CBSE Class 10 Maths HOTs Pair Of Linear Equations In Two Variables Set 02

Very Short Answer Type Questions

Question. Find whether the pair of linear equations \( y = 0 \) and \( y = -5 \) has no solution, unique solution or infinitely many solutions.
Answer: The given variable \( y \) has different values. Therefore the pair of equations \( y = 0 \) and \( y = -5 \) has no solution.

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, \( am = bl \), we have \( \frac{a}{l} = \frac{b}{m} \neq \frac{c}{n} \). Thus, \( ax + by = c \) and \( lx + my = n \) has no solution.

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 solutions.
Answer: Since \( ad \neq bc \) or \( \frac{a}{c} \neq \frac{b}{d} \). Hence, the pair of given linear equations has unique solution.

Question. Two lines are given to be parallel. The equation of one of the lines is \( 4x + 3y = 14 \), then find the equation of the second line.
Answer: The equation of one line is \( 4x + 3y = 14 \). We know that if two lines \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) are parallel, then \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) or \( \frac{4}{a_2} = \frac{3}{b_2} \neq \frac{14}{c_2} \Rightarrow \frac{a_2}{12} = \frac{b_2}{9} \neq \frac{c_2}{5} \). Hence, one of the possible, second parallel line is \( 12x + 9y = 5 \).

Short Answer Type Questions - I

Question. Find whether the lines represented by \( 2x + y = 3 \) and \( 4x + 2y = 6 \) are parallel, coincident or intersecting.
Answer: Here \( a_1 = 2, b_1 = 1, c_1 = -3 \) and \( a_2 = 4, b_2 = 2, c_2 = -6 \). If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), then the lines are parallel. Clearly \( \frac{2}{4} = \frac{1}{2} = \frac{3}{6} \). Hence lines are coincident.

Question. Find whether the following pair of linear equation is consistent or inconsistent: \( 3x + 2y = 8, 6x - 4y = 9 \)
Answer: We have \( \frac{3}{6} \neq \frac{2}{-4} \), i.e., \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Hence, the pair of linear equation is consistent.

Question. Is the system of linear equations \( 2x + 3y - 9 = 0 \) and \( 4x + 6y - 18 = 0 \) consistent? Justify your answer.
Answer: For the equation \( 2x + 3y - 9 = 0 \) we have \( a_1 = 2, b_1 = 3 \) and \( c_1 = -9 \) and for the equation \( 4x + 6y - 18 = 0 \) we have \( a_2 = 4, b_2 = 6 \) and \( c_2 = -18 \). Here \( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2} \). Thus \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). Hence, system is consistent and dependent.

Question. Given the linear equation \( 3x + 4y = 9 \). Write another linear equation in these two variables such that the geometrical representation of the pair so formed is: (1) intersecting lines (2) coincident lines.
Answer: (1) For intersecting lines \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). So, one of the possible equation \( 3x - 5y = 10 \). (2) For coincident lines \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). So, one of the possible equation \( 6x + 8y = 18 \).

Question. For what value of \( p \) does the pair of linear equations given below has unique solution? \( 4x + py + 8 = 0 \) and \( 2x + 2y + 2 = 0 \).
Answer: We have \( 4x + py + 8 = 0 \) and \( 2x + 2y + 2 = 0 \). The condition of unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Hence, \( \frac{4}{2} \neq \frac{p}{2} \) or \( 2 \neq \frac{p}{2} \). Thus \( p \neq 4 \). The value of \( p \) is other than 4 it may be 1, 2, 3, -4 etc.

Question. For what value of \( k \), the pair of linear equations \( kx - 4y = 3 \), \( 6x - 12y = 9 \) has an infinite number of solutions?
Answer: We have \( kx - 4y - 3 = 0 \) and \( 6x - 12y - 9 = 0 \). Where, \( a_1 = k, b_1 = -4, c_1 = -3 \) and \( a_2 = 6, b_2 = -12, c_2 = -9 \). Condition for infinite solutions: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \frac{k}{6} = \frac{-4}{-12} = \frac{-3}{-9} \). Hence, \( k = 2 \).

