equations In 2 Variables Worksheet Set E

Pair of Linear Equations in Two Variables

Points to Remember

• A pair of linear equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) graphically or geometrically represents a pair of straight lines which are,
  • (i) Intersecting, if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
  • (ii) Parallel, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
  • (iii) Coincident, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
• For a pair of linear equations in two variables \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \),
  • (i) If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), then the pair of linear equations represents a consistent and unique solution,
  • (ii) If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), then the pair of linear equations represents an inconsistent solution, and
  • (iii) If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), then the pair of linear equations represents consistent but infinite solutions.

Multiple Choice Questions

Question. The value of \( k \) for which the system of linear equations \( x + 2y = 3, 5x + ky + 7 = 0 \) is inconsistent is 
(a) \( -\frac{14}{3} \)
(b) \( \frac{2}{5} \)
(c) 5
(d) 10
Answer: (d)
Sol. For the system of linear equations to be inconsistent, we have \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Here \( a_1 = 1, b_1 = 2, c_1 = -3 \) and \( a_2 = 5, b_2 = k, c_2 = 7 \). So, \( \frac{1}{5} = \frac{2}{k} \neq \frac{-3}{7} \Rightarrow k = 10 \). \(\therefore\) Option (d) is correct.

Question. The values of \( x \) and \( y \) in \( 2x + 3y = 2 \) and \( x - 2y = 8 \) are :
(a) \( -4, 2 \)
(b) \( -4, -2 \)
(c) \( 4, -2 \)
(d) \( 4, 2 \)
Answer: (c)
Sol. Given, \( 2x + 3y = 2 \) ...(i) and \( x - 2y = 8 \) ...(ii). Multiplying equation (i) with 2 and equation (ii) with 3, we get \( 4x + 6y = 4 \) ...(iii) and \( 3x - 6y = 24 \) ...(iv). Thus, adding the equations (iii) and (iv), we get \( 7x = 28 \Rightarrow x = 4 \). Putting \( x = 4 \) in equation (ii), we get \( 4 - 2y = 8 \Rightarrow -2y = 4 \Rightarrow y = -2 \). Hence, the values of \( x \) and \( y \) are \( 4, -2 \) respectively. So, the correct option is (c).

Question. A number consists of 2 digits. The sum of the digits is 12 and the units digit, when divided by the tens digit gives the result as 3. Find the number.
(a) 84
(b) 48
(c) 93
(d) 39
Answer: (d)
Sol. Let the number be \( 10x + y \). So, \( x + y = 12 \) and \( \frac{y}{x} = 3 \). Thus, \( y = 3x \). Hence, \( x + 3x = 12 \Rightarrow 4x = 12 \Rightarrow x = 3 \) and \( y = 9 \). So, the number is \( 3(10) + 9 = 39 \) and the correct option is (d).

Fill in The Blanks

Question. If the lines \( 3x + 2py = 2 \) and \( 6x + 5y = 5 \) are parallel, then the value of \( p \) is .............
Answer: \( \frac{5}{4} \)

Question. The system of equations \( kx - y - 2 = 0 \) and \( 6x - 2y - 3 = 0 \) has no solution if \( k = \dots\dots\dots \) .
Answer: 3

Question. If \( \frac{a_1}{a_2} \dots\dots \frac{b_1}{b_2} \), then the system of equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) has a unique solution.
Answer: \( \neq \)

True/False

Question. If the pair of linear equations are parallel, then they are inconsistent.
Answer: True

Question. Cross-multiplication is one of the methods to solve two linear equations in two variables algebraically.
Answer: True

Very Short Answer Type Questions

Question. Find the number of solutions of the following pair of linear equations : \( x + 2y - 8 = 0 \) and \( 2x + 4y = 16 \) 
Answer: Given equations are, \( x + 2y - 8 = 0 \) and \( 2x + 4y = 16 \). On comparing the given equations with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) respectively, we get \( a_1 = 1, b_1 = 2, c_1 = -8 \) and \( a_2 = 2, b_2 = 4, c_2 = -16 \). So, \( \frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}, \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \). Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the given pair of linear equations have infinitely many solutions.

Question. If the lines given by \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \) are parallel, then find the value of \( k \). 
Answer: Since, the given lines \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \) are parallel, \( \frac{3}{2} = \frac{2k}{5} \neq \frac{-2}{1} \Rightarrow k = \frac{15}{4} \).

