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Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.3 Question: 1 Solve the system of equations: 11x + 15y + 23 = 0 and 7x – 2y – 20 = 0 Solution: The given system of equation is 11x + 15y + 23 = 0 ……. (i) 7x - 2y - 20 = 0 ….. (ii) From (ii) 2y = 7x - 20 Substituting the value of y in equation (i) we get, 127x = 254 x = 2 Putting the value of x in the equation (iii) y = - 3 The value of x and y are 2 and -3 respectively. Question: 2 Solve the system of equations: 3x – 7y + 10 = 0, y – 2x – 3 = 0 Solution: The given system of equation is 3x - 7y + 10 = 0 …. (i) y - 2x - 3 = 0 ….. (ii) From (ii) y - 2x - 3 = 0 y = 2x + 3 …… (iii) Substituting the value of y in equation (i) we get, = 3x - 7(2x + 3) + 10 = 0 = 3x + 14x - 21 + 10 = 0 = -11x = 11 = x = -1 Putting the value of x in the equation (iii) = y = 2(- 1) + 3 y = 1 The value of x and y are -1 and 1 respectively. Question: 3 Solve the system of equations: 0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8 Solution: The given system of equation is 0.4x + 0.3y = 1.7 0.7x - 0.2y = 0.8 Multiplying both sides by 10 4x + 3y = 17 ….. (i) 7x - 2y = 8 …… (ii) From (ii) 7x - 2y = 8 Substituting the value of y in equation (i) we get, 32 + 29y = 119 29y = 87 y = 3 Putting the value of y in the equation (iii) x = 2 The value of x and y are 2 and 3 respectively. Question: 4 Solution: The given system of equation is Therefore x + 2y = 1.6 x + 2y = 1.6 7 = 10x + 5y Multiplying both sides by 10 10x + 20y = 16 ….. (i) 10x + 5y = 7 …… (ii) Subtracting two equations we get, 15y = 9 The value of x and y are 2/5 and 3/5 respectively. Question: 5 Solve the system of equations: 7(y + 3) - 2(x + 3) = 14 4(y - 2) + 3(x - 3) = 2 Solution: The given system of equation is 7(y + 3) - 2(x + 3) = 14 ……. (i) 4(y - 2) + 3(x - 3) = 2 ….. (ii) From (i) 7y + 21 - 2x - 4 = 14 7y = 14 + 4 - 21 + 2x From (ii) 4y - 8 + 3x - 9 = 2 4y + 3x - 17 - 2 = 0 4y + 3x - 19 = 0 ….. (iii) Substituting the value of y in equation (iii) 8x - 12 + 21x - 133 = 0 29x = 145 x = 5 Putting the value of x in the above equation y = 1 The value of x and y are 5 and 1 respectively. Question: 6 Solve the system of equations: Solution: The given system of equation is From (i) = 9x - 2y = 108 … (iii) Substituting the value of x in equation (iii) we get, 945 - 63y - 6y = 324 945 - 324 = 69y 69y = 621 y = 9 Putting the value of y in the above equation y = 14 The value of x and y are 5 and 14 respectively. Question: 7 Solve the system of equations: Solution: The given system of equation is From (i) 4x + 3y = 132 … (iii) From (ii) 5x - 2y = - 42 …… (iv) Let us eliminate y from the given equations. The co efficient of y in the equation (iii) and (iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y equal to 6 in the two equations. Multiplying equation (iii) 2 and (iv) 3 we get 8x + 6y = 264 …. (v) 15x - 6y = -126 … (vi) Adding equation (v) and (vi) 8x + 15x = 264 - 126 23x = 138 x = 6 Putting the value of x in the equation (iii) 24 + 3y = 132 3y = 108 y = 36 The value of x and y are 36 and 6 respectively. Question: 8 Solve the system of equations: 6x - 4y = - 56x - 4y = - 5 Solution: The new equation becomes 4u + 3y = 8 … (i) 6u - 4y = – 5 …. (ii) From (i) 4u = 8 - 3y u = (8-3y)/4 From (ii) 24 - 17y = - 10 - 17y = - 34 y = 2 x = 2 So the Solution of the given system of equation is x = 2 and y = 2 Question: 9 Solve the system of equations: Solution: The given system of equation is: From (i) we get, 2x + y = 8 y = 8 - 2x From (ii) we get, x + 6y = 15 …… (iii) Substituting y = 8 - 2x in (iii), we get x + 6(8 - 2x) = 15 x + 48 - 12x = 15 - 11x = 15 - 48 - 11x = - 