Question. For what value of \( k \), \( 2x + 3y = 4 \) and \( (k+2)x + 6y = 3k + 2 \) will have infinitely many solutions?
Answer: We have \( 2x + 3y - 4 = 0 \) and \( (k+2)x + 6y - (3k+2) = 0 \). Here \( a_1 = 2, b_1 = 3, c_1 = -4 \) and \( a_2 = k+2, b_2 = 6, c_2 = -(3k+2) \). For infinitely many solutions \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) or, \( \frac{2}{k+2} = \frac{3}{6} = \frac{4}{3k+2} \). From \( \frac{2}{k+2} = \frac{3}{6} \) we have \( 3(k+2) = 2 \times 6 \Rightarrow (k+2) = 4 \Rightarrow k = 2 \). From \( \frac{3}{6} = \frac{4}{3k+2} \) we have \( 3(3k+2) = 4 \times 6 \Rightarrow (3k+2) = 8 \Rightarrow k = 2 \). Thus \( k = 2 \).

Question. For what value of \( k \), the system of equations \( kx + 3y = 1 \), \( 12x + ky = 2 \) has no solution.
Answer: The given equations can be written as \( kx + 3y - 1 = 0 \) and \( 12x + ky - 2 = 0 \). Here, \( a_1 = k, b_1 = 3, c_1 = -1 \) and \( a_2 = 12, b_2 = k, c_2 = -2 \). The equation for no solution if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) or, \( \frac{k}{12} = \frac{3}{k} \neq \frac{-1}{-2} \). From \( \frac{k}{12} = \frac{3}{k} \) we have \( k^2 = 36 \Rightarrow k = \pm 6 \). From \( \frac{3}{k} \neq \frac{-1}{-2} \) we have \( k \neq 6 \). Thus \( k = -6 \).

Short Answer Type Questions - II

Question. 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: (a) intersecting lines (b) parallel lines (c) coincident lines.
Answer: Given, linear equation is \( 2x + 3y - 8 = 0 \). (a) For intersecting lines, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). One of the possible equation may be taken as \( 5x + 2y - 9 = 0 \). (b) For parallel lines, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). One of the possible line parallel to equation (1) may be taken as \( 6x + 9y + 7 = 0 \). (c) For coincident lines, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). To get its coincident line, one of the possible equation may be taken as \( 4x + 6y - 16 = 0 \).

Question. Determine the values of \( m \) and \( n \) so that the following system of linear equation have infinite number of solutions: \( (2m - 1)x + 3y - 5 = 0 \) and \( 3x + (n - 1)y - 2 = 0 \).
Answer: We have \( (2m - 1)x + 3y - 5 = 0 \). Here \( a_1 = 2m - 1, b_1 = 3, c_1 = -5 \). And \( 3x + (n - 1)y - 2 = 0 \). Here \( a_2 = 3, b_2 = n - 1, c_2 = -2 \). For a pair of linear equations to have infinite number of solutions, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \frac{2m - 1}{3} = \frac{3}{n - 1} = \frac{5}{2} \). Or \( 2(2m - 1) = 15 \) and \( 5(n - 1) = 6 \). Hence, \( m = \frac{17}{4}, n = \frac{11}{5} \).

Question. Find the values of \( \alpha \) and \( \beta \) for which the following pair of linear equations has infinite number of solutions: \( 2x + 3y = 7 \); \( 2\alpha x + (\alpha + \beta)y = 28 \).
Answer: We have \( 2x + 3y = 7 \) and \( 2\alpha x + (\alpha + \beta)y = 28 \). For a pair of linear equations to be consistent and having infinite number of solutions, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \frac{2}{2\alpha} = \frac{3}{\alpha + \beta} = \frac{7}{28} \). \( \frac{2}{2\alpha} = \frac{7}{28} \Rightarrow 2 \times 28 = 2\alpha \times 7 \Rightarrow \alpha = 4 \).

