Solve the following systems of equations:
`4/x + 3y = 14`
`3/x - 4y = 23`
`4/x + 3y = 14`
`3/x - 4y = 23`
Let `1/x = p`
The given equations reduce to:
4p + 3y = 14
=> 4p + 3y - 14 = 0 ...(1)
3p - 4y = 23
=> 3p - 4y - 23 = 0 ....(2)
Using cross-multiplication method, we obtain
`p/(-69-56) = y/(-42 - (-92)) = 1/(-16 - 9)`
`p/(-125) = y/50 = (-1)/25`
p = 5 , y = -2
`:. p = 1/x = 5`
`x = 1/5`
Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method
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Page 2
Solve the following systems of equations:
`x+y = 2xy`
`(x - y)/(xy) = 6` x != 0, y != 0
The system of the given equation is
x + y = 2xy .....(i)
And `(x - y)/(xy) = 6`
x - y = 6xy .....(ii)
Adding equation (i) and equation (ii), we get
2x = 2xy + 6xy
=> 2x = 8xy
`=> (2x)/(8x) = y`
`=> y = 1/4`
Putting y = 1/4 in equation (i) we get
`x + 1/4 = 2x xx 1/4`
`=> x + 1/4 = x/2`
`=> x - x/2 = (-1)/4`
`=> (2x - x)/2 = (-1)/4`
`=> x = (-2)/4 = (-1)/2`
Hence, solution of the given system of equation is `x = (-1)/2, y = 1/4`
Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method
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Page 3
The system of the given equation is
2(3u − ν) = 5uν
=> 6u - 2v = 5uv ....(i)
And 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 = 156uv + 5uv
=> 20u = 20uv
`=> (20u)/(20u)= v`
=> v = 1
Putting v = 1in equation (i), we get
6u - 2 x 1 = 5u x 1
=> 6u - 2 = 5u
=> 6u - 5u = 2
=> u = 2
Hence, solution of the given system of equation is u = 2, v=- 1
Page 4
Let `1/(3x + 2y) = u and 1/(3x - 2y) = v` Then, the given system of equation becomes
`2u + 3v = 17/5` ....(i)
5u + v = 2 ....(ii)
Multiplying equation (ii) by 3, we get
`15u - 2u = 6 - 17/5`
`=> 13u = (30 - 17)/5`
`=> 13u = 13/5`
`=> u = 13/(5 xx 13) = 1/5`
Putting u = 1/5 in eqaution (ii) we get
`5 xx 1/5 + v = 2`
=> 1 + v = 2
=> v = 2 - 1
=> v = 1
Now `u = 1/(3x + 2y)`
`=> 1/(3x + 2y) = 1/5`
=> 3x + 2y = 5 ....(iv)
And `v = 1/(3x + 2y)`
`=> 1/(3x + 2y ) = 1/5`
=> 3x + 2y = 5 ....(iv)
And `v = 1/(3x + 2y)`
=> 3x - 2y = 1 ...(v)
Adding equation (iv) and (v), we get
6x = 1 + 5
6x = 6
x = 1
Putting 1 x in equation (v), we get
3 xx 1 + 2y = 5
2y = 5 - 3
2y = 2
y = 2/2 = 1
Hence, solution of the given system of equation is x = 1, y = 1
Page 5
We have,
x − y + z = 4 ...(i)
x + y + z = 2 ....(ii)
2x + y − 3z = 0 ....(iii)
From equation (i), we get
z = 4 - x + y
z = -x + y + 4
Substituting z = -x + y + 4 in equation (ii), we get
x + y + (-x + y + 4) = 2
=> x + y - x + y + 4 = 2
=> 2y + 4 = 2
`=> 2y = 2 - 4 = -2`
=> 2y = -2
`=> y = (-2)/2 = -1`
Substituting the value of z in equation (iii), we get
2x + y -3(-x + y + 4) = 0
=> 2x + y + 3x - 3y - 12 = 0
=> 5x - 2y - 12 = 0
=> 5x - 2y = 12 ....(iv)
Putting y = -1 in equation (iv), we get
`5x - 2xx (-1) = 12`
=> 5x + 2 = 12
=> 5x = 12 - 2 = 10
`=> x = 10/5 = 2`
Putting x = 2 and y = -1 in z = -x + y + 4 we get
z = -2 + (-1) + 4
=-2 - 1 + 4
= -3 + 4
= 1
Hence, solution of the giving system of equation is x = 2, y = -1, z = 1
Page 6
Let `1/(x + y) = u and 1/(x - y) = v`
Then, the system of the given equations becomes
44u + 30v = 10 ....(i)
55u + 40v = 13 ....(ii)
Multiplying equation (i) by 4 and equation (ii) by 3, we get
176u + 120v = 40 ...(iii)
165u + 120v = 39 ...(iv)
Subtracting equation (iv) by equation (iii), we get
