CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables VBQs Set 05

Question. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solution? In case there is a unique solution, find it by using cross multiplication method.
(i) \( x - 3y - 3 = 0 \); \( 3x - 9y - 2 = 0 \)
(ii) \( 2x + y = 5 \); \( 3x + 2y = 8 \)
(iii) \( 3x - 5y = 20 \); \( 6x - 10y = 40 \)
(iv) \( x - 3y - 7 = 0 \); \( 3x - 3y - 15 = 0 \)
[NCERT Textbook]
Answer:
(i) The given system of equations is
\( x - 3y - 3 = 0 \)
and \( 3x - 9y - 2 = 0 \)
These equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
and \( a_2x + b_2y + c_2 = 0 \)
where \( a_1 = 1, b_1 = -3, c_1 = -3 \) and \( a_2 = 3, b_2 = -9, c_2 = -2 \)
We have: \( \frac{a_1}{a_2} = \frac{1}{3} \), \( \frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3} \)
and \( \frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2} \)
Clearly, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) (as \( \frac{1}{3} = \frac{1}{3} \neq \frac{3}{2} \))
So, the given system of equations has no solution, i.e., it is inconsistent.
(ii) The given system of equations may be written as
\( 2x + y - 5 = 0 \) and \( 3x + 2y - 8 = 0 \)
These equations are of the form
\( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \)
where \( a_1 = 2, b_1 = 1, c_1 = -5 \)
and \( a_2 = 3, b_2 = 2, c_2 = -8 \)
We have: \( \frac{a_1}{a_2} = \frac{2}{3} \), \( \frac{b_1}{b_2} = \frac{1}{2} \)
Clearly, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
So, the given system of equations has unique solution.
To find the solution, we use the cross-multiplication method. By cross-multiplication, we have:
\( \frac{x}{1 \times (-8) - 2 \times (-5)} = \frac{y}{-5 \times 3 - (-8) \times 2} = \frac{1}{2 \times 2 - 3 \times 1} \)
\( \implies \frac{x}{-8 + 10} = \frac{y}{-15 + 16} = \frac{1}{4 - 3} \)
\( \implies \frac{x}{2} = \frac{y}{1} = \frac{1}{1} \) or \( x = 2, y = 1 \)
Hence, the given system of equations has a unique solution given by \( x = 2, y = 1 \).
(iii) The given system of equations may be written as
\( 3x - 5y - 20 = 0 \) and \( 6x - 10y - 40 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \)
where \( a_1 = 3, b_1 = -5, c_1 = -20 \) and \( a_2 = 6, b_2 = -10, c_2 = -40 \).
We have: \( \frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} \)
and \( \frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2} \)
Clearly, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) (as \( \frac{1}{2} = \frac{1}{2} = \frac{1}{2} \))
So, the given system of equations has infinitely many solutions.
(iv) The given system of equations is
\( x - 3y - 7 = 0 \) and \( 3x - 3y - 15 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \)
where \( a_1 = 1, b_1 = -3, c_1 = -7 \)
and \( a_2 = 3, b_2 = -3, c_2 = -15 \)
We have: \( \frac{a_1}{a_2} = \frac{1}{3} \), \( \frac{b_1}{b_2} = \frac{-3}{-3} = 1 \)
Clearly, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) (as \( \frac{1}{3} \neq 1 \))
So, the given system of equations has unique solution.
To find the solution, we use cross-multiplication method.
By cross-multiplication, we have:
\( \frac{x}{45 - 21} = \frac{y}{-21 + 15} = \frac{1}{-3 + 9} \)
or \( \frac{x}{24} = \frac{y}{-6} = \frac{1}{6} \)
or \( x = \frac{24}{6} = 4 \), \( y = \frac{-6}{6} = -1 \)
Hence, the given system of equations has a unique solution given by \( x = 4, y = -1 \).

Question. Show that the following system has infinitely many solutions:
\( 2y = 4x - 6 \)
\( 2x = y + 3 \)
Answer:
The given system of equations can be written as:
\( 4x - 2y - 6 = 0 \)
\( 2x - y - 3 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where, \( a_1 = 4, b_1 = -2, c_1 = -6 \)
and \( a_2 = 2, b_2 = -1, c_2 = -3 \)
\( \frac{a_1}{a_2} = \frac{4}{2} = 2 \), \( \frac{b_1}{b_2} = \frac{-2}{-1} = 2 \), \( \frac{c_1}{c_2} = \frac{-6}{-3} = 2 \)
Clearly, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), so the given system of equations has infinitely many solutions. Proved.