Question. For what value of \( p \) will the following system of equations have no solution?
\( (2p - 1)x + (p - 1)y = 2p + 1; y + 3x - 1 = 0 \)
Answer: We have \( (2p - 1)x + (p - 1)y - (2p + 1) = 0 \)
Here \( a_1 = 2p - 1, b_1 = p - 1 \) and \( c_1 = -(2p + 1) \)
Also \( 3x + y - 1 = 0 \)
Here \( a_2 = 3, b_2 = 1 \) and \( c_2 = -1 \)
The condition for no solution is
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \frac{2p-1}{3} = \frac{p-1}{1} \neq \frac{2p+1}{1} \)
From \( \frac{2p-1}{3} = \frac{p-1}{1} \) we have
\( 3p - 3 = 2p - 1 \)
\( 3p - 2p = 3 - 1 \)
\( p = 2 \)
From \( \frac{p-1}{1} \neq \frac{2p+1}{1} \) we have
\( p - 1 \neq 2p + 1 \) or \( 2p - p \neq -1 - 1 \)
\( p \neq -2 \)
From \( \frac{2p-1}{3} \neq \frac{2p+1}{1} \) we have
\( 2p - 1 \neq 6p + 3 \)
\( 4p \neq -4 \)
\( p \neq -1 \)
Hence, system has no solution when \( p = 2 \).

Question. Find the value of \( k \) for which the following pair of equations has no solution:
\( x + 2y = 3, (k - 1)x + (k + 1)y = (k + 2) \).
Answer: For \( x + 2y = 3 \) or \( x + 2y - 3 = 0 \)
\( a_1 = 1, b_1 = 2, c_1 = -3 \)
For \( (k - 1)x + (k + 1)y = (k + 2) \)
or \( (k - 1)x + (k + 1)y - (k + 2) = 0 \)
\( a_2 = (k - 1), b_2 = (k + 1), c_2 = -(k + 2) \)
For no solution, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \frac{1}{k-1} = \frac{2}{k+1} \neq \frac{3}{k+2} \)
From \( \frac{1}{k-1} = \frac{2}{k+1} \) we have
\( k + 1 = 2k - 2 \)
\( 3 = k \)
Thus \( k = 3 \).

Question. Solve the following pair of linear equations by cross multiplication method:
\( x + 2y = 2, x - 3y = 7 \).
Answer: We have \( x + 2y - 2 = 0 \)
\( x - 3y - 7 = 0 \)
Using the formula \( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \)
we have \( \frac{x}{-14 - 6} = \frac{y}{-2 + 7} = \frac{1}{-3 - 2} \)
\( \frac{x}{-20} = \frac{y}{5} = \frac{1}{-5} \)
\( \frac{x}{-20} = \frac{1}{-5} \Rightarrow x = 4 \)
\( \frac{y}{5} = \frac{1}{-5} \Rightarrow y = -1 \)

Question. Solve the following pair of linear equations by substitution method:
\( 3x + 2y - 7 = 0, 4x + y - 6 = 0 \).
Answer: We have \( 3x + 2y - 7 = 0 \) ...(1)
\( 4x + y - 6 = 0 \) ...(2)
From equation (2), \( y = 6 - 4x \) ...(3)
Putting this value of \( y \) in equation (1) we have
\( 3x + 2(6 - 4x) - 7 = 0 \)
\( 3x + 12 - 8x - 7 = 0 \)
\( 5 - 5x = 0 \)
\( 5x = 5 \)
Thus \( x = 1 \)
Substituting this value of \( x \) in (2), we obtain,
\( y = 6 - 4 \times 1 = 2 \)
Hence, values of \( x \) and \( y \) are 1 and 2 respectively.

