CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables VBQs Set 04

Question. Solve for \( x \) and \( y \):
\( \frac{ax}{b} - \frac{by}{a} = a + b \); \( ax - by = 2ab \)
Answer: The given system of equations is
\( \frac{ax}{b} - \frac{by}{a} = a + b \) ...(1)
\( ax - by = 2ab \) ...(2)
Dividing (2) by \( a \), we get
\( x - \frac{by}{a} = 2b \) ...(3)
On subtracting (3) from (1), we get
\( \frac{ax}{b} - x = a - b \)
\( \implies \) \( x \left( \frac{a}{b} - 1 \right) = a - b \)
\( \implies \) \( x \left( \frac{a - b}{b} \right) = a - b \)
\( \implies \) \( x = \frac{(a - b)b}{a - b} = b \)
\( x = b \)
On substituting the value of \( x \) in (3), we get \( b - \frac{by}{a} = 2b \)
\( \implies \) \( b \left( 1 - \frac{y}{a} \right) = 2b \)
\( \implies \) \( 1 - \frac{y}{a} = 2 \)
\( \implies \) \( \frac{y}{a} = 1 - 2 \)
\( \implies \) \( \frac{y}{a} = -1 \)
\( \implies \) \( y = -a \)
Hence, the solution of the given equations is \( x = b, y = -a \). Ans.

Question. Solve for \( x \) and \( y \):
\( x + \frac{6}{y} = 6 \); \( 3x - \frac{8}{y} = 5 \)
Answer: We have,
\( x + \frac{6}{y} = 6 \) ...(1)
\( 3x - \frac{8}{y} = 5 \) ...(2)
Multiplying (1) by 3, we get
\( 3x + \frac{18}{y} = 18 \) ...(3)
Subtracting (2) from (3), we get
\( \frac{1}{y} (18 + 8) = 13 \)
\( \implies \) \( \frac{26}{y} = 13 \)
\( \implies \) \( y = \frac{26}{13} = 2 \)
Putting \( y = 2 \) in (1), we get
\( x + \frac{6}{2} = 6 \)
\( \implies \) \( x + 3 = 6 \)
\( \implies \) \( x = 6 - 3 = 3 \)
Hence, \( x = 3 \) and \( y = 2 \).

Question. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) \( x + y = 5 \) and \( 2x - 3y = 4 \)
(ii) \( 3x + 4y = 10 \) and \( 2x - 2y = 2 \)
(iii) \( 3x - 5y - 4 = 0 \) and \( 9x = 2y + 7 \)
(iv) \( \frac{x}{2} + \frac{2y}{3} = -1 \) and \( x - \frac{y}{3} = 3 \)
Answer: (i) By elimination method:
The given system of equations is
\( x + y = 5 \) ...(1)
and \( 2x - 3y = 4 \) ...(2)
Multiplying (1) by 3, we get
\( 3x + 3y = 15 \) ...(3)
Adding (2) and (3), we get
\( 5x = 19 \)
\( \implies \) \( x = \frac{19}{5} \)
Putting \( x = \frac{19}{5} \) in (1), we get
\( \frac{19}{5} + y = 5 \)
\( \implies \) \( y = 5 - \frac{19}{5} = \frac{25 - 19}{5} = \frac{6}{5} \)
Hence, \( x = \frac{19}{5}, y = \frac{6}{5} \).
By substitution method:
The given system of equations is
\( x + y = 5 \) ...(1)
and \( 2x - 3y = 4 \) ...(2)
From (1), \( y = 5 - x \)
Substituting \( y = 5 - x \) in (2), we get
\( 2x - 3(5 - x) = 4 \)
\( \implies \) \( 2x - 15 + 3x = 4 \)
\( \implies \) \( 5x = 4 + 15 \)
\( \implies \) \( 5x = 19 \)
\( \implies \) \( x = \frac{19}{5} \)
Putting \( x = \frac{19}{5} \) in (1), we get
\( \frac{19}{5} + y = 5 \)
\( \implies \) \( y = 5 - \frac{19}{5} = \frac{25 - 19}{5} = \frac{6}{5} \).
Hence, \( x = \frac{19}{5}, y = \frac{6}{5} \). Ans.