Question. Show that the following system of equations is inconsistent:
\( 2x + 7y = 11 \)
\( 5x + \frac{35}{2}y = 25 \)
Answer:
The given system of equations can be written as
\( 2x + 7y - 11 = 0 \)
\( 5x + \frac{35}{2}y - 25 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where, \( a_1 = 2, b_1 = 7, c_1 = -11 \)
and \( a_2 = 5, b_2 = \frac{35}{2}, c_2 = -25 \)
\( \frac{a_1}{a_2} = \frac{2}{5} \), \( \frac{b_1}{b_2} = \frac{7}{\frac{35}{2}} = \frac{2}{5} \), \( \frac{c_1}{c_2} = \frac{-11}{-25} = \frac{11}{25} \)
Clearly, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
So, the given system of equations has no solution, i.e., it is inconsistent. Proved.

Question. For what value of k the following system of equations has a unique solution:
\( x - ky = 2 \)
\( 3x + 2y = -5 \)
Answer:
The given system of equations can be written as
\( x - ky - 2 = 0 \)
\( 3x + 2y + 5 = 0 \)
The given system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where, \( a_1 = 1, b_1 = -k, c_1 = -2 \)
and \( a_2 = 3, b_2 = 2, c_2 = 5 \)
Clearly, for unique solution \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
\( \implies \frac{1}{3} \neq \frac{-k}{2} \)
\( \implies k \neq \frac{-2}{3} \)
Hence, k can take any value except \( \frac{-2}{3} \).

Question. Find the value of k for which the following system of equations has infinitely many solutions:
\( (k - 1)x + 3y = 7 \)
\( (k + 1)x + 6y = (5k - 1) \)
Answer:
The given system of equations can be written as
\( (k - 1)x + 3y - 7 = 0 \)
\( (k + 1)x + 6y - (5k - 1) = 0 \)
The given system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
and \( a_2x + b_2y + c_2 = 0 \),
where \( a_1 = (k - 1), b_1 = 3, c_1 = -7 \)
\( a_2 = (k + 1), b_2 = 6, c_2 = -(5k - 1) \)
For the system of equations to have infinite number of solutions:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \implies \frac{k - 1}{k + 1} = \frac{3}{6} = \frac{-7}{-(5k - 1)} \)
\( \implies \frac{k - 1}{k + 1} = \frac{1}{2} = \frac{7}{5k - 1} \)
Taking I and II:
\( \frac{k - 1}{k + 1} = \frac{1}{2} \)
\( \implies 2k - 2 = k + 1 \)
\( \implies k = 3 \)
Taking II and III:
\( \frac{1}{2} = \frac{7}{5k - 1} \)
\( \implies 5k - 1 = 14 \)
\( \implies 5k = 15 \)
\( \implies k = 3 \)
Hence, \( k = 3 \).

Question. Find the values of k for which the following system of equations has no solution:
\( x + 2y = 3 \)
\( (k - 1)x + (k + 1)y = k + 2 \)
Answer:
The given system of equations can be written as
\( x + 2y - 3 = 0 \)
\( (k - 1)x + (k + 1)y - (k + 2) = 0 \)
The above system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where \( a_1 = 1, b_1 = 2, c_1 = -3 \)
and \( a_2 = k - 1, b_2 = k + 1, c_2 = -(k + 2) \)
Clearly, for no solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \implies \frac{1}{k - 1} = \frac{2}{k + 1} \) and \( \frac{2}{k + 1} \neq \frac{-3}{-(k + 2)} \)
\( \implies k + 1 = 2k - 2 \)
\( \implies 2k - k = 1 + 2 \)
\( \implies k = 3 \)
And \( \frac{2}{k + 1} \neq \frac{3}{k + 2} \)
\( \implies 2k + 4 \neq 3k + 3 \)
\( \implies 2k - 3k \neq 3 - 4 \)
\( \implies -k \neq -1 \implies k \neq 1 \)
Further, \( \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \) for no solution.
\( \implies \frac{1}{k - 1} \neq \frac{-3}{-(k + 2)} \implies \frac{1}{k - 1} \neq \frac{3}{k + 2} \)
\( \implies k + 2 \neq 3k - 3 \)
\( \implies 2k \neq 5 \)
\( \implies k \neq 5/2 \)
Thus, for no solution \( k \neq 1, k \neq 5/2 \) and \( k = 3 \).