SHORT ANSWER TYPE QUESTIONS - II

Question. Solve : \( 99x + 101y = 499, 101x + 99y = 501 \)
Answer: We have \( 99x + 101y = 499 \) ...(1) \( 101x + 99y = 501 \) ...(2) Adding equation (1) and (2), we have \( 200x + 200y = 1000 \) or, \( x + y = 5 \) ...(3) Subtracting equation (2) from equation (1), we get \( -2x + 2y = -2 \) or, \( x - y = 1 \) ...(4) Adding equations (3) and (4), we have \( 2x = 6 \) \( x = 3 \) Substituting the value of \( x \) in equation (3), we get \( 3 + y = 5 \) \( y = 2 \)

Question. Sum of the ages of a father and the son is 40 years. If father’s age is three times that of his son, then find their respective ages.
Answer: Let age of father and son be \( x \) and \( y \) respectively. \( x + y = 40 \) ...(1) \( x = 3y \) ...(2) By solving equations (1) and (2), we get \( x = 30 \) and \( y = 10 \) Ages are 30 years and 10 years.

Question. Solve the following system of linear equations by substitution method: \( 2x - y = 2 \) \( x + 3y = 15 \)
Answer: We have \( 2x - y = 2 \) ...(1) \( x + 3y = 15 \) ...(2) From equation (1), we get \( y = 2x - 2 \) ...(3) Substituting the value of \( y \) in equation (2), \( x + 3(2x - 2) = 15 \) \( x + 6x - 6 = 15 \) or, \( 7x = 21 \) \( x = 3 \) Substituting this value of \( x \) in (3), we get From equation (1), we have \( y = 2 \times 3 - 2 = 4 \) \( x = 3 \) and \( y = 4 \)

Question. Solve using cross multiplication method: \( 5x + 4y - 4 = 0 \) \( x - 12y - 20 = 0 \)
Answer: We have \( 5x + 4y - 4 = 0 \) ...(1) \( x - 12y - 20 = 0 \) ...(2) By cross-multiplication method, \( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \) \( \frac{x}{-80 - 48} = \frac{y}{-4 + 100} = \frac{1}{-60 - 4} \) \( \frac{x}{-128} = \frac{y}{96} = \frac{1}{-64} \) \( \frac{x}{-128} = \frac{1}{-64} \Rightarrow x = 2 \) and \( \frac{y}{96} = \frac{1}{-64} \Rightarrow y = \frac{-3}{2} \) Hence, \( x = 2 \) and \( y = -\frac{3}{2} \)

Question. Find the value(s) of \( k \) for which the pair of Linear equations \( kx + y = k^2 \) and \( x + ky = 1 \) have infinitely many solutions.
Answer: We have \( kx + y = k^2 \) and \( x + ky = 1 \) \( \frac{a_1}{a_2} = \frac{k}{1} \), \( \frac{b_1}{b_2} = \frac{1}{k} \), \( \frac{c_1}{c_2} = \frac{k^2}{1} \) For infinitely many solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) \( \frac{k}{1} = \frac{1}{k} = \frac{k^2}{1} \Rightarrow k^2 = 1 \) \( k = \pm 1 \)

Question. A part of monthly hostel charge is fixed and the remaining depends on the number of days one has taken food in the mess. When Swati takes food for 20 days, she has to pay Rs. 3,000 as hostel charges whereas Mansi who takes food for 25 days Rs. 3,500 as hostel charges. Find the fixed charges and the cost of food per day.
Answer: Let fixed charge be \( x \) and per day food cost be \( y \) \( x + 20y = 3000 \) ...(1) \( x + 25y = 3500 \) ...(2) Subtracting (1) from (2) we have \( 5y = 500 \Rightarrow y = 100 \) Substituting this value of \( y \) in (1), we get \( x + 20(100) = 3000 \) \( x = 1000 \) Thus \( x = 1000 \) and \( y = 100 \) Fixed charge and cost of food per day are Rs. 1,000 and Rs. 100.