Question. Find the value of \( c \) for which the pair of equations \( cx - y = 2 \) and \( 6x - 2y = 3 \) will have infinitely many solutions. 
Answer: Here, \( \frac{a_1}{a_2} = \frac{c}{6}; \frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}; \frac{c_1}{c_2} = \frac{-2}{-3} = \frac{2}{3} \). Since \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), there is no value of \( c \), for which the given pair of equtions will have infinitely many solutions.

Question. Do the equations \( 4x + 3y - 1 = 5 \) and \( 12x + 9y = 15 \) represent a pair of coincident lines ?
Answer: Here, \( \frac{a_1}{a_2} = \frac{4}{12} = \frac{1}{3}; \frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}; \frac{c_1}{c_2} = \frac{-6}{15} \). i.e., \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). \(\therefore\) Given pair of equations do not represent coincident lines.

Question. Find whether the following pair of linear equations are consistent or inconsistent : \( 3x + 2y = 5 \) and \( 2x - 3y = 7 \)  
Answer: Here, \( \frac{a_1}{a_2} = \frac{3}{2} \); \( \frac{b_1}{b_2} = \frac{2}{-3} = -\frac{2}{3} \); \( \frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7} \)
i.e., \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \therefore \) Given pair of linear equations are consistent.

Question. Form a pair of linear equations in the following problem : 10 students of class X took part in mathematics quiz. The number of girls is 4 more than the number of boys.
Answer: Let the number of girls be \( x \) and number of boys be \( y \).
Then, Total students = 10
\( \Rightarrow x + y = 10 \)
Also, Number of girls = 4 + Number of boys
\( \Rightarrow x = 4 + y \)
or \( x - y = 4 \)
Thus, pair of linear equations are
\( x + y = 10 \) and \( x - y = 4 \)

Question. If \( x = a \) and \( y = b \) is the solution of the pair of equations \( x - y = 2 \) and \( x + y = 4 \), find the values of \( a \) and \( b \). 
Answer: Since, \( x = a \) and \( y = b \) is the solution of given pair of equations
\( \therefore a - b = 2 \) ...(i)
and \( a + b = 4 \) ...(ii)
Adding the above equations, we get
\( 2a = 6 \)
\( \Rightarrow a = 3 \)
Putting \( a = 3 \) in equation (i), we get
\( 3 - b = 2 \)
\( \Rightarrow b = 1 \)
Thus, \( a = 3 \) and \( b = 1 \).

Question. The cost of 2 kg apples and 1 kg grapes was ₹ 160. After a month, the cost of 4 kg apples and 2 kg grapes was ₹ 300. Represent the situation algebraically.  
Answer: Let the cost of 1 kg of apples be ₹ \( a \) and the cost of 1 kg of grapes be ₹ \( g \).
Thus \( 2a + g = 160 \) ...(i)
and \( 4a + 2g = 300 \)
or \( 2a + g = 150 \) ...(ii)

Question. Find the point of intersection of the lines \( x - 6 = 0 \) and \( y + 2 = 0 \).
Answer: Given equations are,
\( x - 6 = 0 \) ..(i)
and \( y + 2 = 0 \) ...(ii)
From (i), we have \( x = 6 \)
From (ii), we have \( y = -2 \)
Thus, the point of intersection of these lines is \( (6, -2) \).

Question. For what value of \( k \), the pair of linear equations \( 3x + 5y = 3 \) and \( 6x + ky = 8 \) do not have a solution ?
Answer: For the given pair of equations to have no solution
\( \frac{3}{6} = \frac{5}{k} \neq \frac{-3}{-8} \)
or \( k = 10 \)

Question. For what value of \( k \), the following pair of linear equations has infinitely many solutions ?
\( 10x + 5y - (k - 5) = 0 \)
\( 20x + 10y - k = 0 \)
Answer: For infinitely many solutions,
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \Rightarrow \frac{10}{20} = \frac{5}{10} = \frac{-(k - 5)}{-k} \)
\( \Rightarrow \frac{1}{2} = \frac{k - 5}{k} \)
\( \Rightarrow 2k - 10 = k \)
\( \Rightarrow k = 10 \)

Question. Does the following pair of equations represent a pair of coincident lines ?
\( \frac{x}{2} + y + \frac{2}{5} = 0 \) and \( 4x + 8y + \frac{5}{16} = 0 \)
Answer: Here, \( \frac{a_1}{a_2} = 1/2 / 4 = 1/8 \); \( \frac{b_1}{b_2} = 1/8 \); \( \frac{c_1}{c_2} = 2/5 / 5/16 = 32/25 \)
i.e., \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \therefore \) The given pair of equations do not represent coincident lines.