Question. Find the values of a and b for which the following system of linear equations has infinite number of solutions:
\( 2x - 3y = 7 \)
\( (a + b)x - (a + b - 3)y = 4a + b \)
Answer:
The given system of equations can be written as
\( 2x - 3y - 7 = 0 \)
\( (a + b)x - (a + b - 3)y - (4a + b) = 0 \)
The above system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
where \( a_1 = 2, b_1 = -3, c_1 = -7 \)
\( a_2 = (a + b), b_2 = -(a + b - 3), c_2 = -(4a + b) \)
Hence \( \frac{a_1}{a_2} = \frac{2}{a + b}, \frac{b_1}{b_2} = \frac{-3}{-(a + b - 3)}, \frac{c_1}{c_2} = \frac{-7}{-(4a + b)} \)
For the given system of equations to have an infinite number of solutions
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \implies \frac{2}{a + b} = \frac{-3}{-(a + b - 3)} = \frac{-7}{-(4a + b)} \)
\( \implies \frac{2}{a + b} = \frac{3}{(a + b - 3)} = \frac{7}{(4a + b)} \)
\( \implies \frac{2}{a + b} = \frac{3}{a + b - 3} \) and \( \frac{3}{a + b - 3} = \frac{7}{4a + b} \)
\( \implies 2a + 2b - 6 = 3a + 3b \)
\( \implies 2a - 3a = 3b - 2b + 6 \)
\( \implies -a = b + 6 \)
\( \implies a + b + 6 = 0 \quad ...(1) \)
And
\( 12a + 3b = 7a + 7b - 21 \)
\( 12a - 7a + 3b - 7b = -21 \)
\( 5a - 4b = -21 \)
\( 5a - 4b + 21 = 0 \quad ...(2) \)
Solving (1) and (2), we get
\( b = -1, a = -5 \)
Hence, the given system of linear equations has an infinite number of solutions when \( a = -5, b = -1 \).

Question. (i) For what values of a and b, the following system of equations have an infinite number of solutions?
\( 2x + 3y = 7 \)
\( (a - b)x + (a + b)y = 3a + b - 2 \)
[NCERT Textbook]
(ii) For what 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:
(i) The given system of linear equations can be written as
\( 2x + 3y - 7 = 0 \quad ...(1) \)
\( (a - b)x + (a + b)y - (3a + b - 2) = 0 \quad ...(2) \)
The above system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \),
where \( a_1 = 2, b_1 = 3, c_1 = -7 \)
\( a_2 = (a - b), b_2 = (a + b), c_2 = -(3a + b - 2) \)
For the given system of equations to have an infinite number of solutions
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here, \( \frac{a_1}{a_2} = \frac{2}{a - b}, \frac{b_1}{b_2} = \frac{3}{a + b} \) and \( \frac{c_1}{c_2} = \frac{-7}{-(3a + b - 2)} = \frac{7}{3a + b - 2} \)
\( \implies \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3a + b - 2} \)
(\( \because \) For infinite solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \))
\( \implies \frac{2}{a - b} = \frac{3}{a + b} \) and \( \frac{3}{a + b} = \frac{7}{3a + b - 2} \)
\( \implies 2a + 2b = 3a - 3b \) and \( 9a + 3b - 6 = 7a + 7b \)
\( \implies 2a - 3a = -3b - 2b \) and \( 9a - 7a = 7b - 3b + 6 \)
\( \implies -a = -5b \) and \( 2a = 4b + 6 \)
\( \implies a = 5b \quad ...(3) \)
and \( a = 2b + 3 \quad ...(4) \)
Solving (3) and (4), we get
\( 5b = 2b + 3 \implies 3b = 3 \implies b = 1 \)
Substituting \( b = 1 \) in (3), we get
\( a = 5 \times 1 = 5 \)
Thus, \( a = 5 \) and \( b = 1 \)
Hence, the given system of equations has infinite number of solutions when \( a = 5, b = 1 \).
(ii) The given system of equations may be written as
\( 3x + y - 1 = 0 \)
\( (2k - 1)x + (k - 1)y - (2k + 1) = 0 \)
The above system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
where \( a_1 = 3, b_1 = 1, c_1 = -1 \)
and \( a_2 = (2k - 1), b_2 = (k - 1), c_2 = -(2k + 1) \)
\( \frac{a_1}{a_2} = \frac{3}{2k - 1}, \frac{b_1}{b_2} = \frac{1}{k - 1}, \frac{c_1}{c_2} = \frac{-1}{-(2k + 1)} = \frac{1}{2k + 1} \)
Clearly, for no solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \implies \frac{3}{2k - 1} = \frac{1}{k - 1} \) and \( \frac{1}{k - 1} \neq \frac{1}{2k + 1} \) and \( \frac{3}{2k - 1} \neq \frac{1}{2k + 1} \)
\( \implies 3k - 3 = 2k - 1 \implies 2k + 1 \neq k - 1 \implies 6k + 3 \neq 2k - 1 \)
\( \implies k = 2 \implies k \neq -2 \implies 4k \neq -4 \implies k \neq -1 \)
Hence, the given system of linear equations has no solution, when \( k = 2 \) and \( k \neq -2 \) and \( k \neq -1 \).

