CBSE Class 10 Pair of Linear Equations in Two Variables Sure Shot Questions Set F

Question. Solve the following system of equations by eliminating \( x \) (by substitution) :
\( x + y = 7 \)
\( 2x - 3y = 11 \)
Answer: We have ;
\( x + y = 7 \) ....(1)
\( 2x - 3y = 11 \) ....(2)
We shall eliminate \( x \) by substituting its value from one equation into the other. From equation (1), we get ;
\( x + y = 7 \Rightarrow x = 7 - y \)
Substituting the value of \( x \) in equation (2), we get ;
\( 2 \times (7 - y) - 3y = 11 \)
\( \Rightarrow 14 - 2y - 3y = 11 \)
\( \Rightarrow -5y = -3 \) or, \( y = 3/5 \).
Now, substituting the value of \( y \) in equation (1), we get;
\( x + 3/5 = 7 \Rightarrow x = 32/5 \).
Hence, \( x = 32/5 \) and \( y = 3/5 \).

Question. Solve the following system of equations by eliminating \( x \) (by substitution) :
\( x + y = 7 \)
\( 12x + 5y = 7 \)
Answer: We have,
\( x + y = 7 \) ....(1)
\( 12x + 5y = 7 \) ....(2)
From equation (1), we have;
\( x + y = 7 \Rightarrow x = 7 - y \)
Substituting the value of \( y \) in equation (2), we get ;
\( \Rightarrow 12(7 - y) + 5y = 7 \)
\( \Rightarrow 84 - 12y + 5y = 7 \)
\( \Rightarrow -7y = -77 \Rightarrow y = 11 \)
Now, Substituting the value of \( y \) in equation (1), we get ;
\( x + 11 = 7 \Rightarrow x = -4 \)
Hence, \( x = -4 \), \( y = 11 \).

Question. Solve the following system of equations by eliminating \( x \) (by substitution) :
\( 2x - 7y = 1 \)
\( 4x + 3y = 15 \)
Answer: We have;
\( 2x - 7y = 1 \) ....(1)
\( 4x + 3y = 15 \) ....(2)
From equation (1), we get
\( 2x - 7y = 1 \Rightarrow x = \frac{7y + 1}{2} \)
Substituting the value of \( x \) in equation (2), we get ;
\( \Rightarrow 4 \times \frac{7y + 1}{2} + 3y = 15 \)
\( \Rightarrow \frac{28y + 4}{2} + 3y = 15 \)
\( \Rightarrow 28y + 4 + 6y = 30 \)
\( \Rightarrow 34y = 26 \Rightarrow y = \frac{26}{34} = \frac{13}{17} \)
Now, substituting the value of \( y \) in equation (1), we get;
\( 2x - 7 \times \frac{13}{17} = 1 \)
\( \Rightarrow 2x = 1 + \frac{91}{17} = \frac{108}{17} \Rightarrow x = \frac{108}{34} = \frac{54}{17} \)
Hence, \( x = \frac{54}{17} \), \( y = \frac{13}{17} \).

Question. Solve the following system of equations by eliminating \( x \) (by substitution) :
\( 3x - 5y = 1 \)
\( 5x + 2y = 19 \)
Answer: We have ;
\( 3x - 5y = 1 \) .... (1)
\( 5x + 2y = 19 \) .... (2)
From equation (1), we get;
\( 3x - 5y = 1 \Rightarrow x = \frac{5y + 1}{3} \)
Substituting the value of \( x \) in equation (2), we get ;
\( \Rightarrow 5 \times \frac{5y + 1}{3} + 2y = 19 \)
\( \Rightarrow 25y + 5 + 6y = 57 \Rightarrow 31y = 52 \)
Thus, \( y = \frac{52}{31} \)
Now, substituting the value of \( y \) in equation (1), we get ;
\( 3x - 5 \times \frac{52}{31} = 1 \Rightarrow 3x - \frac{260}{31} = 1 \Rightarrow 3x = \frac{291}{31} \)
\( \Rightarrow x = \frac{291}{31 \times 3} = \frac{97}{31} \)
Hence, \( x = \frac{97}{31} \), \( y = \frac{52}{31} \).

Question. Solve the following system of equations by eliminating \( x \) (by substitution) :
\( 5x + 8y = 9 \)
\( 2x + 3y = 4 \)
Answer: We have,
\( 5x + 8y = 9 \) ....(1)
\( 2x + 3y = 4 \) ....(2)
From equation (1), we get;
\( 5x + 8y = 9 \Rightarrow x = \frac{9 - 8y}{5} \)
Substituting the value of \( x \) in equation (2), we get ;
\( \Rightarrow 2 \times \frac{9 - 8y}{5} + 3y = 4 \)
\( \Rightarrow 18 - 16y + 15y = 20 \)
\( \Rightarrow -y = 2 \) or \( y = -2 \)
Now, substituting the value of \( y \) in equation (1), we get ;
\( 5x + 8 (-2) = 9 \Rightarrow 5x = 25 \Rightarrow x = 5 \)
Hence, \( x = 5 \), \( y = -2 \).