Question. Solve for \( x \) and \( y \) : \( \frac{x}{2} + \frac{2y}{3} = -1 \) \( x - \frac{y}{3} = 3 \)
Answer: We have \( \frac{x}{2} + \frac{2y}{3} = -1 \) or \( 3x + 4y = -6 \) ...(1) and \( \frac{x}{1} - \frac{y}{3} = 3 \) or \( 3x - y = 9 \) ...(2) Subtracting equation (2) from equation (1), we have \( 5y = -15 \Rightarrow y = -3 \) Substituting \( y = -3 \) in eq (1), we get \( 3x + 4(-3) = -6 \) \( 3x - 12 = -6 \) \( 3x = 12 - 6 \) Thus \( x = 2 \) Hence \( x = 2 \) and \( y = -3 \).

Question. Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions ? In case there is a unique solution, find it by using cross-multiplication method.
(a) \( x - 3y - 3 = 0 \), \( 3x - 9y - 2 = 0 \)
(b) \( 2x + y = 5 \), \( 3x + 2y = 8 \)
(c) \( 3x - 5y = 20 \), \( 6x - 10y = 40 \)
(d) \( x - 3y - 7 = 0 \), \( 3x - 3y - 15 = 0 \)
Answer: (a) We have \( x - 3y - 3 = 0 \) ...(1)
and \( 3x - 9y - 2 = 0 \) ...(2)
Comparing equation (1) and (2) with \( ax + by + c = 0 \)
\( a_1 = 1, b_1 = -3, c_1 = -3 \)
and \( a_2 = 3, b_2 = -9, c_2 = -2 \)
Thus, \( \frac{a_1}{a_2} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}, \frac{c_1}{c_2} = \frac{3}{2} \)
Since, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Hence, no solution exists.
(b) We have \( 2x + y = 5 \) or, \( 2x + y - 5 = 0 \) ...(1)
and \( 3x + 2y = 8 \) or, \( 3x + 2y - 8 = 0 \) ...(2)
Comparing equation (1) and (2) with \( ax + by + c = 0 \),
\( a_1 = 2, b_1 = 1, c_1 = -5 \)
and \( a_2 = 3, b_2 = 2, c_2 = -8 \)
Now, \( \frac{a_1}{a_2} = \frac{2}{3} \) and \( \frac{b_1}{b_2} = \frac{1}{2} \)
Since \( \frac{2}{3} \neq \frac{1}{2} \), \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Hence, a unique solution exists. By cross multiplication method,
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \)
\( \frac{x}{(1)(-8) - (2)(-5)} = \frac{y}{(-5)(3) - (-8)(2)} = \frac{1}{(2)(2) - (3)(1)} \)
\( \frac{x}{-8 + 10} = \frac{y}{-15 + 16} = \frac{1}{4 - 3} \)
\( \frac{x}{2} = \frac{y}{1} = \frac{1}{1} \)
Thus \( \frac{x}{2} = \frac{1}{1} \text{ and } \frac{y}{1} = \frac{1}{1} \)
Hence \( x = 2 \) and \( y = 1 \).
(c) We have \( 3x - 5y = 20 \) or, \( 3x - 5y - 20 = 0 \) ...(1)
and \( 6x - 10y = 40 \) or, \( 6x - 10y - 40 = 0 \) ...(2)
Comparing equation (1) and (2) with \( ax + by + c = 0 \),
\( a_1 = 3, b_1 = -5, c_1 = -20 \)
and \( a_2 = 6, b_2 = -10, c_2 = -40 \)
Now, \( \frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2} \)
Hence, infinitely many solutions exist.
(d) We have \( x - 3y - 7 = 0 \) ...(1)
and \( 3x - 3y - 15 = 0 \) ...(2)
Comparing equation (1) and (2) with \( ax + by + c = 0 \),
\( a_1 = 1, b_1 = -3, c_1 = -7 \)
and \( a_2 = 3, b_2 = -3, c_2 = -15 \)
Here \( \frac{a_1}{a_2} = \frac{1}{3} \) and \( \frac{b_1}{b_2} = \frac{-3}{-3} = 1 \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Hence, a unique solution exists. By cross multiplication,
\( \frac{x}{(-3)(-15) - (-3)(-7)} = \frac{y}{(-7)(3) - (-15)(1)} = \frac{1}{(1)(-3) - (3)(-3)} \)
\( \frac{x}{45 - 21} = \frac{y}{-21 + 15} = \frac{1}{-3 + 9} \)
\( \frac{x}{24} = \frac{y}{-6} = \frac{1}{6} \)
\( \frac{x}{24} = \frac{1}{6} \text{ and } \frac{y}{-6} = \frac{1}{6} \)
\( x = \frac{24}{6} \text{ and } y = \frac{-6}{6} \)
Hence, \( x = 4 \) and \( y = -1 \).