Short Answer Type Questions-I

Question. Find the value(s) of \( k \) for which the pair of the linear equations \( kx + y = k^2 \) and \( x + ky = 1 \) have infinitely many solutions.  
Answer: A pair of linear equations has infinitely many solutions, if
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here, \( a_1 = k; b_1 = 1; c_1 = -k^2 \)
\( a_2 = 1; b_2 = k; c_2 = -1 \)
So, \( \frac{k}{1} = \frac{1}{k} = \frac{-k^2}{-1} \)
\( \Rightarrow \frac{k}{1} = \frac{1}{k} \Rightarrow \frac{1}{k} = \frac{k^2}{1} \)
\( \Rightarrow k^2 = 1 \Rightarrow k^3 = 1 \)
\( \Rightarrow k = \pm 1 \Rightarrow k = 1 \)
Common solution is \( k = 1 \).

Question. Find the value of \( p \) for which the following pair of linear equations have infinitely many solutions :
\( (p - 3)x + 3y = p \)
\( px + py = 12 \) 
Answer: A pair of linear equations has infinitely many solutions, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here, \( a_1 = p - 3; b_1 = 3; c_1 = -p \)
\( a_2 = p; b_2 = p; c_2 = -12 \)
So, \( \frac{p - 3}{p} = \frac{3}{p} = \frac{-p}{-12} \)
\( \Rightarrow \frac{p - 3}{p} = \frac{3}{p} \Rightarrow \frac{3}{p} = \frac{p}{12} \)
\( \Rightarrow 3p = p^2 - 3p \Rightarrow p^2 = 36 \)
\( \Rightarrow p^2 - 6p = 0 \Rightarrow p = \sqrt{36} \)
\( \Rightarrow p(p - 6) = 0 \Rightarrow p = \pm 6 \)
\( \Rightarrow p = 0, 6 \)
Common solution is \( p = 6 \).

Question. For which values of \( a \) and \( b \) does the following pair of linear equations have an infinite number of solutions.
\( 2x + 3y = 7 \)
\( (a - b)x + (a + b)y = 3a + b - 2 \) 
Answer: A pair of linear equations has infinitely many solutions, if
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here, \( a_1 = 2; b_1 = 3; c_1 = -7 \)
\( a_2 = a - b; b_2 = a + b; c_2 = -(3a + b - 2) \)
So, \( \frac{2}{a - b} = \frac{3}{a + b} = \frac{-7}{-(3a + b - 2)} \)
\( \Rightarrow \frac{2}{a - b} = \frac{3}{a + b} \Rightarrow \frac{3}{a + b} = \frac{7}{3a + b - 2} \)
\( \Rightarrow 2a + 2b = 3a - 3b \Rightarrow 9a + 3b - 6 = 7a + 7b \)
\( \Rightarrow a = 5b \Rightarrow 2a = 4b + 6 \)
\( \Rightarrow a = 2b + 3 \)
\( \Rightarrow 2b + 3 = 5b \)
\( \Rightarrow 3b = 3 \)
\( \Rightarrow b = 1 \)
\( \therefore a = 5 \times 1 = 5 \)
\( \therefore a = 5 \) and \( b = 1 \).

Question. For which value of \( k \) will the following pair of linear equations have no solution
\( 3x + y = 1 \)
\( (2k - 1)x + (k - 1)y = 2k + 1 \)  
Answer: A pair of linear equations has no solution, if
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Here, \( a_1 = 3; b_1 = 1; c_1 = -1 \)
\( a_2 = (2k - 1); b_2 = k - 1; c_2 = -(2k + 1) \)
So, \( \frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)} \)
\( \Rightarrow \frac{3}{2k - 1} = \frac{1}{k - 1} \)
\( \Rightarrow 3k - 3 = 2k - 1 \)
\( \Rightarrow k = 2 \).

Question. Find the value(s) of \( k \) so that the pair of equations \( x + 2y = 5 \) and \( 3x + ky + 15 = 0 \) has a unique solution.  
Answer: Given, \( x + 2y = 5 \)
\( 3x + ky + 15 = 0 \)
Comparing above equations with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \),
We get,
\( a_1 = 1, b_1 = 2, c_1 = -5 \)
\( a_2 = 3, b_2 = k, c_3 = 15 \)
Condition for the pair of equations to have unique solution is
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
\( \frac{1}{3} \neq \frac{2}{k} \)
\( k \neq 6 \)
\( k \) can have any value except 6.