Question. Solve the following system of equations by eliminating 'y' (by substitution) :
\( 3x - y = 3 \)
\( 7x + 2y = 20 \)
Answer: We have;
\( 3x - y = 3 \) ....(1)
\( 7x + 2y = 20 \) ....(2)
From equation (1), we get ;
\( 3x - y = 3 \Rightarrow y = 3x - 3 \)
Substituting the value of '\( y \)' in equation (2), we get ;
\( \Rightarrow 7x + 2 \times (3x - 3) = 20 \)
\( \Rightarrow 7x + 6x - 6 = 20 \Rightarrow 13x = 26 \Rightarrow x = 2 \)
Now, substituting \( x = 2 \) in equation (1), we get;
\( 3 \times 2 - y = 3 \Rightarrow y = 3 \)
Hence, \( x = 2 \), \( y = 3 \).

Question. Solve the following system of equations by eliminating 'y' (by substitution) :
\( 7x + 11y - 3 = 0 \)
\( 8x + y - 15 = 0 \)
Answer: We have;
\( 7x + 11y - 3 = 0 \) ....(1)
\( 8x + y - 15 = 0 \) .....(2)
From equation (1), we get;
\( 7x + 11y = 3 \Rightarrow y = \frac{3 - 7x}{11} \)
Substituting the value of '\( y \)' in equation (2), we get;
\( \Rightarrow 8x + \frac{3 - 7x}{11} = 15 \)
\( \Rightarrow 88x + 3 - 7x = 165 \Rightarrow 81x = 162 \Rightarrow x = 2 \)
Now, substituting, \( x = 2 \) in the equation (2), we get ;
\( 8 \times 2 + y = 15 \Rightarrow y = -1 \)
Hence, \( x = 2 \), \( y = -1 \).

Question. Solve the following system of equations by eliminating 'y' (by substitution) :
\( 2x + y - 17 = 0 \)
\( 17x - 11y - 8 = 0 \)
Answer: We have,
\( 2x + y = 17 \) ....(1)
\( 17x - 11y = 8 \) ....(2)
From equation (1), we get;
\( 2x + y = 17 \Rightarrow y = 17 - 2x \)
Substituting the value of '\( y \)' in equation (2), we get ;
\( 17x - 11 (17 - 2x) = 8 \)
\( \Rightarrow 17x - 187 + 22x = 8 \Rightarrow 39x = 195 \Rightarrow x = 5 \)
Now, substituting the value of '\( x \)' in equation (1), we get ;
\( 2 \times 5 + y = 17 \Rightarrow y = 7 \)
Hence, \( x = 5 \), \( y = 7 \).

Question. Solve the following systems of equations,
\( \frac{15}{u} + \frac{2}{v} = 17 \)
\( \frac{1}{u} + \frac{1}{v} = \frac{36}{5} \)
Answer: The given system of equation is ;
\( \frac{15}{u} + \frac{2}{v} = 17 \) ....(1)
\( \frac{1}{u} + \frac{1}{v} = \frac{36}{5} \) ....(2)
Considering \( 1/u = x \), \( 1/v = y \), the above system of linear equations can be written as :
\( 15x + 2y = 17 \) ....(3)
\( x + y = \frac{36}{5} \) ....(4)
Multiplying (4) by 15 and (3) by 1, we get ;
\( 15x + 2y = 17 \) ....(5)
\( 15x + 15y = \frac{36}{5} \times 15 = 108 \) ....(6)
Subtracting (6) form (5), we get;
\( -13y = -91 \Rightarrow y = 7 \)
Substituting \( y = 7 \) in (4), we get ;
\( x + 7 = \frac{36}{5} \Rightarrow x = \frac{36}{5} - 7 = \frac{1}{5} \)
But, \( y = \frac{1}{v} = 7 \Rightarrow v = \frac{1}{7} \) and, \( x = \frac{1}{u} = \frac{1}{5} \Rightarrow u = 5 \)
Hence, the required solution of the given system is \( u = 5 \), \( v = 1/7 \).

Question. Solve the following systems of equations,
\( \frac{11}{v} - \frac{7}{u} = 1 \)
\( \frac{9}{v} - \frac{4}{u} = 6 \)
Answer: The given system of equation is ;
\( \frac{11}{v} - \frac{7}{u} = 1 \); \( \frac{9}{v} - \frac{4}{u} = 6 \)
Taking \( 1/v = x \) and \( 1/u = y \), the above system of equations can be written as ;
\( 11x - 7y = 1 \) ....(1)
\( 9x - 4y = 6 \) ....(2)
Multiplying (1) by 4 and (2) by 7, we get,
\( 44x - 28y = 4 \) ....(3)
\( 63x - 28y = 42 \) ....(4)
Subtracting (4) from (3) we get,
\( -19x = -38 \Rightarrow x = 2 \)
Substituting the above value of \( x \) in (2), we get;
\( 9 \times 2 - 4y = 6 \Rightarrow -4y = -12 \Rightarrow y = 3 \)
But, \( x = \frac{1}{v} = 2 \Rightarrow v = \frac{1}{2} \) and, \( y = \frac{1}{u} = 3 \Rightarrow u = \frac{1}{3} \)
Hence, the required solution of the given system of the equation is ; \( v = 1/2 \), \( u = 1/3 \).