Special Type of Equations i.e., Homogeneous Equations

The system of equations
\( a_1x + b_1y = 0 \)
\( a_2x + b_2y = 0 \)
called homogeneous equations has only one solution \( x = 0, y = 0 \), if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
(i) when \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
The system of equations has only one solution, and the system is consistent.
(ii) if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \)
The system of equations has infinitely many solutions and the system is consistent.

Question. Find the value of k for which the system of equations
\( 4x + 5y = 0 \)
\( kx + 10y = 0 \) has infinitely many solutions.
Answer:
The given system is of the form
\( a_1x + b_1y = 0 \)
\( a_2x + b_2y = 0 \)
\( a_1 = 4, b_1 = 5 \) and
\( a_2 = k, b_2 = 10 \).
If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), the system has infinitely many solutions.
\( \implies \frac{4}{k} = \frac{5}{10} \)
\( \implies k = 8 \)
 

EXERCISE 

Question. Show that the system of equations \( \frac{x}{2} + \frac{y}{3} = \frac{1}{6} \) and \( 3x + 2y = 1 \) has infinite many solutions.
Answer: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)

Question. Show that system of equations \( x - y = 3 \) and \( 2x + 3y = 7 \) has unique solution.
Answer: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)

Question. Show that the system of equations \( ax + by = c \) and \( bx + ay = c \) can have infinitely many solution find the relation between a and b in this condition.
Answer: For \( a = b \)

Question. Show that the system of equations \( 5x - 10y = 0 \) and \( x + y = 3 \) has unique solution. Also find the value of x and y.
Answer: \( x = 2, y = 1 \)

Question. Find k for unique solution if
\( x + y = 2 \)
\( 3x - 2y = k \)
Answer: It is possible for every real value of k.

Question. Find k for unique solution
\( 2x + y = 3 \)
\( ky + x = 8 \).
Answer: \( k \neq \frac{1}{2} \)

Question. Find k for infinitely many solutions
\( 2x + 3y = 5 \)
\( 4x + 6y = k \).
Answer: \( k = 10 \)

Question. Find k for unique solution
\( kx + y = 10 \)
\( ky - x = 7 \).
Answer: Possible for any real value of k

Question. For which value(s) of \( \lambda \), do the pair of linear equations \( \lambda x + y = \lambda^2 \) and \( x + \lambda y = 1 \) have
(i) no solution? (ii) infinitely many solutions? (iii) a unique solution?
Answer: (i) \( \lambda = -1 \) (ii) \( \lambda = 1 \) (iii) All real values except \( \pm 1 \)

Question. For which value(s) of k will the pair of equations
\( kx + 3y = k - 3 \)
\( 12x + ky = k \) have no solution?
Answer: \( k = \pm 6 \)

Question. For which values of a and b, will the following pair of linear equations have infinitely many solutions?
\( x + 2y = 1 \)
\( (a - b) x + (a + b) y = a + b - 2 \)
Answer: \( a = 3, b = 1 \)

Question. Find the value(s) of p in the given pair of equations:
\( 3x - y - 5 = 0 \) and \( 6x - 2y - p = 0 \),
if the lines represented by these equations are parallel.
Answer: All real values of p except 10.

CONCEPT-4 : EQUATIONS REDUCIBLE TO SIMULTANEOUS LINEAR EQUATIONS

SOLVED EXAMPLES

Question. Solve the following system of linear equations by using the method of elimination by equating the coefficients.
\( \frac{2}{x - 1} + \frac{3}{y + 1} = 2 \); \( x \neq 1, y \neq -1 \); \( \frac{3}{x - 1} + \frac{2}{y + 1} = \frac{13}{6} \)
Answer:
The given system of equations is
\( \frac{2}{x - 1} + \frac{3}{y + 1} = 2 \quad ...(1) \)
\( \frac{3}{x - 1} + \frac{2}{y + 1} = \frac{13}{6} \quad ...(2) \)
Taking \( \frac{1}{x - 1} = u \) and \( \frac{1}{y + 1} = v \), the given equations become
\( 2u + 3v = 2 \quad ...(3) \)
\( 3u + 2v = \frac{13}{6} \quad ...(4) \)
Multiplying equation (3) by 2 and equation (4) by 3 and subtracting, we get
\( 5u = \frac{13}{2} - 4 \)
\( \implies 5u = \frac{5}{2} \)
\( \implies u = \frac{1}{2} \)
Putting \( u = \frac{1}{2} \) in equation (3), we get
\( 2 \times \frac{1}{2} + 3v = 2 \)
\( \implies 1 + 3v = 2 \implies 3v = 2 - 1 = 1 \implies v = \frac{1}{3} \)
Now, \( u = \frac{1}{2} \implies \frac{1}{x - 1} = \frac{1}{2} \implies x - 1 = 2 \implies x = 3 \)
\( v = \frac{1}{3} \implies \frac{1}{y + 1} = \frac{1}{3} \implies y + 1 = 3 \implies y = 2 \)
Hence, the solution is \( x = 3, y = 2 \)