33 x = 3 Putting x = 3 in y 8 - 2x, we get y = 8 - (2×3) y = 8 - 6 Y = 2 The Solution of the given system of equation are x = 3 and y = 2 respectively. Question: 10 Solve the system of equations: Solution: The given system of equation is Let us eliminate y from the given equations. The coefficients of y in the given equations are 2 and 1 respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations. Multiplying equation (i)*1 and (ii)*2 4x + 2y = 3 ……. (iv) Subtracting equation (iii) from (iv) Putting x = 1/2in equation (iv) 2 + 2y = 3 y = 1/2 The Solution of the system of equation is x = 1/2 and y = 1/2 Question: 11 Solve the system of equations: Solution: From equation (i) Substituting this value in equation (ii) we obtain y = 0 Substituting the value of y in equation (iii) we obtain x = 0 The value of x and y are 0 and 0 respectively. Question: 12 Solve the system of equations: Solution: The given system of equation is: From equation (i) 33x - y + 15 = 110 33x + 15 - 110 = y y = 33x - 95 From equation (ii) 14y + x + 11 = 70 14y + x = 70 - 11 14y + x = 59 ….. (iii) Substituting y = 33x - 95 in (iii) we get, 14(33x - 95) + x = 59 462x - 1330 + x = 59 463x = 59 + 1330 463x = 1389 x = 1389/463 x = 3 Putting x = 3 in y = 33x - 95 we get, y = 33(3) - 95 99 - 95 = 4 The Solution of the given system of equation is 3 and 4 respectively. Question: 13 Solve the system of equations: Solution: Taking 1/y = u the given equation becomes, 2x - 3u = 9 ….. (iii) 3x + 7u = 2 ….. (iv) From (iii) 2x = 9 + 3u Substituting the valuein equation (iv) we get, 27 + 23u = 4 u = – 1 y = 1/u = – 1 x = 3 The Solution of the given system of equation is 3 and -1 respectively. Question: 14 Solve the system of equations: 0.5x + 0.7y = 0.74 0.3x + 0.5y = 0.5 Solution: The given system of equation is 0.5x + 0.7y = 0.74 …… (i) 0.3x - 0.5y = 0.5 ….. (ii) Multiplying both sides by 100 50x + 70y = 74 ….. (iii) 30x + 50y = 50 … (iv) From (iii) 50x = 74 - 70y Substituting the value of y in equation (iv) we get, 222 - 210y + 250y = 250 40y = 28 y = 0.7 Putting the value of y in the equation (iii) x = 0.5 The value of x and y are 0.5 and 0.7 respectively. Question: 15 Solve the system of equations: Solution: Multiplying (ii) by 1/2 we get, Solving equation (i) and (iii) Adding we get, When, x = 1/14 we get, Using equation (i) The Solution of the given system of equation is x = 1/14 and y = 1/6 respectively. Question: 16 Solve the system of equations: Solution: 3u + 2v = 12 ….. (i) v = 3 1/u = x x = 1/2 1/v = y y = 1/3 Question: 17 Solve the system of equations: Solution: 15x + 2y = 17 ….. (i) x + y = 36/5 …. (ii) From equation (i) we get, 2y = 17 - 15x Substitutingin equation (ii) we get, 5(-13x + 17) = 72 - 65x = –13 x = 1/5 Putting x = 1/5 in equation (ii), we get y = 7 The Solution of the given system of equation is 5 and 1/7 respectively. Question: 18 Solve the system of equations: Solution: 3u - v = - 9 …. (i) 2u + 3v = 5 …. (ii) Multiplying equation (i) 3 and (ii) 1 we get, 9u - 3v = -27 ….. (iii) 2u + 3v = 5 … (iv) Adding equation (i) and equation (iv) we get, 9u + 2u - 3v + 3v = -27 + 5 u = -2 Putting u = -2 in equation (iv) we get, 2(-2) + 3v = 5 3v = 9 v = 3 Question: 19 Solve the system of equations: Solution: Multiplying equation (i) adding equation (ii) we get, 2y + 3x = 9 ….. (iii) 4y + 9x = 21 …. (iv) From (iii) we get, 3x = 9 - 2y Substitutingin equation (iv) we get 4y + 3(9 - 2y) = 21 - 2y = 21 - 27 y = 3 x = 1 Hence the Solutions of the system of equation are 1 and 3 respectively. Question: 20 Solve the system of equations: Solution: 6u + 5v = 360 …. (i) 7u - 9v = 168 …. (ii) Let us eliminate v from the equation (i) and (ii) multiplying equation (i) by 9 and (ii) by 5 54u + 35u = 3240 + 840 89u = 4080 u = 4080/89 Putting u = 4080/89 in equation (i) we get, So, the solution of the given system of equation is x = 89/4080, y = 89/1512 Question: 21 Solve the system of equations: Solution: Then, the given system of equation becomes, 6u = 7v + 3 6u - 7v = 3 ….. (i) 3u = 2v 3u - 2v = 0 … (ii) Multiplying equation (ii) by 2 and (i) 1 6u - 7v = 3 6u - 4v = 0 Subtracting v = - 1 in equation (ii), we get 3u - 2(-1) = 0 3u + 2 = 0 3u = - 2 and v = –1 x - y = –1 … (vi) Adding equation (v) and equation (vi) we get, Putting x = -2/3 in equation (vi) Question: 22 Solve the system of equations: Solution: 5xy = 6x + 6y …. (i) and xy = 6(y - x) xy = 6y - 6x … (ii) Adding equation (i) and equation (ii) we get, 6xy = 6y + 6y 6xy = 12y x = 2 Putting x = 2 in equation (i) we get, 10y = 12 + 6y 10 - 6y = 12 4y = 12 y = 3 The Solution of the given system of equation is 2 and 3 respectively. Question: 23 Solve the system of equations: Solution: Then the given system of equation becomes: 5u - 2v = -1 ….. (i) 15u + 7v = 10 …… (ii) Multiplying equation (i) by 7 and (ii) by 2 35u - 14v = -7 …… (iii) 30u + 14v = 20 …… (iv) Subtracting equation (iv) from equation (iii) , we get - 2v = – 1 – 1 - 2v = – 2 v = 1 Now, x + y = 5 ….. (v) x - y = 1 ….. (vi) Adding equation (v) and (vi) we get, 2x = 6 x = 3 Putting the value of x in equation (v) 3 + y = 5 y = 2 The Solutions of the given system of equation are 3 and 2 respectively. Question: 24 Solve the system of equations: Solution: Then the given system of equation becomes: 3u + 2v = 2 ….. (i) 9u + 4v = 1 …… (ii) Multiplying equation (i) by 3 and (ii) by 1 6u + 4v = 4 …… (iii) 9u - 4v = 1 …… (iv) Adding equation (iii) and (iv) we get, 45u = 5 u = 1/3 Subtracting equation (iv) from equation (iii), we get 2v = 2 - 1 2v = 1 v = 1/2 Now, x + y = 3 ….. (v) x - y = 2 ….. (vi) Adding equation (v) and (vi) we get, 2x = 5 x = 2/5 Putting the value of x in equation (v) 5/2 + y = 11 y = 1/2 The Solutions of the given system of equation are 5/2 and 1/2 respectively. Question: 25 Solve the system of equations: Solution: Then the given system of equation becomes: 3u + 10v = -9 ….. (i) 25u -12v = 61/3 …… (ii) Multiplying equation (i) by 12 and (ii) by 10 36u + 120v = -108 …… (iii) 250u + 120v = 610/3 …… (iv) Adding equation (iv) and equation (iii), we get 36u + 250u = 610/3 - 108 286u =286/3 U = 1/3 Putting u = 61/3 in equation (i) v = -1 Now, 3x - 2y = –1 ….. (vi) Putting x = 1/2 in equation (v) we get, The Solutions of the given system of equation are 1/2 and 5/4 respectively. Question: 26 Solve the system of equations: x + y = 5xy 3x + 2y = 13xy Solution: The given system of equations is: x + y = 5xy ….. (i) 3x + 2y = 13xy …… (ii) Multiplying equation (i) by 2 and equation (ii) 1 we get, 2x ++ 2y = 10xy …… (iii) 3x + 2y = 13xy ……. (iv) Subtracting equation (iii) from equation (iv) we get, 3x - 2x = 13xy - 10xy x = 3xy x/3x = y 1/3 = y Putting y = 1/3 = y in equation (i) we get, Hence Solution of the given system of equation is 1/2 and 1/3 Question: 27 Solve the system of equations: x + y = xy Solution: x + y = xy ….. (i) Adding equation (i) and (ii) we get, 2x = 2xy + 6xy 2x = 6xy y = x + y = xy y = 1/4 Putting= y = 1/4 in equation (i), we get, Hence the Solution of the given system of equation is x = - 1/2 and y = 1/4 respectively. Question: 28 Solve the system of equations: 2(3u – v) = 5uv 2(u + 3v) = 5uv Solution: 2(3u - v) = 5uv 6u - 2v = 5uv …. (i) 2(u + 3v) = 5uv 2u + 6v = 5uv ….. (ii) Multiplying equation (i) by 3 and equation (ii) by 1 we get, 18u - 6v = 15uv ….. (iii) 2u + 6v = 5uv …….. (iv) Adding equation (iii) and equation (iv) we get, 18u + 2u = 15uv + 5uv v = 1 Putting v = 1 in equation (i) we get, 6u - 2 = 5u u = 2 Hence the Solution of the given system of Solution of equation is 2 and 1 respectively. Question: 29 Solve the system of equations: Solution: Then the given system of equation becomes: 5u - v = 2 …… (ii) Multiplying equation (ii) by 3 Adding equation (iv) and equation (iii), we get 13u = 13/5 u = 1/5 Putting u = 1/5 in equation (i) v = 1 Now, 3x + 2y = 5 ….. (iv) 3x - 2y = 1 ….. (v) Adding equation (iv) and (v) we get, 6x = 6 x =1 Putting the value of x in equation (v) we get, 3 + 2y = 5 y = 1 The Solutions of the given system of equation are 1 and 1 respectively. Question: 30 Solve the system of equations: Solution: Then the given system of equation becomes: 44u + 30v = 10 ….. (i) 55u + 40v = 13 … (ii) Multiplying equation (i) by 4 and (ii) by 3 176u + 120v = 40 …… (iii) 165u + 120v = 39 …… (iv) Subtracting equation (iv) from (iii) we get, 176 - 165u = 40 - 39 u = 1/11 Putting the value of u in equation (i) 4 + 30v = 10 30v = 6 x + y = 11 ….. (v) x - y = 5 ….. (vi) Adding equation (v) and (vi) we get, 2x = 16 x = 8 Putting the value of x in equation (v) 8 + y = 11 y = 3 The Solutions of the given system of equation are 8 and 3 respectively. Question: 31 Solve the system of equations: Solution: Then the given system of equation becomes: 10p + 2q = 4 ….. (i) 15p - 5q = - 2 …… (ii) Multiplying equation (i) by 4 and (ii) by 3 176u + 120v = 40 …… (iii) 165u + 120v = 39 …… (iv) Using cross multiplication method we get, x + y = 5 ….. 3 x - y = 1 …..4 Adding equation 3 and 4 we get, x = 3 Substituting the value of x in equation 3 we get, y = 2 The Solution of the given system of Solution is 3 and 2 respectively. Question: 32 Solve the system of equations: Solution: Then the given system of equation becomes: Can be written as 5p + q = 2 …… 3 6p - 3q = 1 ……. 4 Equation 3 and 4 from a pair of linear equation in the general form. Now, we can use any method to solve these equations. We get p = 1/3 q = 1/3 Substituting the 1/(x –1) for p, we have x - 1 = 3 x = 4 y - 2 = 3 y = 5 The Solution of the required pair of equation is 4 and 5 respectively. Question: 33 Solve the system of equations: Solution: The given equation s reduce to: -2p + 7q = 5 -2p + 7q - 5 = 0 …… 3 7p + 8q = 15 7p + 8q - 15 = 0 …… 4 Using cross multiplication method we get, p = 1/x q = 1/y x = 1 and y = 1 Question: 34 Solve the system of equations: 152x - 378y = – 74 - 378x + 152y = – 604 Solution: 152x - 378y = – 74 …. 1 -378x + 152y = – 604 …. 2 Adding the equations 1 and 2, we obtain - 226x - 226y = -678 x + y = 3 ….. 3 Subtracting the equation 2 from equation 1, we obtain 530x + 530y = 530 x - y = 1 … 4 Adding equations 3 and 4 we obtain, 2x = 4 x = 2 Substituting the value of x in equation 3 we obtain y = 1 Question: 35 Solve the system of equations: 99x + 101y = 409 101x + 99y = 501 Solution: The given system of equation are: 99x + 101y = 409 …. 1 101x + 99y = 501 ….. 2 Adding equation 1 and 2 we get, 99x + 101x + 101y + 99y = 49 + 501 200(x + y) = 1000 x + y = 5 ….. 3 Subtracting equation 1 from 2 101x - 99x + 99y - 101y = 501 - 499 2(x - y) = 2 x - y = 1 …. 4 Adding equation 3 and 4 we get, 2x = 6 x = 3 Putting x = 3 in equation 3 we get, 3 + y = 5 y = 2 The Solution of the given system of equation is 3 and 2 respectively. Question: 36 Solve the system of equations: 23x - 29y = 98 29x - 23y = 110 Solution: 23x - 29y = 98 … 1 29x - 23y = 110 …… 2 Adding equation 1 and 2 we get, 6(x + y) = 12 x + y = 2 … 3 Subtracting equation 1 from 2 we get, 52(x-y) = 208 x - y = 4 …. 