Question. Solve the following system of equations by the method of elimination (substitution).
\( (a + b)x + (a - b)y = a^2 + b^2 \)
\( (a - b)x + (a + b)y = a^2 + b^2 \)
Answer: The given system of equations is
\( (a + b)x + (a - b)y = a^2 + b^2 \) ....(1)
\( (a - b)x + (a + b)y = a^2 + b^2 \) ....(2)
From (2), we get \( (a + b)y = a^2 + b^2 - (a - b)x \Rightarrow y = \frac{a^2 + b^2}{a + b} - \frac{a - b}{a + b}x \) ....(3)
Substituting \( y = \frac{a^2 + b^2}{a + b} - \frac{a - b}{a + b}x \) in (1), we get
\( (a + b)x + (a - b)[\frac{a^2 + b^2}{a + b} - \frac{a - b}{a + b}x] = a^2 + b^2 \)
After simplification (steps omitted for brevity), we get \( 4abx = 2b(a^2 + b^2) \Rightarrow x = \frac{a^2 + b^2}{2a} \)
Putting \( x = \frac{a^2 + b^2}{2a} \) in (3), we get \( y = \frac{a^2 + b^2}{2a} \).
Hence, the solution is \( x = \frac{a^2 + b^2}{2a} \), \( y = \frac{a^2 + b^2}{2a} \).

Question. Solve \( 2x + 3y = 11 \) and \( 2x - 4y = -24 \) and hence find the value of ‘m’ for which \( y = mx + 3 \).
Answer: We have,
\( 2x + 3y = 11 \) ....(1)
\( 2x - 4y = -24 \) ....(2)
From (1), we have \( 2x = 11 - 3y \). Substituting \( 2x = 11 - 3y \) in (2), we get
\( 11 - 3y - 4y = -24 \Rightarrow -7y = -35 \Rightarrow y = 5 \)
Putting \( y = 5 \) in (1), we get
\( 2x + 3 \times 5 = 11 \Rightarrow 2x = 11 - 15 \Rightarrow x = -2 \)
Hence, \( x = -2 \) and \( y = 5 \). Again putting \( x = -2 \) and \( y = 5 \) in \( y = mx + 3 \), we get
\( 5 = m(-2) + 3 \Rightarrow -2m = 2 \Rightarrow m = -1 \).

METHOD OF ELIMINATION BY EQUATING THE COEFFICIENTS

• Step I : Let the two equations obtained be \( a_1x + b_1y + c_1 = 0 \) ....(1) and \( a_2x + b_2y + c_2 = 0 \) ....(2).
• Step II : Multiplying the given equation so as to make the co-efficients of the variable to be eliminated equal.
• Step III : Add or subtract the equations so obtained in Step II, as the terms having the same co-efficients may be either of opposite or the same sign.
• Step IV : Solve the equations in one variable so obtained in Step III.
• Step V : Substitute the value found in Step IV in any one of the given equations and then compute the value of the other variable.

Question. Solve the following system of linear equations by applying the method of elimination by equating the co-efficients :
\( 4x - 3y = 4 \)
\( 2x + 4y = 3 \)
Answer: We have,
\( 4x - 3y = 4 \) ....(1)
\( 2x + 4y = 3 \) ....(2)
Multiplying equation (1) with 1 and equation (2) with 2, we get ;
\( 4x - 3y = 4 \) ....(3)
\( 4x + 8y = 6 \) ....(4)
Subtracting equation (4) from (3), we get ;
\( -11y = -2 \Rightarrow y = 2/11 \)
Substituting \( y = 2/11 \) in equation (1), we get;
\( 4x - 3 \times \frac{2}{11} = 4 \Rightarrow 4x = 4 + \frac{6}{11} \Rightarrow 4x = \frac{50}{11} \Rightarrow x = \frac{25}{22} \).
Hence, solution is : \( x = \frac{25}{22} \), \( y = \frac{2}{11} \).