(ii) By elimination method:
The given system of equations is
\( 3x + 4y = 10 \) ...(1)
and \( 2x - 2y = 2 \) ...(2)
Multiplying (2) by 2 and adding to (1), we get
\( 7x = 14 \)
\( \implies \) \( x = 2 \)
Putting \( x = 2 \) in (1), we get
\( 3(2) + 4y = 10 \)
\( \implies \) \( 4y = 10 - 6 = 4 \)
\( \implies \) \( y = 1 \).
Hence, \( x = 2, y = 1 \).
By substitution method:
The given system of equations is
\( 3x + 4y = 10 \) ...(1)
and \( 2x - 2y = 2 \)
\( \implies \) \( x - y = 1 \) ...(2)
From (2), \( y = x - 1 \)
Substituting \( y = x - 1 \) in (1), we get
\( 3x + 4(x - 1) = 10 \)
\( \implies \) \( 3x + 4x - 4 = 10 \)
\( \implies \) \( 7x = 14 \)
\( \implies \) \( x = 2 \)
Putting \( x = 2 \) in (1), we get
\( 3(2) + 4y = 10 \)
\( \implies \) \( 4y = 10 - 6 = 4 \)
\( \implies \) \( y = 1 \).
Hence, \( x = 2, y = 1 \). Ans.

(iii) By elimination method:
The given system of equations is
\( 3x - 5y - 4 = 0 \)
\( \implies \) \( 3x - 5y = 4 \) ...(1)
and \( 9x = 2y + 7 \)
\( \implies \) \( 9x - 2y = 7 \) ...(2)
Multiplying (1) by 3, we get
\( 9x - 15y = 12 \) ...(3)
Subtracting (2) from (3), we get
\( -13y = 5 \)
\( \implies \) \( y = -\frac{5}{13} \).
Putting \( y = -\frac{5}{13} \) in (1), we get
\( 3x - 5\left(-\frac{5}{13}\right) = 4 \)
\( \implies \) \( 3x + \frac{25}{13} = 4 \)
\( \implies \) \( 3x = 4 - \frac{25}{13} \)
\( \implies \) \( 3x = \frac{52 - 25}{13} = \frac{27}{13} \)
\( \implies \) \( x = \frac{9}{13} \).
Hence, \( x = \frac{9}{13}, y = -\frac{5}{13} \).
By substitution method:
The given system of equations is
\( 3x - 5y = 4 \) ...(1)
and \( 9x - 2y = 7 \) ...(2)
From (2), \( 2y = 9x - 7 \)
\( \implies \) \( y = \frac{9x - 7}{2} \)
Substituting \( y = \frac{9x - 7}{2} \) in (1), we get
\( 3x - 5\left(\frac{9x - 7}{2}\right) = 4 \)
\( \implies \) \( 6x - 45x + 35 = 8 \)
\( \implies \) \( -39x = 8 - 35 \)
\( \implies \) \( -39x = -27 \)
\( \implies \) \( x = \frac{-27}{-39} = \frac{9}{13} \)
Putting \( x = \frac{9}{13} \) in (1), we get
\( 3\left(\frac{9}{13}\right) - 5y = 4 \)
\( \implies \) \( \frac{27}{13} - 5y = 4 \)
\( \implies \) \( 5y = \frac{27}{13} - 4 \)
\( \implies \) \( 5y = \frac{27 - 52}{13} = -\frac{25}{13} \)
\( \implies \) \( y = -\frac{5}{13} \).
Hence, \( x = \frac{9}{13}, y = -\frac{5}{13} \). Ans.

(iv) By elimination method:
The given system of equations is
\( \frac{x}{2} + \frac{2y}{3} = -1 \)
\( \implies \) \( 3x + 4y = -6 \) ...(1)
and \( x - \frac{y}{3} = 3 \)
\( \implies \) \( 3x - y = 9 \) ...(2)
Multiplying (2) by 4 and adding to (1), we get
\( 15x = 30 \)
\( \implies \) \( x = 2 \)
Putting \( x = 2 \) in (2), we get
\( 3(2) - y = 9 \)
\( \implies \) \( -y = 9 - 6 = 3 \)
\( \implies \) \( y = -3 \).
Hence, \( x = 2, y = -3 \).
By substitution method:
The given system of equations is
\( 3x + 4y = -6 \) ...(1)
and \( 3x - y = 9 \) ...(2)
From (2), \( y = 3x - 9 \)
Substituting \( y = 3x - 9 \) in (1), we get
\( 3x + 4(3x - 9) = -6 \)
\( \implies \) \( 3x + 12x - 36 = -6 \)
\( \implies \) \( 15x = 30 \)
\( \implies \) \( x = 2 \)
Putting \( x = 2 \) in (2), we get
\( 3(2) - y = 9 \)
\( \implies \) \( -y = 9 - 6 = 3 \)
\( \implies \) \( y = -3 \).
Hence, \( x = 2, y = -3 \). Ans.