4 Adding equation 3 and 4 we get, 2x = 6 x = 3 Putting the value of x in equation 4 3 + y = 2 y = – 1 The Solution of the given system of equation is 3 and -1 respectively. Question: 37 Solve the system of equations: x - y + z = 4 x - 2y - 2z = 9 2x + y + 3z = 1 Solution: x - y + z = 4 ….. 1 x - 2y - 2z = 9 …… 2 2x + y + 3z = 1 … 3 From equation 1 z = 4 - x + y z = -x + y + 4 Subtracting the value of the z in equation 2 we get, x - 2y - 2(- x + y + 4) = 9 x - 2y + 2x - 2y - 8 = 8 3x - 4y = 17 ….. 4 Subtracting the value of z in equation 3, we get, 2x + y + 3(-x + y + 4) = 1 2x + y + 3x +3y + 12 =1 - x + 4y = -11 Adding equation 4 and 5 we get, 3x - x - 4y + 4y = 17 - 11 2x = 6 x = 3 Putting x = 3 in equation 4, we get, 9 - 4y = 17 - 4y = 17 - 9 y = -2 Putting x = 3 and y = -2 in z = -x + y + 4, we get, Z = -3 - 2 + 4 = -1 The Solution of the given system of equation are 3, – 2 and –1 respectively. Question: 38 Solve the system of equations: x - y + z = 4 x + y + z = 2 2x + y - 3z = 0 Solution: x - y + z = 4 …… 1 x + y + z = 2 …. 2 2x + y - 3z = 0 …… 3 From equation 1 z = - x + y + 4 Substituting z = -x + y + 4 in equation 2, we get, x + y + (-x + y + 4) = 2 x + y - x + y + 4 = 2 2y = 2 y = 1 Substituting the value of z in equation 3 2x + y - 3(-x + y + 4) = 0 2x + y + 3x - 3y -12 = 0 5x - 2y = 12 …… 4 Putting the y = - 1 in equation 4 5x - 2(-1) = 12 5x = 10 x = 2 Putting x = 2 and y = -1 in z = -x + y + 4 z = -2 - 1 + 4 = 1 The Solution of the given system of equations are 2, -1 and 1 respectively.
Solve the system of equations:
11x + 15y + 23 = 0 and 7x – 2y – 20 = 0
The given system of equation is
11x + 15y + 23 = 0 ……. (i)
7x - 2y - 20 = 0 ….. (ii)
From (ii)
2y = 7x - 20
Substituting the value of y in equation (i) we get,
127x = 254
x = 2
Putting the value of x in the equation (iii)
y = - 3
The value of x and y are 2 and -3 respectively.
3x – 7y + 10 = 0, y – 2x – 3 = 0
3x - 7y + 10 = 0 …. (i)
y - 2x - 3 = 0 ….. (ii)
y - 2x - 3 = 0
y = 2x + 3 …… (iii)
= 3x - 7(2x + 3) + 10 = 0
= 3x + 14x - 21 + 10 = 0
= -11x = 11
= x = -1
= y = 2(- 1) + 3
y = 1
The value of x and y are -1 and 1 respectively.
0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8
0.4x + 0.3y = 1.7
0.7x - 0.2y = 0.8
Multiplying both sides by 10
4x + 3y = 17 ….. (i)
7x - 2y = 8 …… (ii)
7x - 2y = 8
32 + 29y = 119
29y = 87
y = 3
Putting the value of y in the equation (iii)
The value of x and y are 2 and 3 respectively.
Therefore x + 2y = 1.6
x + 2y = 1.6
7 = 10x + 5y
10x + 20y = 16 ….. (i)
10x + 5y = 7 …… (ii)
Subtracting two equations we get,
15y = 9
The value of x and y are 2/5 and 3/5 respectively.
7(y + 3) - 2(x + 3) = 14
4(y - 2) + 3(x - 3) = 2
7(y + 3) - 2(x + 3) = 14 ……. (i)
4(y - 2) + 3(x - 3) = 2 ….. (ii)
From (i)
7y + 21 - 2x - 4 = 14
7y = 14 + 4 - 21 + 2x
4y - 8 + 3x - 9 = 2
4y + 3x - 17 - 2 = 0
4y + 3x - 19 = 0 ….. (iii)
Substituting the value of y in equation (iii)
8x - 12 + 21x - 133 = 0
29x = 145
x = 5
Putting the value of x in the above equation
The value of x and y are 5 and 1 respectively.
= 9x - 2y = 108 … (iii)
Substituting the value of x in equation (iii) we get,
945 - 63y - 6y = 324
945 - 324 = 69y
69y = 621
y = 9
Putting the value of y in the above equation
y = 14
The value of x and y are 5 and 14 respectively.
4x + 3y = 132 … (iii)
5x - 2y = - 42 …… (iv)
Let us eliminate y from the given equations. The co efficient of y in the equation (iii) and (iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y equal to 6 in the two equations.
Multiplying equation (iii) 2 and (iv) 3 we get
8x + 6y = 264 …. (v)
15x - 6y = -126 … (vi)
Adding equation (v) and (vi)
8x + 15x = 264 - 126
23x = 138
x = 6
24 + 3y = 132
3y = 108
y = 36
The value of x and y are 36 and 6 respectively.