176 - 165u = 40 - 39
=> 11u = 1
`=> u = 1/11`
Putting u = 1/11 in equation (i) we get
`44 xx 1/11 + 30v = 10`
4 + 30v = 10
=> 30v = 10 - 4
=> 30v = 6
`=> v = 6/30 = 1/5`
Now `u = 1/(x + y)`
`=> 1/(x + y) = 1/11`
=> x + y = 11 ...(v)
Adding equation (v) and (vi), we get
2x = 11 + 5
=> 2x = 16
`=> x = 16/2 = 8`
Putting x = 8 in equation (v) we get
8 + y = 11
=> y = 11 - 8 - 3
Hence, solution of the given system of equations is x = 8, y = 3
Page 7
`10/(x + y) + 2/(x - y) = 4`
`15/(x + y) - 5/(x - y) = -2`
Let `1/(x + y) = p and 1/(x - y) = q`
The given equations reduce to:
10p + 2q = 4
=> 10p + 2q - 4 = 0 ....(1)
15p - 5q = -2
=> 15p - 5q + 2 = 0 ...(2)
Using cross-multiplication method, we obtain:
`p.(4- 20) = q/(-60-20) = 1/(-50-30)`
`p/(-16) = q/(-80) = 1/(-80)`
p = 1/5 and q =1
p = 1/(x + y) = 1/5 and `q = 1/(x - y) = 1`
x + y = 5 .....(3)
x - y = 1 ....(4)
Adding equation (3) and (4), we obtain:
2x = 6
x = 3
Substituting the value of x in equation (3), we obtain:
y = 2
∴ x = 3, y = 2
Page 8
The given system of equation is
`4/x + 15y = 21` ....(i)
`3/x + 4y = 5` ....(ii)
Multiplying equation (i) by 3 and equation (ii) by 4, we get
`12/x + 15y = 21` .....(iii)
`12/x + 16y = 20` .....(iv)
Subtracting equation (iii) from equation (iv), we get
`12/x - 12/x + 16y - 15y = 20 - 21`
y = -1
Putting y = -1 in equation (i) we get
`4/x + 5xx (-1) = 7`
`=> 4/x - 5 = 7`
`=> 4/x = 7 + 5`
`=> 4/x = 12`
=> 4 = 12x
=> 4/12 = x
`=> x = 4/12`
`=> x = 1/3`
Hence, solution of the given system of equation x = 1/3, y =- -1
Page 9
Let us write the given pair of equation as
`2(1/x) + 3(1/y) = 13` ....(1)
`5(1/x) - 4(1/y) = -2` ....(2)
These equation are not in the form ax + by + c = 0 However, if we substitute
`1/x = p` and `1/y = q` in equation (1) and (2) we get
2p + 3q = 13
5p - 4q = -2
So, we have expressed the equations as a pair of linear equations. Now, you can use any method to solve these equations, and get p = 2, q = 3
You know that `p =1/x and q =1/y`
Substitute the values of p and q to get
`1/x = 2 i.e x = 1/2 and 1/y = 3 " i.e " y = 1/3`
Page 10
The given equation is
`1/(3x + y) + 1/(3x - y) = 3/4`
`1/(2(3x + y)) - 1/(2(3x - y)) = -1/8`
Let `1/(3x + y) = u and 1/(3x - y) = v` then equation are
`u + v = 3/4` ...(i)
`u/2 - v/2 = 1/8` ....(ii)
Multiply equation (ii) by 2 and add both equations, we get
`u + v = 3/4`
`u - v = -1/4`
`2u = 1/2`
`u = 1/4`
Put the value of u in equation (i) we get
`1 xx 1/4 + v = 3/4`
`v= 1/2`
Then
`1/(3x + y) = 1/4` ....(iii)
3x + y = 4
`1/(3x -y) = 1/2` ...(iv)
3x - y = 2
Add both equation we get
3x + y = 4
3x - y = 2
_________
6x = 6
x = 1
Put the value of x in equation (iii) we get
3 x 1 + y = 4
y = 1
Hence value of x = 1 and y = 1
Page 11
Solve the following systems of equations:
`5/(x - 1) + 1/(y - 2) = 2`
Let us put` 1/(x - 1) = p and 1/(y - 2) = q` The the given equation
`5(1/(x - 1)) + 1/(y - 2) = 2` .....(1)
`6(1/(x - 1)) -3 (1/(y - 2) ) = 1` ....(2)
Can be written as
5p + q = 2 ....(3)
6p - 3q = 1 ...(4)
Equation 3 and 4 from a pair of linear equations in the genera; form. Now you can use any method to solve these equation we get p = 1/3 and q = 1/3 now
substituting `1/(x - 1)` fpr p we have
`1(x - 1) = 1/3`
i.e x - 1 = 3 i.e x = 4
Similary substituting 1/(y -2) for q we get