Question. Find the values of \( x \) and \( y \) in \( 2x + 3y = 2 \) and \( x - 2y = 8 \).
Answer: Given equations are,
\( 2x + 3y = 2 \) ...(i)
\( x - 2y = 8 \) ...(ii)
Multiplying equation (i) with 2 and equation (ii) with 3, we get
\( 4x + 6y = 4 \) ...(iii)
and \( 3x - 6y = 24 \) ...(iv)
Thus, adding the equations (iii) and (iv), we get
\( 7x = 28 \)
\( \Rightarrow x = 4 \)
Replacing \( x = 4 \) in equation (ii), we get
\( 4 - 2y = 8 \)
\( \Rightarrow -2y = 4 \)
\( \Rightarrow y = -2 \)
Hence, the values of \( x \) and \( y \) are 4, -2 respectively.

Question. The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.  
Answer: Let two angles A and B are supplementary.
\( \therefore A + B = 180^\circ \) ...(i)
Given, \( A = B + 18^\circ \)
On putting \( A = B + 18^\circ \) in equation (i), we get
\( B + 18^\circ + B = 180^\circ \)
\( 2B + 18^\circ = 180^\circ \)
\( 2B = 162^\circ \)
\( B = 81^\circ \)
\( A = B + 18^\circ \)
\( \Rightarrow A = 99^\circ \)

Question. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?  
Answer: Let age of Sumit be \( x \) years and age of his son be \( y \) years. Then, according to question we have,
\( x = 3y \) ...(i)
Five years later,
\( x + 5 = 2 \frac{1}{2}(y + 5) \) ...(ii)
On putting \( x = 3y \) in equation (ii)
\( 3y + 5 = \frac{5}{2}(y + 5) \)
\( 3y + 5 = \frac{5y}{2} + \frac{25}{2} \)
\( 3y - \frac{5y}{2} = \frac{25}{2} - 5 \)
\( \frac{6y - 5y}{2} = \frac{25 - 10}{2} = \frac{15}{2} \)
\( \frac{y}{2} = \frac{15}{2} \)
\( y = 15 \) years
Then, age of sumit is
\( 3 \times y = 3 \times 15 \)
\( = 45 \) years

Question. Peter secured x marks in Mathematics and y marks in Physics. Find the values of x and y, if x – y = 2 and xy = 2600.
Answer: Given, \( x - y = 2 \)
\( \Rightarrow x = y + 2 \) ...(i)
and \( xy = 2600 \) ...(ii)
Substituting the value of \( x \) from equation (i) to equation (ii), we get
\( (y + 2)y = 2600 \)
\( \Rightarrow y^2 + 2y - 2600 = 0 \)
\( \Rightarrow y^2 + 52y - 50y - 2600 = 0 \)
\( \Rightarrow y(y + 52) - 50(y + 52) = 0 \)
\( \Rightarrow (y + 52)(y - 50) = 0 \)
\( \Rightarrow y = 50, -52 \)
But as marks cannot be negative, so \( y = 50 \) and from equation (i)
\( x = 2 + 50 = 52 \)
\( \therefore \) Marks secured in Mathematics = 52 and in Physics = 50.

Question. A number consists of 2 digits. The sum of the digits is 12 and the units digit, when divided by the tens digit gives the result as 3. Find the number.
Answer: Let the number be \( 10x + y \)
Given, \( x + y = 12 \) ...(i)
and \( \frac{y}{x} = 3 \)
\( \Rightarrow y = 3x \) ...(ii)
Hence, \( x + 3x = 12 \) [From (i)]
\( \Rightarrow 4x = 12 \)
\( \Rightarrow x = 3 \)
and \( y = 9 \) [From (ii)]
So the number is \( 3(10) + 9 = 39 \)

Question. Cost of a burger is ₹ 20 more than the cost of juice of one glass of orange. If cost of one burger and one glass of orange juice is ₹ 60. Find the cost of each.
Answer: Let the cost of burger be \( ₹ x \) and the cost of orange juice be \( ₹ y \).
Then \( x = 20 + y \)
\( \Rightarrow x - y = 20 \) ...(i)
and \( x + y = 60 \) ...(ii)
On adding equations (i) and (ii), we get
\( 2x = 80 \)
\( x = 40 \)
Replacing the value of \( x \) in eq. (ii), we get
\( 40 + y = 60 \)
\( \Rightarrow y = 20 \)
Thus, the cost of burger is ₹ 40 and that of orange juice is ₹ 20.