6x - 4y = - 56x - 4y = - 5
The new equation becomes
4u + 3y = 8 … (i)
6u - 4y = – 5 …. (ii)
4u = 8 - 3y
u = (8-3y)/4
24 - 17y = - 10
- 17y = - 34
y = 2
So the Solution of the given system of equation is x = 2 and y = 2
The given system of equation is:
From (i) we get,
2x + y = 8
y = 8 - 2x
From (ii) we get,
x + 6y = 15 …… (iii)
Substituting y = 8 - 2x in (iii), we get
x + 6(8 - 2x) = 15
x + 48 - 12x = 15
- 11x = 15 - 48
- 11x = - 33
x = 3
Putting x = 3 in y 8 - 2x, we get
y = 8 - (2×3)
y = 8 - 6
Y = 2
The Solution of the given system of equation are x = 3 and y = 2 respectively.
Let us eliminate y from the given equations. The coefficients of y in the given equations are 2 and 1 respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations.
Multiplying equation (i)*1 and (ii)*2
4x + 2y = 3 ……. (iv)
Subtracting equation (iii) from (iv)
Putting x = 1/2in equation (iv)
2 + 2y = 3
y = 1/2
The Solution of the system of equation is x = 1/2 and y = 1/2
From equation (i)
Substituting this value in equation (ii) we obtain
y = 0
Substituting the value of y in equation (iii) we obtain
x = 0
The value of x and y are 0 and 0 respectively.
33x - y + 15 = 110
33x + 15 - 110 = y
y = 33x - 95
From equation (ii)
14y + x + 11 = 70
14y + x = 70 - 11
14y + x = 59 ….. (iii)
Substituting y = 33x - 95 in (iii) we get,
14(33x - 95) + x = 59
462x - 1330 + x = 59
463x = 59 + 1330
463x = 1389
x = 1389/463
Putting x = 3 in y = 33x - 95 we get,
y = 33(3) - 95
99 - 95 = 4
The Solution of the given system of equation is 3 and 4 respectively.
Taking 1/y = u the given equation becomes,
2x - 3u = 9 ….. (iii)
3x + 7u = 2 ….. (iv)
From (iii)
2x = 9 + 3u
Substituting the valuein equation (iv) we get,
27 + 23u = 4
u = – 1
y = 1/u = – 1
The Solution of the given system of equation is 3 and -1 respectively.
0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5
0.5x + 0.7y = 0.74 …… (i)
0.3x - 0.5y = 0.5 ….. (ii)
Multiplying both sides by 100
50x + 70y = 74 ….. (iii)
30x + 50y = 50 … (iv)
50x = 74 - 70y
Substituting the value of y in equation (iv) we get,
222 - 210y + 250y = 250
40y = 28
y = 0.7
x = 0.5
The value of x and y are 0.5 and 0.7 respectively.
Multiplying (ii) by 1/2 we get,
Solving equation (i) and (iii)
Adding we get,
When, x = 1/14 we get,
Using equation (i)
The Solution of the given system of equation is x = 1/14 and y = 1/6 respectively.
3u + 2v = 12 ….. (i)
v = 3
1/u = x
x = 1/2
1/v = y
y = 1/3
15x + 2y = 17 ….. (i)
x + y = 36/5 …. (ii)
From equation (i) we get,
2y = 17 - 15x
Substitutingin equation (ii) we get,
5(-13x + 17) = 72
- 65x = –13
x = 1/5
Putting x = 1/5 in equation (ii), we get
y = 7
The Solution of the given system of equation is 5 and 1/7 respectively.
3u - v = - 9 …. (i)
2u + 3v = 5 …. (ii)
Multiplying equation (i) 3 and (ii) 1 we get,
9u - 3v = -27 ….. (iii)
2u + 3v = 5 … (iv)
Adding equation (i) and equation (iv) we get,
9u + 2u - 3v + 3v = -27 + 5
u = -2
Putting u = -2 in equation (iv) we get,
2(-2) + 3v = 5
3v = 9
Multiplying equation (i) adding equation (ii) we get,
2y + 3x = 9 ….. (iii)
4y + 9x = 21 …. (iv)
From (iii) we get,
3x = 9 - 2y
Substitutingin equation (iv) we get
4y + 3(9 - 2y) = 21
- 2y = 21 - 27
x = 1
Hence the Solutions of the system of equation are 1 and 3 respectively.