`1/(y -2) = 1/3`
i.e 3 = y - 2 i.e y = 5
Hence x = 4, y = 5 is required solution of the given pair of euation
Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method
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Page 12
Solve the following systems of equations:
`(7x - 2y)/"xy" = 5`
`(8x + 7y)/"xy" = 15`
`(7x - 2y)/"xy" = 5`
`=> 7/y - 2/x = 5` .....(1)
`(8x + 7y)/(xy) = 15`
`=> 8/y + 7/x = 15` ..........(2)
Let `1/x = p and 1/y = q`
The given equations reduce to:
-2p + 7q = 5
=> -2p + 7q - 5 = 0 ...(3)
7p + 8q = 15
=> 7p + 8q - 15 = 0 ....(4)
Using cross multiplication method, we obtain:
`p/(-105-(-40)) = q/(-35-30) = 1/(-16-49)`
`p/(-65) = 1/(-65), q/(-65) = 1/(-65)`
p = 1, q= 1
`p = 1/x = 1, q= 1/y =1`
x = 1 , y = 1
Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method
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Page 13
Solve the following systems of equations:
152x − 378y = −74
−378x + 152y = −604
152x − 378y = −74 ...(1)
−378x + 152y = −604 ...(2)
Adding the equations (1) and (2), we obtain:
-226x - 226uy = -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
Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method
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Page 14
The given system of equation is
99x + 101y = 499 ...(i)
101x + 99y = 501 ...(ii)
Adding equation (i) and equation (ii), we get
99x + 101x + 101y + 99y = 499 + 501
=> 200x + 200y = 1000
`=> 200(x + y) = 1000`
`=> x + y = 1000/200 = 5`
=> x + y = 5 ....(iii)
Subtracting equation (i) by equation (ii), we get
101x - 99x + 99y - 101y = 501 - 499
=> 2x - 2y = 2
=> 2(x - y) = 2
=> x - y = 2/2
=> x - y = 1 ......(iv)
Adding equation (iii) and equation (iv), we get
2x = 5 + 1
`=> x = 6/2 = 3`
Putting x = 3 in equation (iii), we get
3 + y = 5
=> y = 5 - 3 = 2
Hence, solution of the given system of equation is x = 3, y = 2
Page 15
The given system of equation is
23x − 29y = 98 ....(i)
29x − 23y = 110 ...(ii)
Adding equation (i) and equation (ii), we get
23x + 29x - 29y - 23y = 98 + 110
=> 52x - 52y = 208
=> 52 (x - y) = 208
`=> x - y = 208/52 = 4`
=> x - y - 4 ...(iii)
Subtracting equation (i) by equation (ii), we get
29x - 23x - 23y + 29y = 110 - 98
=> 6x + 6y = 12
=> 6(x + y) = 12
`=> x + y = 12/6 = 2`
=> x = y = 2 ...(iv)
Adding equation (iii) and equation (iv), we get
2x = 2 + 4 = 6
Putting x = 3 in equation (iv), we get
3 + y = 2
=>? y = 2- 3 = -1
Hence, solution of the given system of equation is x = 3 , y = -1
Page 16
We have,
x − y + z = 4 .....(i)
x − 2y − 2z = 9 ....(ii)
2x + y + 3z = 1 .....(iii)
From equation (i), we get
z = 4 - x + y
=> z = -x + y + 4
Subtracting the value of z in equation (ii), we get
x - 2y - 2(-x + y + 4) = 9
`=> x - 2y + 2x - 2y - 8 = 8`
=> 3x - 4x = 9 + 8
=> 3x - 4y = 17 ......(iv)
Subtracting the value of z in equation (iii), we get
2x + y + 3(-x + y + 4) = 1
=> 2x + y + 3x + 3y + 12 = 1
=> -x + 4y =1 - 12
=> -x + 4y = -11 .....(v)
Adding equations (iv) and (v), we get
3x - x - 4y + 4y = 17 - 11
=> 2x = 6
=> x = 6/2 = 3
Putting x = 3 in equation (iv), we get
`3 xx 2 - 4y = 17`
=> 9 - 4y = 17
=> -4y = 17 - 9
=> -4y = 8
`=> y = 8/(-4) = -2`
Putting x = 3 and y = -2 in z = -x + y + 4 we get
z = -3 -2 + 4
=> x = -5 + 4
=> z= -1
Hence, solution of the giving system of equation is x = 3, y = -2, z = -1