Question. Solve \( 4x + \frac{6}{y} = 15 \); \( 6x - \frac{8}{y} = 14 \) and hence, find 'P' if \( y = Px - 2 \).
Answer:
The given system of equations is
\( 4x + \frac{6}{y} = 15 \quad ...(1) \)
\( 6x - \frac{8}{y} = 14 \quad ...(2) \)
To eliminate y: The coefficients of \( \frac{1}{y} \) are 6 and -8. The L.C.M. of 6 and 8 is 24. So, we make the coefficients of \( \frac{1}{y} \) as 24 and -24.
Multiplying equation (1) by 4 and equation (2) by 3, we get
\( 16x + \frac{24}{y} = 60 \quad ...(3) \)
\( 18x - \frac{24}{y} = 42 \quad ...(4) \)
Adding equation (3) and equation (4), we get
\( 34x = 102 \implies x = \frac{102}{34} = 3 \)
Putting \( x = 3 \) in equation (1), we get
\( 4(3) + \frac{6}{y} = 15 \implies 12 + \frac{6}{y} = 15 \)
\( \implies \frac{6}{y} = 15 - 12 = 3 \)
\( \implies 3y = 6 \implies y = \frac{6}{3} = 2 \)
Hence, the solution is \( x = 3 \) and \( y = 2 \).
Now, we have \( y = Px - 2 \quad ...(5) \)
On putting \( y = 2 \) and \( x = 3 \) in (5), we get
\( 2 = 3P - 2 \implies 3P = 4 \implies P = \frac{4}{3} \)