Question. Solve the following pair of linear equations by the substitution and cross - multiplication method :
\( 8x + 5y = 9 \)
\( 3x + 2y = 4 \)
Answer: We have \( 8x + 5y = 9 \) or, \( 8x + 5y - 9 = 0 \) ...(1)
and \( 3x + 2y = 4 \) or, \( 3x + 2y - 4 = 0 \) ...(2)
Comparing equation (1) and (2) with \( ax + by + c = 0 \),
\( a_1 = 8, b_1 = 5, c_1 = -9 \)
and \( a_2 = 3, b_2 = 2, c_2 = -4 \)
By cross-multiplication method,
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \)
\( \frac{x}{(5)(-4) - (2)(-9)} = \frac{y}{(-9)(3) - (-4)(8)} = \frac{1}{(8)(2) - (3)(5)} \)
or, \( \frac{x}{-2} = \frac{1}{1} \text{ and } \frac{y}{5} = \frac{1}{1} \)
\( x = -2 \) and \( y = 5 \)
We use substitution method. From equation (2), we have
\( 3x = 4 - 2y \)
or, \( x = \frac{4 - 2y}{3} \) ...(3)
Substituting this value of \( y \) in equation (3) in (1), we get
\( 8 \left( \frac{4 - 2y}{3} \right) + 5y = 9 \)
\( 32 - 16y + 15y = 27 \)
\( -y = 27 - 32 \)
Thus \( y = 5 \)
Substituting this value of \( y \) in equation (3)
\( x = \frac{4 - 2(5)}{3} = \frac{4 - 10}{3} = -2 \)
Hence, \( x = -2 \) and \( y = 5 \).