Question. Solve the following system of equations by the method of cross-multiplication.
\( 2x - 6y + 10 = 0 \)
\( 3x - 7y + 13 = 0 \)
Answer: The given system of equations is
\( 2x - 6y + 10 = 0 \)
\( 3x - 7y + 13 = 0 \)
By cross-multiplication, we have
\( \frac{x}{(-6 \times 13) - (-7 \times 10)} = \frac{-y}{(2 \times 13) - (3 \times 10)} = \frac{1}{(2 \times -7) - (3 \times -6)} \)
\( \implies \) \( \frac{x}{-78 + 70} = \frac{-y}{26 - 30} = \frac{1}{-14 + 18} \)
\( \implies \) \( \frac{x}{-8} = \frac{-y}{-4} = \frac{1}{4} \)
\( \implies \) \( \frac{x}{-8} = \frac{1}{4} \implies x = -2 \)
\( \implies \) \( \frac{-y}{-4} = \frac{1}{4} \implies y = 1 \).
Hence, the solution is \( x = -2, y = 1 \).

Question. Solve the following system of equations by the method of cross-multiplication.
\( 11x + 15y = -23 \)
\( 7x - 2y = 20 \)
Answer: The given system of equations is
\( 11x + 15y + 23 = 0 \)
\( 7x - 2y - 20 = 0 \)
Now, by cross-multiplication method, we have
\( \frac{x}{(15 \times -20) - (-2 \times 23)} = \frac{-y}{(11 \times -20) - (7 \times 23)} = \frac{1}{(11 \times -2) - (7 \times 15)} \)
\( \implies \) \( \frac{x}{-300 + 46} = \frac{-y}{-220 - 161} = \frac{1}{-22 - 105} \)
\( \implies \) \( \frac{x}{-254} = \frac{-y}{-381} = \frac{1}{-127} \)
\( \implies \) \( \frac{x}{-254} = \frac{1}{-127} \implies x = 2 \)
and \( \frac{-y}{-381} = \frac{1}{-127} \implies y = -3 \).
Hence, \( x = 2, y = -3 \) is the required solution.

Question. Solve the following pair of linear equations:
(i) \( px + qy = p - q \) and \( qx - py = p + q \)
(ii) \( ax + by = c \) and \( bx + ay = 1 + c \)
(iii) \( \frac{x}{a} - \frac{y}{b} = 0 \) and \( ax + by = a^2 + b^2 \)
(iv) \( (a - b)x + (a + b)y = a^2 - 2ab - b^2 \) and \( (a + b)(x + y) = a^2 + b^2 \)
(v) \( 152x - 378y = -74 \) and \( -378x + 152y = -604 \)
Answer: (i) The given pair of equations is
\( px + qy = p - q \implies px + qy - (p - q) = 0 \)
and \( qx - py = p + q \implies qx - py - (p + q) = 0 \)
By cross-multiplication, we get
\( \frac{x}{-q(p + q) - p(p - q)} = \frac{y}{-q(p - q) + p(p + q)} = \frac{1}{-p^2 - q^2} \)
\( \implies \) \( \frac{x}{-pq - q^2 - p^2 + pq} = \frac{y}{-pq + q^2 + p^2 + pq} = \frac{1}{-(p^2 + q^2)} \)
\( \implies \) \( \frac{x}{-(p^2 + q^2)} = \frac{y}{p^2 + q^2} = \frac{1}{-(p^2 + q^2)} \)
\( \implies \) \( x = \frac{-(p^2 + q^2)}{-(p^2 + q^2)} = 1 \)
and \( y = \frac{(p^2 + q^2)}{-(p^2 + q^2)} = -1 \).
Hence, the required solution is \( x = 1 \) and \( y = -1 \).