Question. Solve the following pairs of equations by reducing them to a pair of linear equations :
\( \frac{1}{2x} + \frac{1}{3y} = 2 \)
\( \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \)
Answer: Taking \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \), the given system of equations becomes
\( \frac{1}{2}u + \frac{1}{3}v = 2 \)
\( \implies 3u + 2v = 12 \) ...(1)
\( \frac{1}{3}u + \frac{1}{2}v = \frac{13}{6} \)
\( \implies 2u + 3v = 13 \) ...(2)
Multiplying (1) by 3 and (2) by 2, we have :
\( 9u + 6v = 36 \) ...(3)
and \( 4u + 6v = 26 \) ...(4)
Subtracting (4) from (3), we get
\( 5u = 10 \)
\( \implies u = 2 \)
Putting \( u = 2 \) in (3), we get
\( 18 + 6v = 36 \)
\( \implies 6v = 18 \)
\( \implies v = 3 \)
Now, \( u = 2 \) gives \( \frac{1}{x} = 2 \)
\( \implies x = \frac{1}{2} \)
and \( v = 3 \) gives \( \frac{1}{y} = 3 \)
\( \implies y = \frac{1}{3} \)
Hence, the solution is \( x = \frac{1}{2}, y = \frac{1}{3} \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations :
\( \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \) and \( \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1 \)
Answer: Putting \( u = \frac{1}{\sqrt{x}} \) and \( v = \frac{1}{\sqrt{y}} \), the given equations becomes
\( 2u + 3v = 2 \) ...(1)
and \( 4u - 9v = -1 \) ...(2)
Multiplying (1) by 3, we get
\( 6u + 9v = 6 \) ...(3)
Adding (2) and (3), we get
\( 10u = 5 \)
\( \implies u = \frac{5}{10} = \frac{1}{2} \)
Putting \( u = \frac{1}{2} \) in (1), we get
\( 2 \times \frac{1}{2} + 3v = 2 \)
\( \implies 3v = 1 \)
\( \implies v = \frac{1}{3} \)
Now, \( u = \frac{1}{2} \) gives \( \frac{1}{\sqrt{x}} = \frac{1}{2} \)
\( \implies \sqrt{x} = 2 \)
\( \implies x = 4 \)
and \( v = \frac{1}{3} \) gives \( \frac{1}{\sqrt{y}} = \frac{1}{3} \)
\( \implies \sqrt{y} = 9 \)
\( \implies y = 9 \)
Hence, the solution is \( x = 4, y = 9 \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations :
\( \frac{4}{x} + 3y = 14 \)
\( \frac{3}{x} - 4y = 23 \)
Answer: Multiplying (1) by 4 and (2) by 3, we get
\( \frac{16}{x} + 12y = 56 \) ...(3)
and \( \frac{9}{x} - 12y = 69 \) ...(4)
Adding (3) and (4), we get
\( \frac{25}{x} = 125 \)
\( \implies x = \frac{25}{125} = \frac{1}{5} \)
Putting \( x = \frac{1}{5} \) in (1), we get
\( 4 \times 5 + 3y = 14 \)
\( 3y = 14 - 20 \)
\( \implies 3y = -6 \)
\( \implies y = -2 \)
Hence, the solution is \( x = \frac{1}{5}, y = -2 \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations :
\( \frac{5}{x - 1} + \frac{1}{y - 2} = 2 \)
\( \frac{6}{x - 1} - \frac{3}{y - 2} = 1 \)
Answer: Let \( u = \frac{1}{x - 1} \) and \( v = \frac{1}{y - 2} \). Then, the given system of equations becomes
\( 5u + v = 2 \) ...(1)
and \( 6u - 3v = 1 \) ...(2)
Multiplying (1) by 3, we get
\( 15u + 3v = 6 \) ...(3)
Adding (2) and (3), we get
\( 21u = 7 \)
\( \implies u = \frac{1}{3} \) ...(4)
Putting \( u = \frac{1}{3} \) in (1), we get
\( \frac{5}{3} + v = 2 \)
\( \implies v = 2 - \frac{5}{3} = \frac{6 - 5}{3} = \frac{1}{3} \)
Now, \( u = \frac{1}{3} \) gives
\( \frac{1}{x - 1} = \frac{1}{3} \)
\( \implies x - 1 = 3 \)
\( \implies x = 4 \)
\( v = \frac{1}{3} \) gives
\( \frac{1}{y - 2} = \frac{1}{3} \)
\( \implies y - 2 = 3 \)
\( \implies y = 5 \)
Hence, the solution is \( x = 4, y = 5 \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations :
\( \frac{7x - 2y}{xy} = 5 \)
\( \frac{8x + 7y}{xy} = 15 \)
Answer: The given system of equations is
\( \frac{7x - 2y}{xy} = 5 \)
\( \implies \frac{7}{y} - \frac{2}{x} = 5 \)
and \( \frac{8x + 7y}{xy} = 15 \)
\( \implies \frac{8}{y} + \frac{7}{x} = 15 \)
Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Then, the above equations becomes
\( 7v - 2u = 5 \) ...(1)
and \( 8v + 7u = 15 \) ...(2)
Multiplying (1) by 7 and (2) by 2, we get
\( 49v - 14u = 35 \) ...(3)
and \( 16v + 14u = 30 \) ...(4)
Adding (3) and (4), we get
\( 65v = 65 \)
\( \implies v = 1 \)
Putting \( v = 1 \) in (1), we get
\( 7 - 2u = 5 \)
\( \implies -2u = -2 \)
\( \implies u = 1 \)
Now, \( u = 1 \) gives \( \frac{1}{x} = 1 \)
\( \implies x = 1 \)
and \( v = 1 \) gives \( \frac{1}{y} = 1 \)
\( \implies y = 1 \)
Hence, the solution is \( x = 1, y = 1 \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations :
\( 6x + 3y = 6xy \)
\( 2x + 4y = 5xy \)
Answer: The given system of equations is \( 6x + 3y = 6xy \) and \( 2x + 4y = 5xy \), where \( x \) and \( y \) are non-zero.
Since \( x \neq 0, y \neq 0 \), we have \( xy \neq 0 \).
On dividing each one of the given equations by \( xy \), we get
\( \frac{3}{x} + \frac{6}{y} = 6 \) and \( \frac{4}{x} + \frac{2}{y} = 5 \)
Taking \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \), the above equations becomes
\( 3u + 6v = 6 \) ...(1)
and \( 4u + 2v = 5 \) ...(2)
Multiplying (2) by 3, we get
\( 12u + 6v = 15 \) ...(3)
Subtracting (1) from (3), we get
\( 9u = 15 - 6 = 9 \)
\( \implies u = 1 \)
Putting \( u = 1 \) in (1), we get
\( 3 \times 1 + 6v = 6 \)
\( \implies 6v = 6 - 3 = 3 \)
\( \implies v = \frac{1}{2} \)
Now, \( u = 1 \) gives \( \frac{1}{x} = 1 \)
\( \implies x = 1 \)
and \( v = \frac{1}{2} \) gives \( \frac{1}{y} = \frac{1}{2} \)
\( \implies y = 2 \)
Hence, the given system of equations has one solution \( x = 1, y = 2 \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations :
\( \frac{10}{x + y} + \frac{2}{x - y} = 4 \)
\( \frac{15}{x + y} - \frac{5}{x - y} = -2 \)
Answer: Putting \( u = \frac{1}{x + y} \) and \( v = \frac{1}{x - y} \), the given equations become
\( 10u + 2v = 4 \)
\( \implies 5u + v = 2 \) ...(1)
and \( 15u - 5v = -2 \) ...(2)
\( 25u + 5v = 10 \) ...(3) [From (1)]
Adding (2) and (3), we get
\( 40u = 8 \)
\( \implies u = \frac{1}{5} \)
Putting \( u = \frac{1}{5} \) in (1), we get
\( 5\left(\frac{1}{5}\right) + v = 2 \)
\( \implies v = 2 - 1 = 1 \)
Now, \( u = \frac{1}{5} \) gives \( \frac{1}{x + y} = \frac{1}{5} \)
\( \implies x + y = 5 \) ...(4)
and \( v = 1 \) gives \( \frac{1}{x - y} = 1 \)
\( \implies x - y = 1 \) ...(5)
Adding (4) and (5), we get
\( 2x = 6 \)
\( \implies x = 3 \)
When \( x = 3 \), then from (4), we get
\( 3 + y = 5 \)
\( \implies y = 5 - 3 = 2 \)
Hence, the given system of equations has one solution \( x = 3, y = 2 \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations :
\( \frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \)
\( \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = -\frac{1}{8} \)
Answer: Taking \( u = \frac{1}{3x + y} \) and \( v = \frac{1}{3x - y} \), the given system of equations becomes
\( u + v = \frac{3}{4} \) ...(1)
and \( \frac{1}{2}u - \frac{1}{2}v = \frac{-1}{8} \)
\( \implies u - v = \frac{-1}{4} \) ...(2)
Adding (1) and (2), we get
\( 2u = \frac{3}{4} - \frac{1}{4} \)
\( \implies 2u = \frac{2}{4} = \frac{1}{2} \)
\( \implies u = \frac{1}{4} \)
Putting \( u = \frac{1}{4} \) in (1), we get
\( \frac{1}{4} + v = \frac{3}{4} \)
\( \implies v = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
Now, \( u = \frac{1}{4} \) gives \( \frac{1}{3x + y} = \frac{1}{4} \)
\( \implies 3x + y = 4 \) ...(3)
and \( v = \frac{1}{2} \) gives \( \frac{1}{3x - y} = \frac{1}{2} \)
\( \implies 3x - y = 2 \) ...(4)
Adding (3) and (4), we get
\( 6x = 6 \) or \( x = 1 \)
Putting \( x = 1 \) in (3), we get
\( 3 + y = 4 \) or \( y = 4 - 3 = 1 \)
Hence, the solution is \( x = 1, y = 1 \).