Question. The age of the father is twice the sum of the ages of his 2 children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Answer: Let the sum of ages of the 2 children be \( x \) and the age of the father be \( y \).
According to the question,
\( y = 2x \)
\( \Rightarrow 2x - y = 0 \) ...(i)
and \( x + 40 = y + 20 \)
\( \Rightarrow x - y = -20 \) ...(ii)
Subtracting equations (i) and (ii)
\( 2x - y = 0 \)
\( x - y = -20 \)
\( - \quad + \quad + \)
\( x = 20 \)
\( \therefore y = 2 \times 20 = 40 \)
Thus, Father’s age = 40 years.

Short Answer Type Questions-II

Question. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and \( \left(\frac{y}{x} - 2\right) \). 
Answer: Given : \( 2x + y = 23 \) ...(i)
and \( 4x - y = 19 \) ...(ii)
From equation (i), we have
\( y = 23 - 2x \)
Putting this value in equation (ii), we get
\( 4x - (23 - 2x) = 19 \)
\( \Rightarrow 4x - 23 + 2x = 19 \)
\( \Rightarrow 6x = 19 + 23 \)
\( \Rightarrow 6x = 42 \)
\( x = 7 \)
So, \( y = 23 - 2 \times 7 = 23 - 14 = 9 \)
So, \( 5y - 2x = 5 \times 9 - 2 \times 7 = 45 - 14 = 31 \)
And \( \frac{y}{x} - 2 = \frac{9}{7} - 2 = \frac{9 - 14}{7} = -\frac{5}{7} \)

Question. Find whether the following pair of linear equations has a unique solution. If yes, find the solution. 7x – 4y = 49 and 5x – 6y = 57
Answer: Given equations are,
\( 7x - 4y = 49 \) ...(i)
and \( 5x - 6y = 57 \) ...(ii)
For a system of linear equations to have unique solution
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Here, \( a_1 = 7, b_1 = -4, c_1 = -49 \)
\( a_2 = 5, b_2 = -6, c_2 = -57 \)
So, \( \frac{a_1}{a_2} = \frac{7}{5}; \frac{b_1}{b_2} = \frac{-4}{-6} = \frac{2}{3}; \frac{c_1}{c_2} = \frac{-49}{-57} = \frac{49}{57} \)
i.e., \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \therefore \) The given pair of linear equations has unique solution.
Now, multiplying equation (i) by 3 and equation (ii) by 2, we get
\( 21x - 12y = 147 \) ...(iii)
\( 10x - 12y = 114 \) ...(iv)
Subtracting equation (iv) from equation (iii)
\( 21x - 12y = 147 \)
\( 10x - 12y = 114 \)
\( - \quad + \quad - \)
\( 11x = 33 \)
\( \Rightarrow x = \frac{33}{11} = 3 \)
Putting \( x = 3 \) in equation (i), we get
\( 7(3) - 4y = 49 \)
\( \Rightarrow -4y = 49 - 21 = 28 \)
\( \Rightarrow y = \frac{28}{-4} = -7 \)
\( \therefore x = 3 \) and \( y = -7 \).

Question. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of m for which y = mx + 3.  
Answer: Given linear equations are,
\( 2x + 3y = 11 \) ...(i)
\( 2x - 4y = -24 \) ...(ii)
Subtracting equation (ii) from equation (i),
\( 2x + 3y = 11 \)
\( 2x - 4y = -24 \)
\( - \quad + \quad + \)
\( 7y = 35 \)
\( \Rightarrow y = \frac{35}{7} = 5 \)
Putting \( y = 5 \) in equation (i), we get
\( 2x + 3(5) = 11 \)
\( \Rightarrow 2x = 11 - 15 = -4 \)
\( \Rightarrow x = \frac{-4}{2} = -2 \)
\( \therefore x = -2 \) and \( y = 5 \)
Now, we have \( y = mx + 3 \)
or \( 5 = m(-2) + 3 \)
\( \Rightarrow -2m = 2 \)
\( \Rightarrow m = -1 \)