Question. Solve the following system of linear equations by applying the method of elimination by equating the co-efficients :
\( 5x - 6y = 8 \)
\( 3x + 2y = 6 \)
Answer: We have;
\( 5x - 6y = 8 \) ....(1)
\( 3x + 2y = 6 \) ....(2)
Multiplying both sides of equation (1) with 1 and equation (2) with 3, we get ;
\( 5x - 6y = 8 \) ....(3)
\( 9x + 6y = 18 \) ....(4)
Adding equation (3) and (4), we get ;
\( 14x = 26 \Rightarrow x = 13/7 \)
Putting \( x = 13/7 \) in equation (1), we get ;
\( 5 \times \frac{13}{7} - 6y = 8 \Rightarrow \frac{65}{7} - 8 = 6y \Rightarrow \frac{9}{7} = 6y \Rightarrow y = \frac{3}{14} \).
Hence, the solution is \( x = 13/7 \), \( y = 3/14 \).

Question. Solve the following system of equations by using the method of elimination by equating the co-efficients.
\( \frac{x}{2} + \frac{2y}{5} + 2 = 10 \); \( \frac{2x}{7} - \frac{y}{2} + 1 = 9 \)
Answer: The given system of equations simplifies to:
\( 5x + 4y = 80 \) ....(3)
\( 4x - 7y = 112 \) ....(4)
Multiplying (3) by 4 and (4) by 5, we get:
\( 20x + 16y = 320 \) ....(7)
\( 20x - 35y = 560 \) ....(8)
Subtracting (7) from (8), we get: \( y = -240/51 = -80/17 \).
Substituting \( y \) in (3), we get: \( 5x + 4(-80/17) = 80 \Rightarrow 5x = 80 + 320/17 \Rightarrow x = 336/17 \).
Hence, solution is \( x = 336/17 \), \( y = -80/17 \).

Question. Solve the following system of linear equations by usnig the method of elimination by equating the coefficients :
\( 3x + 4y = 25 \) ; \( 5x - 6y = -9 \)
Answer: The given system of equations is
\( 3x + 4y = 25 \) ....(1)
\( 5x - 6y = -9 \) ....(2)
Multiplying equation (1) by 3 and equation (2) by 2, we get
\( 9x + 12y = 75 \) ....(3)
\( 10x - 12y = -18 \) ....(4)
Adding equation (3) and equation (4), we get \( 19x = 57 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (1), we get, \( 3 \times 3 + 4y = 25 \Rightarrow 4y = 16 \Rightarrow y = 4 \).
Hence, the solution is \( x = 3, y = 4 \).

Question. Solve the following system of equations :
\( 15x + 4y = 61 \); \( 4x + 15y = 72 \)
Answer: The given system of equation is
\( 15x + 4y = 61 \) ....(1)
\( 4x + 15y = 72 \) ....(2)
Multiplying (1) by 15 and (2) by 4, we get
\( 225x + 60y = 915 \) ....(3)
\( 16x + 60y = 288 \) ....(4)
Substracting (4) from (3), we get \( 209x = 627 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (1), we get \( 15 \times 3 + 4y = 61 \Rightarrow 4y = 16 \Rightarrow y = 4 \).
Hence, the solution is \( x = 3, y = 4 \).

Question. Solve the following system of linear equatoins by using the method of elimination by equating the coefficients
\( \sqrt{3}x - \sqrt{2}y = \sqrt{3} \) ; \( \sqrt{5}x + \sqrt{3}y = \sqrt{2} \)
Answer: Multiplying eq (1) by \( \sqrt{3} \) and (2) by \( \sqrt{2} \), we get:
\( 3x - \sqrt{6}y = 3 \) ....(3)
\( \sqrt{10}x + \sqrt{6}y = 2 \) ....(4)
Adding: \( (3 + \sqrt{10})x = 5 \Rightarrow x = 5(\sqrt{10} - 3) \).
Substituting \( x \) back, we get \( y = 5\sqrt{15} - 8\sqrt{6} \).

Question. Solve for \( x \) and \( y \) :
\( \frac{ax}{b} - \frac{by}{a} = a + b \); \( ax - by = 2ab \)
Answer: From eq (2), \( x - by/a = 2b \) ....(3). Subtracting (3) from (1): \( x(a/b - 1) = a - b \Rightarrow x = b \). Substituting \( x = b \) in (3): \( b - by/a = 2b \Rightarrow y = -a \). Hence \( x = b, y = -a \).

Question. Solve the following system of linear equations :
\( 2(ax - by) + (a + 4b) = 0 \)
\( 2(bx + ay) + (b - 4a) = 0 \)
Answer: We get \( x = -1/2 \) and \( y = 2 \).

Question. Solve \( (a - b)x + (a + b)y = a^2 - 2ab - b^2 \)
\( (a + b)(x + y) = a^2 + b^2 \)
Answer: Solving the equations, we get \( x = a + b \) and \( y = -2ab/(a + b) \).

Type II : Solving a system of equations which is reducible to a system of simultaneous linear equations

Question. Solve the following system of equations
\( \frac{1}{2x} - \frac{1}{y} = -1 \); \( \frac{1}{x} + \frac{1}{2y} = 8 \)
Answer: Let \( 1/x = u \) and \( 1/y = v \). Equations become \( u/2 - v = -1 \) and \( u + v/2 = 8 \). Solving, we find \( u = 6 \) and \( v = 4 \). Thus \( x = 1/6, y = 1/4 \).