(ii) The given system of equations may be written as
\( ax + by - c = 0 \)
and \( bx + ay - (1 + c) = 0 \)
By cross-multiplication, we have :
\( \frac{x}{-b(1 + c) + ac} = \frac{y}{-bc + a(1 + c)} = \frac{1}{a^2 - b^2} \)
\( \implies \) \( \frac{x}{-b - bc + ac} = \frac{y}{-bc + a + ac} = \frac{1}{a^2 - b^2} \)
\( \implies \) \( \frac{x}{c(a - b) - b} = \frac{y}{c(a - b) + a} = \frac{1}{(a - b)(a + b)} \)
\( \implies \) \( x = \frac{c(a - b) - b}{(a - b)(a + b)} = \frac{c}{a + b} - \frac{b}{(a - b)(a + b)} \)
and \( y = \frac{c(a - b) + a}{(a - b)(a + b)} = \frac{c}{a + b} + \frac{a}{(a - b)(a + b)} \).

(iii) The given system of equations is
\( \frac{x}{a} - \frac{y}{b} = 0 \implies bx - ay = 0 \)
and \( ax + by - (a^2 + b^2) = 0 \)
By cross-multiplication, we have :
\( \frac{x}{a(a^2 + b^2) - 0} = \frac{y}{0 + b(a^2 + b^2)} = \frac{1}{b^2 + a^2} \)
\( \implies \) \( \frac{x}{a(a^2 + b^2)} = \frac{y}{b(a^2 + b^2)} = \frac{1}{a^2 + b^2} \)
\( \implies \) \( x = \frac{a(a^2 + b^2)}{a^2 + b^2} = a \) and \( y = \frac{b(a^2 + b^2)}{a^2 + b^2} = b \).
Hence, the required solution is \( x = a \) and \( y = b \).

(iv) The given pair of equations may be rewritten as
\( (a - b)x + (a + b)y - (a^2 - 2ab - b^2) = 0 \)
and \( (a + b)x + (a + b)y - (a^2 + b^2) = 0 \)
By cross-multiplication, we have :
\( \frac{x}{-(a + b)(a^2 + b^2) + (a + b)(a^2 - 2ab - b^2)} = \frac{y}{-(a + b)(a^2 - 2ab - b^2) + (a - b)(a^2 + b^2)} = \frac{1}{(a - b)(a + b) - (a + b)^2} \)
\( \implies \) \( \frac{x}{(a + b)(-a^2 - b^2 + a^2 - 2ab - b^2)} = \frac{y}{-a^3 + 2a^2b + ab^2 - a^2b + 2ab^2 + b^3 + a^3 + ab^2 - a^2b - b^3} = \frac{1}{a^2 - b^2 - a^2 - b^2 - 2ab} \)
\( \implies \) \( \frac{x}{(a + b)(-2ab - 2b^2)} = \frac{y}{4ab^2} = \frac{1}{-2b^2 - 2ab} \)
\( \implies \) \( \frac{x}{(a + b)(-2b)(a + b)} = \frac{y}{4ab^2} = \frac{1}{-2b(a + b)} \)
\( \implies \) \( x = a + b \) and \( y = \frac{-2ab}{a + b} \).

(v) We have :
\( 152x - 378y = -74 \) ...(1)
and \( -378x + 152y = -604 \) ...(2)
Adding (1) and (2), we get
\( -226x - 226y = -678 \)
\( \implies \) \( x + y = 3 \) ...(3)
Subtracting (1) from (2), we get
\( -530x + 530y = -530 \)
\( \implies \) \( x - y = 1 \) ...(4)
Adding (3) and (4), we get
\( 2x = 4 \)
\( \implies \) \( x = 2 \)
Putting \( x = 2 \) in (1), we get
\( 2 \times 2 = 4 \)
\( \implies \) \( y = 1 \).
Hence, the required solution is \( x = 2 \) and \( y = 1 \).

Question. Solve the following system of equations by cross-multiplication method.
\( ax + by = a - b \)
\( bx - ay = a + b \)
Answer: Rewriting the given system of equations, we get
\( ax + by - (a - b) = 0 \)
\( bx - ay - (a + b) = 0 \)
By cross-multiplication method, we have
\( \frac{x}{b\{-(a + b)\} - (-a)\{-(a - b)\}} = \frac{-y}{a\{-(a + b)\} - b\{-(a - b)\}} = \frac{1}{a \times (-a) - b \times b} \)
\( \implies \) \( \frac{x}{-b(a + b) - a(a - b)} = \frac{-y}{-a(a + b) + b(a - b)} = \frac{1}{-a^2 - b^2} \)
\( \implies \) \( \frac{x}{-ab - b^2 - a^2 + ab} = \frac{-y}{-a^2 - ab + ab - b^2} = \frac{1}{-(a^2 + b^2)} \)
\( \implies \) \( \frac{x}{-(a^2 + b^2)} = \frac{-y}{-(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \)
\( \implies \) \( \frac{x}{-(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \implies x = 1 \)
and \( \implies \) \( \frac{-y}{-(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \implies y = -1 \).
Hence, the solution is \( x = 1, y = -1 \).