Question. Solve for \(x\) and \(y\) : \(ax + by - a + b = 0\) and \(bx - ay - a - b = 0\).
Answer: The equations \(ax + by - a + b = 0\) and \(bx - ay - a - b = 0\) can be written as
\(ax + by = a - b\) ...(i)
and \(bx - ay = a + b\) ...(ii)
Multiplying equation (i) with \(a\) and equation (ii) with \(b\), we have
\(a^2x + aby = a^2 - ab\) ...(iii)
\(b^2x - aby = ab + b^2\) ...(iv)
On adding equations (iii) and (iv), we get
\((a^2 + b^2)x = a^2 + b^2\)
\(\Rightarrow x = 1\)
Substituting \(x = 1\) in equation (i), we get
\(a + by = a - b\)
\(\Rightarrow by = - b\)
\(\Rightarrow y = - 1\)
Thus, \(x = 1\) and \(y = - 1\). Ans.

Question. Solve for \(x\) and \(y\) : \(47x + 31y = 63\) and \(31x + 47y = 15\).
Answer: Given, \(47x + 31y = 63\) ...(i)
and \(31x + 47y = 15\) ...(ii)
Adding equations (i) and (ii), we get
\(78x + 78y = 78\)
\(\Rightarrow 78(x + y) = 78\)
\(\Rightarrow x + y = 1\) ...(iii)
Also, subtracting equation (ii) from equation (i), we get
\(16x - 16y = 48\)
\(\Rightarrow 16(x - y) = 48\)
\(\Rightarrow x - y = 3\) ...(iv)
Adding equations (iii) and (iv), we get
\(2x = 4\)
\(\Rightarrow x = 2\)
Substituting \(x = 2\) in equation (iii), we get
\(2 + y = 1\)
\(\Rightarrow y = 1 - 2 = - 1\)
Thus \(x = 2\) and \(y = - 1\). Ans.

Question. In \(\triangle ABC\), \(\angle A = x\), \(\angle B = 3x\) and \(\angle C = y\). If \(3y - 5x = 30^\circ\), show that the triangle is right angled.
Answer: Given : \(\angle A = x\), \(\angle B = 3x\) and \(\angle C = y\)
Also, \(3y - 5x = 30^\circ\) ...(i)
We know, sum of interior angles of a triangle is \(180^\circ\).
\(\therefore \angle A + \angle B + \angle C = 180^\circ\)
\(\Rightarrow x + 3x + y = 180^\circ\)
\(\Rightarrow 4x + y = 180^\circ\) ...(ii)
Multiplying equation (ii) by 3, we get
\(12x + 3y = 540^\circ\)
Subtracting equations (i) and (iii), we get
\(3y - 5x = 30^\circ\)
\(3y + 12x = 540^\circ\)
— — —
\(- 17x = - 510^\circ\)
\(\Rightarrow x = \frac{- 510^\circ}{- 17} = 30^\circ\)
Putting \(x = 30^\circ\) in equation (i), we get
\(3y - 5(30^\circ) = 30^\circ\)
\(3y = 180^\circ\)
\(\Rightarrow y = \frac{180^\circ}{3} = 60^\circ\)
So, \(\angle A = x = 30^\circ\)
\(\angle B = 3x = 3 \times 30^\circ = 90^\circ\)
and \(\angle C = y = 60^\circ\).
\(\because \angle B = 90^\circ\)
\(\therefore \triangle ABC\) is a right angled at B. Hence Proved.

Question. Solve the following pair of linear equations by the cross-multiplication method :
\(x + 2y = 2\)
\(x - 3y = 7\)
Answer: Given, \(x + 2y - 2 = 0\)
and \(x - 3y - 7 = 0\)
\(\frac{x}{2(-7) - (-3)(-2)} = \frac{y}{-2(1) - (-7)(1)} = \frac{1}{1(-3) - (1)(2)}\)
\(\Rightarrow \frac{x}{-14 - 6} = \frac{y}{-2 + 7} = \frac{1}{-3 - 2}\)
\(\Rightarrow \frac{x}{-20} = \frac{y}{5} = \frac{1}{-5}\)
\(\Rightarrow \frac{x}{-20} = \frac{1}{-5}, \frac{y}{5} = \frac{1}{-5}\)
\(\Rightarrow x = 4, y = -1\)
Thus \(x = 4\) and \(y = -1\) Ans.