Question. Solve \( \frac{2}{x} + \frac{1}{3y} = \frac{1}{5} \); \( \frac{3}{x} + \frac{2}{3y} = 2 \) and also find ‘a’ for which \( y = ax - 2 \).
Answer: Let \( 1/x = u, 1/y = v \). Solving the equations gives \( u = -8/5, v = 51/5 \). Thus \( x = -5/8, y = 5/51 \). Putting these in \( y = ax - 2 \), we find \( a = -856/255 \).

Question. Solve \( \frac{2}{x + 2y} + \frac{6}{2x - y} = 4 \); \( \frac{5}{2(x + 2y)} + \frac{1}{3(2x - y)} = 1 \)
Answer: Let \( 1/(x + 2y) = u \) and \( 1/(2x - y) = v \). Solving gives \( u = 14/43, v = 144/43 \). This leads to \( 14x + 28y = 43 \) and \( 288x - 144y = 43 \). Solving these gives \( x \approx 1.3, y \approx 1.6 \).

Question. Solve \( \frac{1}{x + y} + \frac{2}{x - y} = 2 \); \( \frac{2}{x + y} - \frac{1}{x - y} = 3 \)
Answer: Let \( 1/(x + y) = u, 1/(x - y) = v \). Solving gives \( u = 8/5, v = 1/5 \). This yields \( 8x + 8y = 5 \) and \( x - y = 5 \). Solving these gives \( x = 45/16, y = -35/16 \).

Type-III : Equation of the form, \( ax + by = c \) and \( bx + ay = d \) where \( a \neq b \).

Steps:

• Step I : Let us write the equations in the form \( ax + by = c \) and \( bx + ay = d \).
• Step II : Adding and subtracting the above type of two equations, we find \( x + y = \frac{c + d}{a + b} \) and \( x - y = \frac{c - d}{a - b} \).
• Step III : We get the values of \( x \) and \( y \) after adding or subtracting the simplified equations.

Question. Solve the following equations.
\( 156x + 112y = 580 \); \( 112x + 156y = 492 \)
Answer: Adding gives \( x + y = 4 \). Subtracting gives \( x - y = 2 \). Solving these yields \( x = 3, y = 1 \).

Question. Solve the following system of equations.
\( 43x + 35y = 207 \); \( 35x + 43y = 183 \)
Answer: Adding gives \( x + y = 5 \). Subtracting gives \( x - y = 3 \). Solving these yields \( x = 4, y = 1 \).

Question. Solve the following system of equations.
\( 43x + 35y = 207; 35x + 43y = 183 \)
Answer: The given system of equations is ;
\( 43x + 35y = 207 \) ....(1)
\( 35x + 43y = 183 \) ....(2)
Adding equation (1) and (2), we get;
\( 78x + 78y = 390 \)
\( \Rightarrow 78(x + y) = 390 \)
\( \Rightarrow x + y = 5 \) ....(3)
Subtracting equation (2) from the equation (1), we get ;
\( 8x - 8y = 24 \)
\( \Rightarrow x - y = 3 \) ....(4)
Adding equation (3) and (4), we get;
\( 2x = 8 \Rightarrow x = 4 \)
Putting \( x = 4 \) in equation (3), we get;
\( 4 + y = 5 \Rightarrow y = 1 \)
Hence, the solution of the system of equation is ; \( x = 4, y = 1 \).

Type IV : Equation of the form, \( a_1x + b_1y + c_1z = d_1 \), \( a_2x + b_2y + c_2z = d_2 \), \( a_3x + b_3y + c_3z = d_3 \)

We may use the following method to solve the above type of equations.

Steps :

• Step I : Consider any one of the three given equations.
• Step II : Find the value of one of the variable, say \( z \), from it.
• Step III : Substitute the value of \( z \) found in Step II in the other two equations to get two linear equatons in \( x, y \).
• Step IV : Taking the help of elimination method, solve the equations in \( x, y \) obtained in Step III.
• Step V : Substitute the values of \( x, y \) found in Step IV and Step II to get the value of \( z \).

Question. Solve the following system of equations.
\( x - z = 5 \)
\( y + z = 3 \)
\( x - y = 2 \)
Answer: The given system of equations to ;
\( x - z = 5 \) ....(1)
\( y + z = 3 \) ....(2)
\( x - y = 2 \) .....(3)
From equation (1), we have;
\( z = x - 5 \)
Putting \( z = x - 5 \) in equation (2), we get ;
\( y + x - 5 = 3 \)
\( \Rightarrow x + y = 8 \) ....(4)
Adding equations (3) and (4), we get;
\( 2x = 10 \)
\( \Rightarrow x = 5 \)
Again putting \( x = 5 \) in equation (1), we get;
\( 5 - z = 5 \)
\( \Rightarrow z = 0 \)
Hence, the solution of the given system of equation is \( x = 5, y = 3, z = 0 \).
Other method :
adding (1), (2) & (3)
\( 2x = 10 \)
\( x = 5 \)
now put this value in equation (1) & (3), we get \( z = 0, y = 3 \) respectively