Question. Solve the following system of equations by cross-multiplication method.
\( x + y = a - b \)
\( ax - by = a^2 + b^2 \)
Answer: The given system of equations can be rewritten as:
\( x + y - (a - b) = 0 \)
\( ax - by - (a^2 + b^2) = 0 \)
By cross-multiplication method, we have
\( \frac{x}{1 \times -(a^2 + b^2) - (-b) \times -(a - b)} = \frac{-y}{1 \times -(a^2 + b^2) - a \times -(a - b)} = \frac{1}{1 \times (-b) - a \times 1} \)

\( \implies \) \( \frac{x}{-(a^2 + b^2) - b(a - b)} = \frac{-y}{-(a^2 + b^2) + a(a - b)} = \frac{1}{-(b + a)} \)

\( \implies \) \( \frac{x}{-a^2 - b^2 - ab + b^2} = \frac{-y}{-a^2 - b^2 + a^2 - ab} = \frac{1}{-(a + b)} \)

\( \implies \) \( \frac{x}{-a(a + b)} = \frac{-y}{-b(a + b)} = \frac{1}{-(a + b)} \)

\( \implies \) \( \frac{x}{-a(a + b)} = \frac{1}{-(a + b)} \)

\( \implies \) \( x = a \)
and

\( \implies \) \( \frac{-y}{-b(a + b)} = \frac{1}{-(a + b)} \)

\( \implies \) \( y = -b \)
Hence, the solution is \( x = a, y = -b \). Ans.

Question. Solve the following system of equations by the method of cross-multiplication :
\( \frac{x}{a} + \frac{y}{b} = a + b \); \( \frac{x}{a^2} + \frac{y}{b^2} = 2 \)
Answer: The given system of equations is rewritten as :
\( \frac{x}{a} + \frac{y}{b} - (a + b) = 0 \) ...(1)
\( \frac{x}{a^2} + \frac{y}{b^2} - 2 = 0 \) ...(2)
Multiplying equation (1) by \( ab \), we get \( bx + ay - ab(a + b) = 0 \) ...(3)
Multiplying equation (2) by \( a^2 b^2 \), we get \( b^2 x + a^2 y - 2 a^2 b^2 = 0 \) ...(4)
By cross multiplication method, we have
\( \frac{x}{a(-2a^2b^2) - a^2[-ab(a + b)]} = \frac{-y}{b(-2a^2b^2) - b^2[-ab(a + b)]} = \frac{1}{ba^2 - b^2a} \)

\( \implies \) \( \frac{x}{-2a^3b^2 + a^3b(a + b)} = \frac{-y}{-2a^2b^3 + ab^3(a + b)} = \frac{1}{a^2b - ab^2} \)

\( \implies \) \( \frac{x}{-2a^3b^2 + a^4b + a^3b^2} = \frac{-y}{-2a^2b^3 + a^2b^3 + ab^4} = \frac{1}{ab(a - b)} \)

\( \implies \) \( \frac{x}{a^4b - a^3b^2} = \frac{-y}{ab^4 - a^2b^3} = \frac{1}{ab(a - b)} \)

\( \implies \) \( \frac{x}{a^3b(a - b)} = \frac{y}{a^2b^3(a - b)} = \frac{1}{ab(a - b)} \)

\( \implies \) \( x = \frac{a^3b(a - b)}{ab(a - b)} = a^2 \)

\( \implies \) \( y = \frac{a^2b^3(a - b)}{ab(a - b)} = ab^2 \)
Wait, correcting the algebra from the image steps:
\( \frac{y}{ab^3(a-b)} = \frac{1}{ab(a-b)} \implies y = b^2 \).
Hence, the solution is \( x = a^2, y = b^2 \). Ans.