Question. Solve the following pairs of linear equations by the substitution method:
(a) \( x + y = 14, x - y = 4 \)
(b) \( s - t = 3, \frac{s}{3} + \frac{t}{2} = 6 \)
(c) \( 3x - y = 3, 9x - 3y = 9 \)
(d) \( 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3 \)
(e) \( \sqrt{2}x + \sqrt{3}y = 0, \sqrt{3}x - \sqrt{8}y = 0 \)
(f) \( \frac{3x}{2} - \frac{5y}{3} = -2, \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)
Answer: (a) We have \( x + y = 14 \) ...(1) and \( x - y = 4 \) ...(2)
or, \( y = x - 4 \) ...(3)
Substituting the value of \( y \) from equation (3) in equation (1), we get
\( x + x - 4 = 14 \)
\( 2x = 18 \)
Thus \( x = 9 \)
Substituting this value of \( x \) in equation (3), we get
\( y = 9 - 4 \Rightarrow y = 5 \)
Hence, \( x = 9 \) and \( y = 5 \)
(b) We have, \( s - t = 3 \) ...(1) and \( \frac{s}{3} + \frac{t}{2} = 6 \) ...(2)
From equation (1), \( s = t + 3 \) ...(3)
Substituting \( s = t + 3 \) in equation (2), we get
\( \frac{t+3}{3} + \frac{t}{2} = 6 \)
\( 2(t + 3) + 3t = 36 \)
\( 5t + 6 = 36 \Rightarrow 5t = 30 \Rightarrow t = 6 \)
Substituting this value of \( t \) in (1) we get \( s = 6 + 3 = 9 \)
Hence, \( s = 9, t = 6 \)
(c) We have \( 3x - y = 3 \) ...(1) and \( 9x - 3y = 9 \) ...(2)
From equation (1) \( y = 3x - 3 \) ...(3)
Substituting this value of \( y \) in equation (2),
\( 9x - 3(3x - 3) = 9 \)
\( 9 = 9 \)
Hence, \( x \) and \( y \) both have infinitely many solutions
(d) We have \( 0.2x + 0.3y = 1.3 \) ...(1) and \( 0.4x + 0.5y = 2.3 \) ...(2)
From equation (1) we have \( 3y = 13 - 2x \Rightarrow y = \frac{13 - 2x}{3} \) ...(3)
Substituting this value of \( y \) in equation (2),
\( \frac{4}{10}x + \frac{5}{10} \times \left( \frac{13 - 2x}{3} \right) = \frac{23}{10} \)
\( 4x + \frac{5}{3}(13 - 2x) = 23 \)
\( 12x + 5(13 - 2x) = 3 \times 23 \)
\( 12x + 65 - 10x = 69 \Rightarrow 2x = 69 - 65 = 4 \Rightarrow x = 2 \)
Substituting \( x = 2 \) in equation (3), we get \( y = \frac{13 - 2 \times 2}{3} = \frac{9}{3} = 3 \)
Hence, \( x = 2, y = 3 \)
(e) We have \( \sqrt{2}x + \sqrt{3}y = 0 \) ...(1) and \( \sqrt{3}x - \sqrt{8}y = 0 \) ...(2) or \( y = \frac{\sqrt{3}x}{\sqrt{8}} \) ...(3)
Substituting \( y \) from equation (3) in equation (1),
\( \sqrt{2}x + \sqrt{3} \times \left[ \frac{\sqrt{3}x}{\sqrt{8}} \right] = 0 \Rightarrow \sqrt{2}x + \frac{3x}{\sqrt{8}} = 0 \)
\( \sqrt{2}x \times \sqrt{8} + 3x = 0 \Rightarrow \sqrt{16}x + 3x = 0 \)
\( 4x + 3x = 0 \Rightarrow 7x = 0 \Rightarrow x = 0 \)
Substituting \( x = 0 \) in equation (3), we have \( y = \frac{\sqrt{3} \times 0}{\sqrt{8}} = 0 \)
Hence, \( x = 0, y = 0 \)
(f) We have \( \frac{3x}{2} - \frac{5y}{3} = -2 \) ...(1) and \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \) ...(2)
From equation (2), \( \frac{y}{2} = \frac{13}{6} - \frac{x}{3} = \frac{13 - 2x}{6} \)
\( y = 2 \times \frac{(13 - 2x)}{6} = \frac{(13 - 2x)}{3} \) ...(3)
Substituting this value of \( y \) in equation (1),
\( \frac{3x}{2} - \frac{5}{3} \times \frac{(13 - 2x)}{3} = -2 \)
\( \frac{3x}{2} - \frac{5}{9}(13 - 2x) = -2 \)
\( 27x - 10(13 - 2x) = -36 \)
\( 27x - 130 + 20x = -36 \Rightarrow 47x = 94 \Rightarrow x = 2 \)
Substituting \( x = 2 \) in equation (3), we have \( y = \frac{13 - 2 \times 2}{3} = 3 \)
Hence, \( x = 2, y = 3 \)