Question. Solve,
\( x + 2y + z = 12 \)
\( 2x - z = 4 \)
\( x - 2y = 4 \)
Answer: We have,
\( x + 2y + z = 12 \) ....(1)
\( 2x - z = 4 \) ....(2)
\( x - 2y = 4 \) ....(3)
From equation (1), we have \( z = 12 - x - 2y \).
Putting, \( z = 12 - x - 2y \) in the equation (2), we get;
\( 2x - (12 - x - 2y) = 4 \)
\( \Rightarrow 2x - 12 + x + 2y = 4 \)
\( \Rightarrow 3x + 2y = 16 \) ....(4)
Adding equations (3) and (4), we get;
\( 4x = 20 \)
\( \Rightarrow x = 5 \)
Putting the value of \( x = 5 \) in equation (2), we get
\( 2 \times 5 - z = 4 \)
\( \Rightarrow z = 10 - 4 = 6 \)
Again putting the value of \( x = 5 \) in equation (3), we get
\( 5 - 2y = 4 \Rightarrow y = 1/2 \)
Hence, the solution of the given system of equations is ; \( x = 5, y = 1/2, z = 6 \)

CROSS-MULTIPLICATION METHOD

By the method of elimination by substitution, only those equations can be solved, which have unique solution. But the method of cross multiplication discussed below is applicable in all the cases; whether the system has a unique solution, no solution or infinitely many solutions.
Let us solve the following system of equations
\( a_1x + b_1y + c_1 = 0 \) ....(1)
\( a_2x + b_2y + c_2 = 0 \) ....(2)
Multiplying equation (1) by \( b_2 \) and equation (2) by \( b_1 \), we get
\( a_1b_2x + b_1b_2y + b_2c_1 = 0 \) ....(3)
\( a_2b_1x + b_1b_2y + b_1c_2 = 0 \) ....(4)
Subtracting equation (4) from equation (3), we get
\( (a_1b_2 - a_2b_1) x + (b_2c_1 - b_1c_2) = 0 \)
\( \Rightarrow x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) \( \left[ a_1b_2 - a_2b_1 \neq 0, \text{ i.e., } \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \right] \)
Similarly, \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)
These values of \( x \) and \( y \) can also be written as
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{-y}{a_1c_2 - a_2c_1} = \frac{1}{a_1b_2 - a_2b_1} \)

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
\( \frac{x}{3 \times 14 - 5 \times 8} = \frac{-y}{2 \times 14 - 4 \times 8} = \frac{1}{2 \times 5 - 4 \times 3} \)
\( \Rightarrow \frac{x}{42 - 40} = \frac{-y}{28 - 32} = \frac{1}{10 - 12} \)
\( \Rightarrow \frac{x}{2} = \frac{-y}{-4} = \frac{1}{-2} \)
\( \Rightarrow \frac{x}{2} = -\frac{1}{2} \Rightarrow x = -1 \)
and \( \frac{-y}{-4} = -\frac{1}{2} \Rightarrow y = -2 \).
Hence, the solution is \( x = -1, y = -2 \)

Question. Solve the follownig system of equations by the method of cross-multiplication.
\( 2x - 6y + 10 = 0 \)
\( 3x - 7y + 13 = 0 \)
Answer: The given system of equations is
\( 2x - 6y + 10 = 0 \) ....(1)
\( 3x - 7y + 13 = 0 \) ....(2)
By cross-multiplication, we have
\( \frac{x}{-6 \times 13 - (-7) \times 10} = \frac{-y}{2 \times 13 - 3 \times 10} = \frac{1}{2 \times (-7) - 3 \times (-6)} \)
\( \Rightarrow \frac{x}{-78 + 70} = \frac{-y}{26 - 30} = \frac{1}{-14 + 18} \)
\( \Rightarrow \frac{x}{-8} = \frac{-y}{-4} = \frac{1}{4} \)
\( \Rightarrow \frac{x}{-8} = \frac{1}{4} \Rightarrow x = -2 \)
\( \Rightarrow \frac{-y}{-4} = \frac{1}{4} \Rightarrow y = 1 \)
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
\( \frac{x}{15 \times (-20) - (-2) \times 23} = \frac{-y}{11 \times (-20) - 7 \times 23} = \frac{1}{11 \times (-2) - 7 \times 15} \)
\( \Rightarrow \frac{x}{-300 + 46} = \frac{-y}{-220 - 161} = \frac{1}{-22 - 105} \)
\( \Rightarrow \frac{x}{-254} = \frac{-y}{-381} = \frac{1}{-127} \)
\( \Rightarrow \frac{x}{-254} = \frac{1}{-127} \Rightarrow x = 2 \)
and \( \frac{-y}{-381} = \frac{1}{-127} \Rightarrow y = -3 \)
Hence, \( x = 2, y = -3 \) is the required solution.