Question. Solve the following system of equations by cross-multiplication method:
\( ax + by = 1 \)
\( bx + ay = \frac{(a + b)^2}{a^2 + b^2} - 1 \)
Answer: The given system of equations can be written as.
\( ax + by - 1 = 0 \) ...(1)
\( bx + ay = \frac{(a + b)^2}{a^2 + b^2} - 1 \)

\( \implies \) \( bx + ay = \frac{a^2 + 2ab + b^2 - a^2 - b^2}{a^2 + b^2} \)

\( \implies \) \( bx + ay = \frac{2ab}{a^2 + b^2} \)

\( \implies \) \( bx + ay - \frac{2ab}{a^2 + b^2} = 0 \) ...(2)
Rewriting the equations (1) and (2), we have
\( ax + by - 1 = 0 \)
\( bx + ay - \frac{2ab}{a^2 + b^2} = 0 \)
Now, by cross-multiplication method, we have
\( \frac{x}{b(-\frac{2ab}{a^2 + b^2}) - a(-1)} = \frac{-y}{a(-\frac{2ab}{a^2 + b^2}) - b(-1)} = \frac{1}{a \times a - b \times b} \)

\( \implies \) \( \frac{x}{\frac{-2ab^2}{a^2 + b^2} + a} = \frac{-y}{\frac{-2a^2b}{a^2 + b^2} + b} = \frac{1}{a^2 - b^2} \)

\( \implies \) \( \frac{x}{\frac{-2ab^2 + a^3 + ab^2}{a^2 + b^2}} = \frac{-y}{\frac{-2a^2b + a^2b + b^3}{a^2 + b^2}} = \frac{1}{a^2 - b^2} \)

\( \implies \) \( \frac{x}{\frac{a(a^2 - b^2)}{a^2 + b^2}} = \frac{-y}{\frac{b(b^2 - a^2)}{a^2 + b^2}} = \frac{1}{a^2 - b^2} \)

\( \implies \) \( x = \frac{a(a^2 - b^2)}{a^2 + b^2} \times \frac{1}{a^2 - b^2} \)

\( \implies \) \( x = \frac{a}{a^2 + b^2} \)
and

\( \implies \) \( y = \frac{b(a^2 - b^2)}{a^2 + b^2} \times \frac{1}{a^2 - b^2} \)

\( \implies \) \( y = \frac{b}{a^2 + b^2} \)
Hence, the solution is \( x = \frac{a}{a^2 + b^2}, y = \frac{b}{a^2 + b^2} \). Ans.

Question. Solve the following system of equations by cross-multiplication method.
\( a(x + y) + b(x - y) = a^2 - ab + b^2 \)
\( a(x + y) - b(x - y) = a^2 + ab + b^2 \)
Answer: The given system of equations can be rewritten as
\( ax + bx + ay - by - (a^2 - ab + b^2) = 0 \)

\( \implies \) \( (a + b)x + (a - b)y - (a^2 - ab + b^2) = 0 \) ...(1)
And \( ax - bx + ay + by - (a^2 + ab + b^2) = 0 \)

\( \implies \) \( (a - b)x + (a + b)y - (a^2 + ab + b^2) = 0 \) ...(2)
Now, by cross-multiplication method, we have
\( \frac{x}{(a - b)[-(a^2 + ab + b^2)] - (a + b)[-(a^2 - ab + b^2)]} = \frac{-y}{(a + b)[-(a^2 + ab + b^2)] - (a - b)[-(a^2 - ab + b^2)]} = \frac{1}{(a + b)^2 - (a - b)^2} \)

\( \implies \) \( \frac{x}{-(a^3 - b^3) + (a^3 + b^3)} = \frac{-y}{-a^3 - 2a^2b - 2ab^2 - b^3 + a^3 - 2a^2b + 2ab^2 - b^3} = \frac{1}{4ab} \)

\( \implies \) \( \frac{x}{2b^3} = \frac{-y}{-4a^2b - 2b^3} = \frac{1}{4ab} \)

\( \implies \) \( \frac{x}{2b^3} = \frac{1}{4ab} \)

\( \implies \) \( x = \frac{2b^3}{4ab} = \frac{b^2}{2a} \)
And

\( \implies \) \( \frac{-y}{-2b(2a^2 + b^2)} = \frac{1}{4ab} \)

\( \implies \) \( y = \frac{2b(2a^2 + b^2)}{4ab} = \frac{2a^2 + b^2}{2a} \)
Hence, the solution is \( x = \frac{b^2}{2a}, y = \frac{2a^2 + b^2}{2a} \). Ans.