Question. Solve :
\( 3(2u + v) = 7uv \)
\( 3(u + 3v) = 11uv \)
Answer: The given equations are
\( 3(2u + v) = 7uv \) ...(1)
\( 3(u + 3v) = 11uv \) ...(2)
The given equations will be reduced to linear equations on dividing them both sides by \( uv \).
\( 3\left(\frac{2}{v} + \frac{1}{u}\right) = 7 \) ...(3)
\( 3\left(\frac{1}{v} + \frac{3}{u}\right) = 11 \) ...(4)
Putting \( \frac{1}{u} = x \) and \( \frac{1}{v} = y \), in equations (3) and (4), we get
\( 3(2y + x) = 7 \)
\( \implies 3x + 6y = 7 \) ...(5)
\( 3(y + 3x) = 11 \)
\( \implies 9x + 3y = 11 \) ...(6)
Multiplying equation (6) by 2, we get
\( 18x + 6y = 22 \) ...(7)
Subtracting equation (7) from equation (5), we get
\( -15x = -15 \)
\( \implies x = 1 \)
\( \implies \frac{1}{u} = 1 \)
\( \implies u = 1 \)
On putting \( x = 1 \) in equation (5), we get \( 3 + 6y = 7 \)
\( \implies 6y = 4 \)
\( \implies y = \frac{4}{6} = \frac{2}{3} \)
\( \implies \frac{1}{v} = \frac{2}{3} \)
\( \implies v = \frac{3}{2} \)
Hence, \( u = 1, v = \frac{3}{2} \).