Question. Solve for \(x\) and \(y\) : \(\frac{4}{x} + 3y = 8\) and \(\frac{6}{x} - 4y = -5\).  
Answer: Given,
\(\frac{4}{x} + 3y = 8\) ...(i)
and \(\frac{6}{x} - 4y = - 5\) ...(ii)
Multiplying equation (i) with 4 and equation (ii) with 3, we get
\(\frac{16}{x} + 12y = 32\) ...(iii)
\(\frac{18}{x} - 12y = - 15\) ...(iv)
Adding equations (iii) and (iv), we get
\(\frac{34}{x} = 17\)
\(\Rightarrow x = 2\)
Substituting \(x = 2\) in equation (i), we get
\(\frac{4}{2} + 3y = 8\)
\(\Rightarrow 2 + 3y = 8\)
\(\Rightarrow 3y = 6\)
\(\Rightarrow y = 2\)
Thus \(x = y = 2\). Ans.

Question. Solve for \(x\) and \(y\) : \(\frac{x+1}{2} + \frac{y-1}{3} = 8\) and \(\frac{x-1}{3} + \frac{y+1}{2} = 9\).
Answer: Given, \(\frac{x+1}{2} + \frac{y-1}{3} = 8\)
\(\Rightarrow \frac{3(x + 1) + 2(y - 1)}{6} = 8\)
\(\Rightarrow 3(x + 1) + 2(y - 1) = 48\)
\(\Rightarrow 3x + 3 + 2y - 2 = 48\)
\(\Rightarrow 3x + 2y + 1 = 48\)
\(\Rightarrow 3x + 2y = 47\) ...(i)
and \(\frac{x-1}{3} + \frac{y+1}{2} = 9\)
\(\Rightarrow \frac{2(x - 1) + 3(y + 1)}{6} = 9\)
\(\Rightarrow 2(x - 1) + 3(y + 1) = 54\)
\(\Rightarrow 2x - 2 + 3y + 3 = 54\)
\(\Rightarrow 2x + 3y + 1 = 54\)
\(\Rightarrow 2x + 3y = 53\) ...(ii)
Adding equations (i) and (ii), we get
\(5x + 5y = 100\)
\(\Rightarrow 5(x + y) = 100\)
\(\Rightarrow x + y = 20\) ...(iii)
Also subtracting equation (ii) from equation (i), we get
\(x - y = - 6\) ...(iv)
Adding equations (iii) and (iv), we get
\(2x = 14\)
\(\Rightarrow x = 7\)
Substituting \(x = 7\) in equation (iii), we get
\(7 + y = 20\)
\(\Rightarrow y = 20 - 7 = 13\)
Thus \(x = 7\) and \(y = 13\). Ans.

Question. Determine the values of \(p\) and \(q\) for which the given system of equations has infinitely many solutions \(2x + 3y = 9\) and \((p + q)x + (2p - q)y = 3(p + q + 1)\).
Answer: Given, \(2x + 3y = 9\) ...(i)
and \((p + q)x + (2p - q)y = 3(p + q + 1)\) ...(ii)
For this pair of equations to have infinite number of solutions, it must satisfy the condition of
\(\frac{2}{p+q} = \frac{3}{2p-q} = \frac{-9}{-3(p+q+1)} = \frac{3}{(p+q+1)}\)
\(\Rightarrow 2(2p - q) = 3(p + q)\)
\(\Rightarrow 4p - 2q = 3p + 3q\)
\(\Rightarrow p = 5q\)
\(\Rightarrow p - 5q = 0\) ...(iii)
and \(2(p + q + 1) = 3(p + q)\)
\(\Rightarrow 2p + 2q + 2 = 3p + 3q\)
\(\Rightarrow p + q = 2\) ...(iv)
Subtracting equation (iv) from equation (iii), we get
\(- 6q = - 2\)
\(\Rightarrow 3q = 1\)
\(\Rightarrow q = \frac{1}{3}\)
Substituting \(q = \frac{1}{3}\) in equation (iv), we get
\(p + \frac{1}{3} = 2\)
\(\Rightarrow p = 2 - \frac{1}{3} = \frac{5}{3}\)
Thus, the pair of equations will have infinite number of solutions for \(p = \frac{5}{3}\) and \(q = \frac{1}{3}\). Ans.