6u + 5v = 360 …. (i)
7u - 9v = 168 …. (ii)
Let us eliminate v from the equation (i) and (ii) multiplying equation (i) by 9 and (ii) by 5
54u + 35u = 3240 + 840
89u = 4080
u = 4080/89
Putting u = 4080/89 in equation (i) we get,
So, the solution of the given system of equation is x = 89/4080, y = 89/1512
Then, the given system of equation becomes,
6u = 7v + 3
6u - 7v = 3 ….. (i)
3u = 2v
3u - 2v = 0 … (ii)
Multiplying equation (ii) by 2 and (i) 1
6u - 7v = 3
6u - 4v = 0
Subtracting v = - 1 in equation (ii), we get
3u - 2(-1) = 0
3u + 2 = 0
3u = - 2
and v = –1
x - y = –1 … (vi)
Adding equation (v) and equation (vi) we get,
Putting x = -2/3 in equation (vi)
5xy = 6x + 6y …. (i) and
xy = 6(y - x)
xy = 6y - 6x … (ii)
Adding equation (i) and equation (ii) we get,
6xy = 6y + 6y
6xy = 12y
Putting x = 2 in equation (i) we get,
10y = 12 + 6y
10 - 6y = 12
4y = 12
The Solution of the given system of equation is 2 and 3 respectively.
Then the given system of equation becomes:
5u - 2v = -1 ….. (i)
15u + 7v = 10 …… (ii)
Multiplying equation (i) by 7 and (ii) by 2
35u - 14v = -7 …… (iii)
30u + 14v = 20 …… (iv)
Subtracting equation (iv) from equation (iii) , we get
- 2v = – 1 – 1
- 2v = – 2
v = 1
Now,
x + y = 5 ….. (v)
x - y = 1 ….. (vi)
Adding equation (v) and (vi) we get,
2x = 6
Putting the value of x in equation (v)
3 + y = 5
The Solutions of the given system of equation are 3 and 2 respectively.
3u + 2v = 2 ….. (i)
9u + 4v = 1 …… (ii)
Multiplying equation (i) by 3 and (ii) by 1
6u + 4v = 4 …… (iii)
9u - 4v = 1 …… (iv)
Adding equation (iii) and (iv) we get,
45u = 5
u = 1/3
Subtracting equation (iv) from equation (iii), we get
2v = 2 - 1
2v = 1
v = 1/2
x + y = 3 ….. (v)
x - y = 2 ….. (vi)
2x = 5
x = 2/5
5/2 + y = 11
The Solutions of the given system of equation are 5/2 and 1/2 respectively.
3u + 10v = -9 ….. (i)
25u -12v = 61/3 …… (ii)
Multiplying equation (i) by 12 and (ii) by 10
36u + 120v = -108 …… (iii)
250u + 120v = 610/3 …… (iv)
Adding equation (iv) and equation (iii), we get
36u + 250u = 610/3 - 108
286u =286/3
U = 1/3
Putting u = 61/3 in equation (i)
v = -1
3x - 2y = –1 ….. (vi)
Putting x = 1/2 in equation (v) we get,
The Solutions of the given system of equation are 1/2 and 5/4 respectively.
x + y = 5xy
3x + 2y = 13xy
The given system of equations is:
x + y = 5xy ….. (i)
3x + 2y = 13xy …… (ii)
Multiplying equation (i) by 2 and equation (ii) 1 we get,
2x ++ 2y = 10xy …… (iii)
3x + 2y = 13xy ……. (iv)
Subtracting equation (iii) from equation (iv) we get,
3x - 2x = 13xy - 10xy
x = 3xy
x/3x = y
1/3 = y
Putting y = 1/3 = y in equation (i) we get,
Hence Solution of the given system of equation is 1/2 and 1/3
x + y = xy
x + y = xy ….. (i)
Adding equation (i) and (ii) we get,
2x = 2xy + 6xy
2x = 6xy
y = x + y = xy
y = 1/4
Putting= y = 1/4 in equation (i), we get,
Hence the Solution of the given system of equation is x = - 1/2 and y = 1/4 respectively.
2(3u – v) = 5uv
2(u + 3v) = 5uv
2(3u - v) = 5uv
6u - 2v = 5uv …. (i)
2u + 6v = 5uv ….. (ii)
Multiplying equation (i) by 3 and equation (ii) by 1 we get,
18u - 6v = 15uv ….. (iii)
2u + 6v = 5uv …….. (iv)
Adding equation (iii) and equation (iv) we get,
18u + 2u = 15uv + 5uv
Putting v = 1 in equation (i) we get,
6u - 2 = 5u
u = 2
Hence the Solution of the given system of Solution of equation is 2 and 1 respectively.
5u - v = 2 …… (ii)
Multiplying equation (ii) by 3
13u = 13/5
u = 1/5
Putting u = 1/5 in equation (i)
3x + 2y = 5 ….. (iv)
3x - 2y = 1 ….. (v)
Adding equation (iv) and (v) we get,
6x = 6
x =1
Putting the value of x in equation (v) we get,
3 + 2y = 5
The Solutions of the given system of equation are 1 and 1 respectively.