Question. Solve : \( \frac{1}{2(2x + 3y)} + \frac{12}{7(3x - 2y)} = \frac{1}{2} \) and \( \frac{7}{2x + 3y} + \frac{4}{3x - 2y} = 2 \), where \( 2x + 3y \neq 0 \) and \( 3x - 2y \neq 0 \).
Answer: Putting \( \frac{1}{2x + 3y} = p \) and \( \frac{1}{3x - 2y} = q \), the given equations become
\( \frac{p}{2} + \frac{12q}{7} = \frac{1}{2} \)
\( \implies 7p + 24q = 7 \) ...(1)
and \( 7p + 4q = 2 \) ...(2)
Subtracting equation (2) from equation (1), we get
\( 20q = 5 \)
\( \implies q = \frac{5}{20} = \frac{1}{4} \)
Putting \( q = \frac{1}{4} \) in equation (2), we get
\( 7p + 4\left(\frac{1}{4}\right) = 2 \)
\( \implies 7p + 1 = 2 \)
\( \implies 7p = 2 - 1 = 1 \)
\( \implies p = \frac{1}{7} \)
Now, \( p = \frac{1}{2x + 3y} = \frac{1}{7} \)
\( \implies 2x + 3y = 7 \) ...(3)
\( q = \frac{1}{3x - 2y} = \frac{1}{4} \)
\( \implies 3x - 2y = 4 \) ...(4)
Multiplying equation (3) by 2 and equation (4) by 3, we get
\( 4x + 6y = 14 \) ...(5)
\( 9x - 6y = 12 \) ...(6)
On adding equation (5) and equation (6), we get
\( 13x = 26 \)
\( \implies x = 2 \)
On putting \( x = 2 \) in equation (3), we get
\( 2(2) + 3y = 7 \)
\( \implies 3y = 7 - 4 = 3 \)
\( \implies y = 1 \)
Hence, the solution is \( x = 2, y = 1 \).

Special Type of Equation

\( ax + by = c_1 \)
\( bx + ay = c_2 \) where \( a \neq b \).

Working Rule.

The given equations are
\( ax + by = c_1 \) ...(1)
\( bx + ay = c_2 \) ...(2)

Step 1. Add these equations to get
\( (a + b)x + (a + b)y = c_1 + c_2 \)
\( \implies x + y = \frac{c_1 + c_2}{a + b} \)
\( \implies x + y = p \) (say) ...(3)

Step 2. Subtract these equations to get
\( (a - b)x - (a - b)y = c_1 - c_2 \)
\( \implies x - y = \frac{c_1 - c_2}{a - b} \)
\( \implies x - y = q \) (say) ...(4)

Step 3. Add equations (3) and (4) to get the value of \( x \).
Step 4. Subtract equation (4) from equation (3) to get the value of \( y \).

Question. Solve for \( x \) and \( y \):
47x + 31y = 63
31x + 47y = 15
Answer: The given equations are
\( 47x + 31y = 63 \) ...(1)
\( 31x + 47y = 15 \) ...(2)
Adding (1) and (2), we get
\( 78x + 78y = 78 \)
\( 78(x + y) = 78 \)
\( \implies x + y = 1 \) ...(3)
Subtracting equation (1) from equation (2), we get
\( 16x - 16y = 48 \)
\( 16(x - y) = 48 \)
\( \implies x - y = \frac{48}{16} = 3 \)
\( \implies x - y = 3 \) ...(4)
Adding equations (3) and (4), we get
\( 2x = 4 \)
\( \implies x = 2 \)
Subtracting equation (4) from equation (3), we get
\( 2y = -2 \)
\( \implies y = -1 \)
Hence, the solution is \( x = 2, y = -1 \).

Question. Solve \( 217x + 131y = 913 \) and \( 131x + 217y = 827 \)
Answer: The given equations are
\( 217x + 131y = 913 \) ...(1)
\( 131x + 217y = 827 \) ...(2)
Adding equation (1) and equation (2), we get
\( 348x + 348y = 1740 \)
\( 348(x + y) = 1740 \)
\( \implies x + y = \frac{1740}{348} = 5 \) ...(3)
Subtracting equation (2) from equation (1), we get
\( 86x - 86y = 86 \)
\( 86(x - y) = 86 \)
\( \implies x - y = 1 \) ...(4)
Adding equations (3) and (4), we get
\( 2x = 6 \)
\( \implies x = 3 \)
Subtracting equation (4) from equation (3), we get
\( 2y = 4 \)
\( \implies y = 2 \)
Hence, the solution is \( x = 3, y = 2 \).