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
\( \frac{x}{b \times \{-(a + b)\} - (-a) \times \{-(a - b)\}} = \frac{-y}{a \times \{-(a + b)\} - b \times \{-(a - b)\}} = \frac{1}{a \times (-a) - b \times b} \)
\( \Rightarrow \frac{x}{-ab - b^2 - a^2 + ab} = \frac{-y}{-a^2 - ab + ab - b^2} = \frac{1}{-a^2 - b^2} \)
\( \Rightarrow \frac{x}{-(a^2 + b^2)} = \frac{-y}{-(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \)
\( \Rightarrow \frac{x}{-(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \Rightarrow x = 1 \)
and \( \frac{-y}{-(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \Rightarrow y = -1 \)
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
\( \frac{x}{1 \times \{-(a^2 + b^2)\} - (-b) \times \{-(a - b)\}} = \frac{-y}{1 \times \{-(a^2 + b^2)\} - a \times \{-(a - b)\}} = \frac{1}{1 \times (-b) - a \times 1} \)
\( \Rightarrow \frac{x}{-(a^2 + b^2) - b(a - b)} = \frac{-y}{-(a^2 + b^2) + a(a - b)} = \frac{1}{-b - a} \)
\( \Rightarrow \frac{x}{-a^2 - b^2 - ab + b^2} = \frac{-y}{-a^2 - b^2 + a^2 - ab} = \frac{1}{-(a + b)} \)
\( \Rightarrow \frac{x}{-a(a + b)} = \frac{-y}{-b(a + b)} = \frac{1}{-(a + b)} \)
\( \Rightarrow \frac{x}{-a(a + b)} = \frac{1}{-(a + b)} \Rightarrow x = a \)
and \( \frac{-y}{-b(a + b)} = \frac{1}{-(a + b)} \Rightarrow y = -b \)
Hence, the solution is \( x = a, y = -b \).

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 \( a^2b^2 \), we get
\( b^2x + a^2y - 2a^2b^2 = 0 \) ....(4)
By cross multiplication method, we have
\( \frac{x}{a \times (-2a^2b^2) - a^2 \times \{-ab(a + b)\}} = \frac{-y}{b \times (-2a^2b^2) - b^2 \times \{-ab(a + b)\}} = \frac{1}{b \times a^2 - b^2 \times a} \)
\( \Rightarrow \frac{x}{-2a^3b^2 + a^3b(a + b)} = \frac{-y}{-2a^2b^3 + ab^3(a + b)} = \frac{1}{a^2b - ab^2} \)
\( \Rightarrow \frac{x}{-2a^3b^2 + a^4b + a^3b^2} = \frac{-y}{-2a^2b^3 + a^2b^3 + ab^4} = \frac{1}{ab(a - b)} \)
\( \Rightarrow \frac{x}{a^4b - a^3b^2} = \frac{-y}{ab^4 - a^2b^3} = \frac{1}{ab(a - b)} \)
\( \Rightarrow \frac{x}{a^3b(a - b)} = \frac{y}{ab^3(a - b)} = \frac{1}{ab(a - b)} \)
\( \Rightarrow x = \frac{a^3b(a - b)}{ab(a - b)} = a^2 \)
And \( y = \frac{ab^3(a - b)}{ab(a - b)} = b^2 \)
Hence, the solution \( x = a^2, y = b^2 \)

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)
\( bx + ay = \frac{(a + b)^2}{a^2 + b^2} - 1 \)
\( \Rightarrow bx + ay = \frac{a^2 + 2ab + b^2 - a^2 - b^2}{a^2 + b^2} \)
\( \Rightarrow bx + ay = \frac{2ab}{a^2 + b^2} \)
\( \Rightarrow bx + ay - \frac{2ab}{a^2 + b^2} = 0 \) ....(2)
Now, by cross-multiplication method, we have
\( \frac{x}{b \times \left(-\frac{2ab}{a^2 + b^2}\right) - a \times (-1)} = \frac{-y}{a \times \left(-\frac{2ab}{a^2 + b^2}\right) - b \times (-1)} = \frac{1}{a \times a - b \times b} \)
\( \Rightarrow \frac{x}{-\frac{2ab^2}{a^2 + b^2} + a} = \frac{-y}{-\frac{2a^2b}{a^2 + b^2} + b} = \frac{1}{a^2 - b^2} \)
\( \Rightarrow \frac{x}{\frac{-2ab^2 + a^3 + ab^2}{a^2 + b^2}} = \frac{-y}{\frac{-2a^2b + a^2b + b^3}{a^2 + b^2}} = \frac{1}{a^2 - b^2} \)
\( \Rightarrow \frac{x}{\frac{a(a^2 - b^2)}{a^2 + b^2}} = \frac{1}{a^2 - b^2} \Rightarrow x = \frac{a}{a^2 + b^2} \)
and \( \frac{-y}{\frac{b(b^2 - a^2)}{a^2 + b^2}} = \frac{1}{a^2 - b^2} \Rightarrow y = \frac{b}{a^2 + b^2} \)
Hence, the solution is \( x = \frac{a}{a^2 + b^2} \), \( y = \frac{b}{a^2 + b^2} \)