Question. Seven times a two-digit number is equal to four times the number obtained by reversing the order of its digits. If the difference of the digits is 3, determine the number. 
Answer: Let the two-digit number be \(10x + y\).
Then, according to the question
\(7(10x + y) = 4(10y + x)\) ...(i)
Also, \(\pm (x - y) = 3\) ...(ii)
Now, from equation (i), we have
\(7 (10x + y) = 4 (10y + x)\)
\(\Rightarrow 70x + 7y = 40y + 4x\)
\(\Rightarrow 66x = 33y\)
\(\Rightarrow 2x = y\)
Substituting the value of \(y\) in equation (ii), we get
\(\pm (x - 2x) = 3\)
\(\Rightarrow x = 3\)
\(\therefore y = 2x = 2 \times 3 = 6\)
Thus, the number \(10x + y = 10 \times 3 + 6 = 36\). Ans.

Question. The sum of the digits of a two-digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number. 
Answer: Let Unit’s digit = \(x\)
Ten’s digit = \(y\)
So, original number = Unit’s digit + \(10 \times\) Ten’s digit = \(x + 10y\)
According to question,
Sum of digits = 8
\(\Rightarrow x + y = 8\) ...(i)
On reversing the digits, unit’s digit = \(y\)
Ten’s digit = \(x\)
So, New number = \(10x + y\)
According to question,
Difference = 18
\(\Rightarrow x + 10y - (10x + y) = 18\)
\(\Rightarrow x + 10y - 10x - y = 18\)
\(\Rightarrow 9y - 9x = 18\)
\(\Rightarrow y - x = 2\) ...(ii)
By adding equations (i) and (ii), we get
\(2y = 10\)
\(y = \frac{10}{2}\)
\(\Rightarrow y = 5\)
Substituting \(y = 5\) in equation (i), we get
\(x + 5 = 8\)
\(\Rightarrow x = 8 - 5\)
\(\Rightarrow x = 3\)
\(\therefore\) Original number = \(10y + x\)
\(= 10 \times 5 + 3\)
\(= 50 + 3\)
\(= 53\) Ans.

Question. A father‘s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.  
Answer: Let the present age of father be \(x\) years and sum of ages of his two children be \(y\) years
According to question
\(x = 3y\) ...(i)
After 5 years
Father‘s age = \((x + 5)\) years
Sum of ages of two children = \((y + 5 + 5)\) years
\(= (y + 10)\) years
In 2nd case
According to question
\(x + 5 = 2 (y + 10)\)
or \(x + 5 = 2y + 20\)
or \(x - 2y = 15\)
or \(3y - 2y = 15\) {Using equations (i)}
\(y = 15\)
Now from equation (i)
\(x = 3y\) (Put \(y = 15\))
or \(x = 3 \times 15\)
\(x = 45\)
So, Present age of father = 45 years. Ans.

Question. The well monitored total expenditure per month of a household consists of a fixed rent of the house and mess charges depending upon the number of people sharing the house. The total monthly expenditure is ₹ 3900 for 2 people and ₹ 7500 for 5 people. Find the rent of the house.
Answer: Let the monthly rent of the house be ₹ \(x\) and the mess charges per head per month be ₹ \(y\).
According to the question,
\(x + 2y = 3900\) ...(i)
and \(x + 5y = 7500\) ...(ii)
Subtracting equation (ii) from equation (i), we get
\(- 3y = - 3600\)
\(\Rightarrow y = 1200\)
Substituting \(y = 1200\) in equation (i), we get
\(x + 2 (1200) = 3900\)
\(\Rightarrow x + 2400 = 3900\)
\(\Rightarrow x = 1500\)
Hence, monthly rent = ₹ 1500 Ans.

Question. The ratio of income of two persons is 11 : 7 and the ratio of their expenditures is 9 : 5. If each of them manages to save ₹ 400 per month, find their monthly incomes.
Answer: Let the income of two persons be \(11x\) and \(7x\).
Also, the expenditures of two persons be \(9y\) and \(5y\), respectively.
According to the question,
\(11x - 9y = 400\) ...(i)
and \(7x - 5y = 400\) ...(ii)
From equations (i) and (ii),
\(11x - 9y = 7x - 5y\)
\(\Rightarrow 4x = 4y\)
\(\Rightarrow x = y\) ...(iii)
Substituting \(x = y\) in equation (ii), we get
\(7y - 5y = 400\)
\(\Rightarrow 2y = 400\)
\(\Rightarrow y = 200\)
\(\therefore\) From equation (iii),
\(x = y = 200\)
Thus, their monthly incomes are
\(11 \times 200 = 2200\)
and \(7 \times 200 = 1400\). Ans.