Question. Solve the following system of equations by the method of cross-multiplication.
\( \frac{a}{x} - \frac{b}{y} = 0 \)
\( \frac{ab^2}{x} + \frac{a^2b}{y} = a^2 + b^2 \); where \( x \neq 0, y \neq 0 \)
Answer: The given system of equations is
\( \frac{a}{x} - \frac{b}{y} = 0 \) ...(1)
\( \frac{ab^2}{x} + \frac{a^2b}{y} - (a^2 + b^2) = 0 \) ...(2)
Putting \( \frac{a}{x} = u \) and \( \frac{b}{y} = v \) in equations (1) and (2) the system of equations reduces to
\( u - v + 0 = 0 \)
\( b^2u + a^2v - (a^2 + b^2) = 0 \)
By the method of cross-multiplication, we have
\( \frac{u}{-1 \times -(a^2 + b^2) - a^2 \times 0} = \frac{-v}{1 \times -(a^2 + b^2) - b^2 \times 0} = \frac{1}{1 \times a^2 - b^2 \times (-1)} \)

\( \implies \) \( \frac{u}{a^2 + b^2} = \frac{-v}{-(a^2 + b^2)} = \frac{1}{a^2 + b^2} \)

\( \implies \) \( \frac{u}{a^2 + b^2} = \frac{1}{a^2 + b^2} \implies u = 1 \)
and

\( \implies \) \( \frac{-v}{-(a^2 + b^2)} = \frac{1}{a^2 + b^2} \implies v = 1 \)
Now, \( u = \frac{a}{x} = 1 \implies x = a \)
\( v = \frac{b}{y} = 1 \implies y = b \)
Hence, the solution of the given system of equations is \( x = a, y = b \). Ans.

Question. Solve the following system of equations by cross-multiplication method.
\( 2x + 3y + 8 = 0 \)
\( 4x + 5y + 14 = 0 \)
Answer: The given system of equations is
\( 2x + 3y + 8 = 0 \)
\( 4x + 5y + 14 = 0 \)
By cross-multiplication, we get
\( \frac{x}{3 \times 14 - 5 \times 8} = \frac{-y}{2 \times 14 - 4 \times 8} = \frac{1}{2 \times 5 - 4 \times 3} \)

\( \implies \) \( \frac{x}{42 - 40} = \frac{-y}{28 - 32} = \frac{1}{10 - 12} \)

\( \implies \) \( \frac{x}{2} = \frac{-y}{-4} = \frac{1}{-2} \)

\( \implies \) \( \frac{x}{2} = -\frac{1}{2} \implies x = -1 \)
and

\( \implies \) \( \frac{-y}{-4} = \frac{1}{-2} \implies y = -2 \)
Hence, the solution is \( x = -1, y = -2 \). Ans.

EXERCISE 

TYPE - 1: Substitution Method

Question. Solve for 'x' and 'y' by using method of substitution: \( 3x - 4y = 10 \); \( 4x + 3y = 5 \)
Answer: \( x = 2, y = -1 \)

Question. Solve for 'x' and 'y' by using method of substitution: \( \frac{x}{2} + y = 0.8 \); \( x + \frac{y}{2} = \frac{7}{10} \)
Answer: \( x = 0.4, y = 0.6 \)

Question. Solve for 'x' and 'y' by using method of substitution: \( x + y = a - b \); \( ax - by = a^2 + b^2 \)
Answer: \( x = a, y = -b \)

Question. Solve for 'x' and 'y' by using method of substitution: \( 0.2x + 0.3y = 1.3 \); \( 0.4x + 0.5y = 2.3 \)
Answer: \( x = 2, y = 3 \)

Question. Solve for 'x' and 'y' by using method of substitution: \( \sqrt{2}x + \sqrt{3}y = 0 \); \( \sqrt{3}x - \sqrt{8}y = 0 \)
Answer: \( x = 0, y = 0 \)

Question. Solve for 'x' and 'y' by using method of substitution: \( 6x + 5y = 7x + 3y + 1 = 2(x + 6y - 1) \)
Answer: \( x = 3, y = 2 \)

Question. Solve for 'x' and 'y' by using method of substitution: \( bx + ay = a + b \); \( ax \left[ \frac{1}{a-b} - \frac{1}{a+b} \right] + by \left[ \frac{1}{b-a} - \frac{1}{b+a} \right] = 2 \)
Answer: \( x = \frac{a}{b}, y = \frac{b}{a} \)