Question. Solve the following system of equations in \( x \) and \( y \) by cross-multiplication method
\( (a - b) x + (a + b) y = a^2 - 2ab - b^2 \)
\( (a + b) (x + y) = a^2 + b^2 \)
Answer: The given system of equations can be rewritten as :
\( (a - b) x + (a + b) y - (a^2 - 2ab - b^2) = 0 \)
\( (a + b) x + (a + b) y - (a^2 + b^2) = 0 \)
By cross-multiplication method, we have
\( \frac{x}{(a + b) \{ -(a^2 + b^2) \} - (a + b) \{ -(a^2 - 2ab - b^2) \}} = \frac{-y}{(a - b) \{ -(a^2 + b^2) \} - (a + b) \{ -(a^2 - 2ab - b^2) \}} = \frac{1}{(a - b) (a + b) - (a + b) (a + b)} \)
\( \Rightarrow \frac{x}{(a + b) [ -(a^2 + b^2) + (a^2 - 2ab - b^2) ]} = \frac{-y}{(a - b) (a^2 + b^2) - (a + b) (a^2 - 2ab - b^2)} = \frac{1}{(a + b) ( a - b - a - b)} \)
\( \Rightarrow \frac{x}{(a + b) (-2ab - 2b^2)} = \frac{-y}{a^3 - a^2b - 3ab^2 - b^3 - a^3 + a^2b + a^2b + a^2b + b^3} = \frac{1}{(a + b) (-2b)} \)
\( \Rightarrow \frac{x}{-2(a + b) (a + b) b} = \frac{1}{-2b(a + b)} \Rightarrow x = a + b \)
and \( \frac{-y}{-4ab^2} = \frac{1}{-2b(a + b)} \Rightarrow y = -\frac{2ab}{a + b} \)
Hence, the solution of the given system of equations is \( x = a + b, y = -\frac{2ab}{a + b} \).

Question. Solve the following system of equations by cross-multiplications 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
\( (a + b) x + (a - b) y - (a^2 - ab + b^2) = 0 \) ....(1)
\( (a - b) x + (a + b) y - (a^2 + ab + b^2) = 0 \) ...(2)
Now, by cross-multiplication method, we have
\( \frac{x}{(a - b) \{ -(a^2 + ab + b^2) \} - (a + b) \{ -(a^2 - ab + b^2) \}} = \frac{-y}{(a + b) \{ -(a^2 + ab + b^2) \} - (a - b) \{ -(a^2 - ab + b^2) \}} = \frac{1}{(a + b) (a + b) - (a - b) (a - b)} \)
\( \Rightarrow \frac{x}{-(a^3 - b^3) + (a^3 + b^3)} = \frac{-y}{- (a^3 + 2a^2b + 2ab^2 + b^3) + (a^3 - 2a^2b + 2ab^2 - b^3)} = \frac{1}{a^2 + 2ab + b^2 - a^2 + 2ab - b^2} \)
\( \Rightarrow \frac{x}{2b^3} = \frac{-y}{-4a^2b - 2b^3} = \frac{1}{4ab} \)
\( \Rightarrow \frac{x}{2b^3} = \frac{1}{4ab} \Rightarrow x = \frac{b^2}{2a} \)
And \( \frac{-y}{-2b(2a^2 + b^2)} = \frac{1}{4ab} \Rightarrow y = \frac{2a^2 + b^2}{2a} \)
Hence, the solution is \( x = \frac{b^2}{2a}, y = \frac{2a^2 + b^2}{2a} \)

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 equatinos (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
\( \frac{u}{(-1)(-(a^2 + b^2)) - a^2 \times 0} = \frac{-v}{1 \times (-(a^2 + b^2)) - b^2 \times 0} = \frac{1}{1 \times a^2 - b^2 \times (-1)} \)
\( \Rightarrow \frac{u}{a^2 + b^2} = \frac{-v}{-(a^2 + b^2)} = \frac{1}{a^2 + b^2} \)
\( \Rightarrow \frac{u}{a^2 + b^2} = \frac{1}{a^2 + b^2} \Rightarrow u = 1 \)
and \( \frac{-v}{-(a^2 + b^2)} = \frac{1}{a^2 + b^2} \Rightarrow v = 1 \)
and \( u = \frac{a}{x} = 1 \Rightarrow x = a \)
\( v = \frac{b}{y} = 1 \Rightarrow y = b \)
Hence, the solution of the given system of equations is \( x = a, y = b \).