44u + 30v = 10 ….. (i)
55u + 40v = 13 … (ii)
Multiplying equation (i) by 4 and (ii) by 3
176u + 120v = 40 …… (iii)
165u + 120v = 39 …… (iv)
Subtracting equation (iv) from (iii) we get,
176 - 165u = 40 - 39
u = 1/11
Putting the value of u in equation (i)
4 + 30v = 10
30v = 6
x + y = 11 ….. (v)
x - y = 5 ….. (vi)
2x = 16
x = 8
8 + y = 11
The Solutions of the given system of equation are 8 and 3 respectively.
10p + 2q = 4 ….. (i)
15p - 5q = - 2 …… (ii)
Using cross multiplication method we get,
x + y = 5 ….. 3
x - y = 1 …..4
Adding equation 3 and 4 we get,
Substituting the value of x in equation 3 we get,
The Solution of the given system of Solution is 3 and 2 respectively.
Can be written as 5p + q = 2 …… 3
6p - 3q = 1 ……. 4
Equation 3 and 4 from a pair of linear equation in the general form. Now, we can use any method to solve these equations.
We get
p = 1/3
q = 1/3
Substituting the 1/(x –1) for p, we have
x - 1 = 3
x = 4
y - 2 = 3
y = 5
The Solution of the required pair of equation is 4 and 5 respectively.
The given equation s reduce to:
-2p + 7q = 5
-2p + 7q - 5 = 0 …… 3
7p + 8q = 15
7p + 8q - 15 = 0 …… 4
p = 1/x
q = 1/y
x = 1 and y = 1
152x - 378y = – 74
- 378x + 152y = – 604
152x - 378y = – 74 …. 1
-378x + 152y = – 604 …. 2
Adding the equations 1 and 2, we obtain
- 226x - 226y = -678
x + y = 3 ….. 3
Subtracting the equation 2 from equation 1, we obtain
530x + 530y = 530
x - y = 1 … 4
Adding equations 3 and 4 we obtain,
2x = 4
Substituting the value of x in equation 3 we obtain y = 1
99x + 101y = 409
101x + 99y = 501
The given system of equation are:
99x + 101y = 409 …. 1
101x + 99y = 501 ….. 2
Adding equation 1 and 2 we get,
99x + 101x + 101y + 99y = 49 + 501
200(x + y) = 1000
Subtracting equation 1 from 2
101x - 99x + 99y - 101y = 501 - 499
2(x - y) = 2
x - y = 1 …. 4
Putting x = 3 in equation 3 we get,
The Solution of the given system of equation is 3 and 2 respectively.
23x - 29y = 98
29x - 23y = 110
23x - 29y = 98 … 1
29x - 23y = 110 …… 2
6(x + y) = 12
x + y = 2 … 3
Subtracting equation 1 from 2 we get,
52(x-y) = 208
x - y = 4 …. 4
Putting the value of x in equation 4
3 + y = 2
y = – 1
x - y + z = 4
x - 2y - 2z = 9
2x + y + 3z = 1
x - y + z = 4 ….. 1
x - 2y - 2z = 9 …… 2
2x + y + 3z = 1 … 3
From equation 1
z = 4 - x + y
z = -x + y + 4
Subtracting the value of the z in equation 2 we get,
x - 2y - 2(- x + y + 4) = 9
x - 2y + 2x - 2y - 8 = 8
3x - 4y = 17 ….. 4
Subtracting the value of z in equation 3, we get,
2x + y + 3(-x + y + 4) = 1
2x + y + 3x +3y + 12 =1
- x + 4y = -11
Adding equation 4 and 5 we get,
3x - x - 4y + 4y = 17 - 11
Putting x = 3 in equation 4, we get,
9 - 4y = 17
- 4y = 17 - 9
y = -2
Putting x = 3 and y = -2 in z = -x + y + 4, we get,
Z = -3 - 2 + 4
= -1
The Solution of the given system of equation are 3, – 2 and –1 respectively.
x + y + z = 2
2x + y - 3z = 0
x - y + z = 4 …… 1
x + y + z = 2 …. 2
2x + y - 3z = 0 …… 3
z = - x + y + 4
Substituting z = -x + y + 4 in equation 2, we get,
x + y + (-x + y + 4) = 2
x + y - x + y + 4 = 2
2y = 2
Substituting the value of z in equation 3
2x + y - 3(-x + y + 4) = 0
2x + y + 3x - 3y -12 = 0
5x - 2y = 12 …… 4
Putting the y = - 1 in equation 4
5x - 2(-1) = 12
5x = 10
Putting x = 2 and y = -1 in z = -x + y + 4
z = -2 - 1 + 4
= 1
The Solution of the given system of equations are 2, -1 and 1 respectively.
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Chapter 3: Pair of Linear Equations in Two...