TYPE - 2: Elimination Method

Question. Solve the followings for x and y by using method of elimination: \( 8x - 3y = 13 \); \( 3x + 2y = 8 \)
Answer: \( x = 2, y = 1 \)

Question. Solve the followings for x and y by using method of elimination: \( x - y = 0.9 \); \( \frac{11}{x + y} = 2(x \neq -y) \)
Answer: \( x = 3.2, y = 2.3 \)

Question. Solve the followings for x and y by using method of elimination: \( \frac{2x}{a} + \frac{y}{b} = 2 \); \( \frac{x}{a} - \frac{y}{b} = 4 \)
Answer: \( x = 2a, y = -2b \)

Question. Solve the followings for x and y by using method of elimination: \( \frac{x}{3} + \frac{y}{4} = 4 \); \( \frac{5x}{6} - \frac{y}{8} = 4 \)
Answer: \( x = 6, y = 8 \)

Question. Solve the followings for x and y by using method of elimination: \( \frac{x}{a} + \frac{y}{b} = 2 \); \( ax - by = a^2 - b^2 \)
Answer: \( x = a, y = b \)

Question. Solve the followings for x and y by using method of elimination: \( (a + 2b)x + (2a - b)y = 2 \); \( (a - 2b)x + (2a + b)y = 3 \)
Answer: \( x = \frac{5b - 2a}{10ab}, y = \frac{a + 10b}{10ab} \)

Question. Solve the followings for x and y by using method of elimination: \( 2^x + 3^y = 17 \); \( 2^{x+2} - 3^{y+1} = 5 \)
Answer: \( x = 3, y = 2 \)

TYPE - 3: Cross-multiplication Method

Question. Solve for x and y by using method of cross multiplication: \( x + y = 3 \); \( 2x + y = -2 \)
Answer: \( x = -5, y = 8 \)

Question. Solve for x and y by using method of cross multiplication: \( \frac{x}{6} + \frac{y}{15} = 4 \); \( \frac{x}{3} - \frac{y}{12} = 4 \frac{3}{4} \)
Answer: \( x = 18, y = 15 \)

Question. Solve for x and y by using method of cross multiplication: \( 4x - 0.5y = 12.5 \); \( 3x + 0.8y = 8.2 \)
Answer: \( x = 3, y = -1 \)

Question. Solve for x and y by using method of cross multiplication: \( \frac{2}{x} + \frac{3}{y} = 2 \); \( \frac{1}{x} - \frac{1}{2y} = \frac{1}{3} \)
Answer: \( x = 2, y = 3 \)

Question. Solve for x and y by using method of cross multiplication: \( \frac{2x}{3} + \frac{3y}{5} = 17 \); \( \frac{3x}{4} + \frac{2y}{3} = 19 \)
Answer: \( x = 12, y = 15 \)

Question. Solve for x and y by using method of cross multiplication: \( \frac{2}{x-1} + \frac{3}{y+1} = 2 \); \( \frac{3}{x-1} + \frac{2}{y+1} = \frac{13}{6} \); \( x \neq 1, y \neq -1 \)
Answer: \( x = 3, y = 2 \)

Question. Solve for x and y by using method of cross multiplication: \( ax + by = a^2 \); \( bx + ay = b^2 \)
Answer: \( x = \frac{a^2 + ab + b^2}{a+b}, y = \frac{-ab}{a+b} \)

Question. Solve for x and y by using method of cross multiplication: \( \frac{5}{x+y} - \frac{2}{x-y} = -1 \); \( \frac{15}{x+y} + \frac{7}{x-y} = 10 \); \( x+y \neq 0, x-y \neq 0 \)
Answer: \( x = 3, y = 2 \)

Question. Solve for x and y by using method of cross multiplication: \( x - y = a + b \); \( ax + by = a^2 - b^2 \)
Answer: \( x = a, y = -b \)

Question. Solve for x and y by using method of cross multiplication: \( \frac{x}{a} + \frac{y}{b} = 2 \); \( ax - by = a^2 - b^2 \); \( a \neq 0, b \neq 0 \)
Answer: \( x = a, y = b \)

Question. Solve for x and y by using method of cross multiplication: \( ax - ay = 2 \); \( (a - 1)x + (a + 1)y = 2(a^2 + 1) \)
Answer: \( x = \frac{a^3 + 2a + 1}{a^2}, y = \frac{a^3 + 1}{a^2} \)