Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Solve the system of equations:
x + 2y + 1 = 0 and 2x – 3y – 12 = 0
x + 2y + 1 = 0 …. (i)
2x - 3y - 12 = 0 …….. (ii)
Here a1 = 1, b1 = 2, c1 = 1
a2 = 2, b2 = -3, c2 = – 12
By cross multiplication method,
Now,
x = 3
And,
= y = - 2
The solution of the given system of equation is 3 and – 2 respectively.
3x + 2y + 25 = 0, 2x + y + 10 = 0
3x + 2y + 25 = 0 …. (i)
2x + y + 10 = 0 ….. (ii)
Here a1 = 3, b1 = 2, c1 = 25
a2 = 2, b2 = 1, c2 = 10
x = 5
y = – 20
The solution of the given system of equation is 5 and -20 respectively.
2x + y = 35, 3x + 4y = 65
2x + y = 35 …. (i)
3x + 4y = 65 ….. (ii)
Here a1 = 2, b1 = 1, c1 = 35
a2 = 3, b2 = 4, c2 = 65
x = 15
y = 5
The solution of the given system of equation is 15 and 5 respectively.
2x – y – 6 = 0, x – y – 2 = 0
2x - y = 6 …. (i)
x - y = 2 ….. (ii)
Here a1 = 2, b1 = -1, c1 = 6
a 2= 1, b2 = -1, c2 = 2
x = 4
y = 2
The solution of the given system of equation is 4 and 2 respectively
Taking 1/x = u
Taking 1/y = v
u + v = 2 …… (iii)
u - v = 6 …. (iv)
u/4 = 1/-2
u = - 2
- v/ – 8 = 1/ – 2
v = – 4
1/u = x
X = – 1/2
y = –1/4
The solution of the given system of equation is -1/2 and -1/4 respectively.
ax + by = a – b, bx - ay = a + b
ax + by = a - b …. (i)
bx - ay = a + b ….. (ii)
Here a1 = a, b1 = b, c1 = a - b
a2 = b, b2 = -a, c2 = a + b
x = 1
y = – 1
The solution of the given system of equation is 1 and -1 respectively.
x + ay - b = 0, ax - by - c = 0
x + ay - b = 0 ….. (i)
ax - by - c = 0 ……. (ii)
Here a1 = 1, b1 = a, c1 = - b
a2 = a, b2 = - b , c2 = - c
The solution of the given system of equation isrespectively.
ax + by = a2
bx + ay = b2
ax + by = a2 …. (i)
bx + ay = b2 ….. (ii)
Here a1 = a, b1 = b, c1 = a2
a2 = b, b2 = a, c2 = b2
The solution of the given system of equation is respectively.
The given system of equations are:
5u - 2v = -1
15u + 7v = -10
Here a1 = 5, b1 = – 2, c1 = 1
a2 = 15, b2 = 7, c2 = 10
x + y = 5 …. (i)
v = 1
x - y = 1 …… (ii)
Adding equation (i) and (ii)
2x = 6
Putting the value of x in equation (i)
3 + y = 5
The solution of the given system of equation is 3 and 2 respectively.
Let 1/x = u
Let 1/y = v
The given system of equations becomes:
2u + 3v = 13 …… (i)
5u - 4v = – 2 …. (ii)
u = 2
v = 3
The solutions of the given system of equations are 1/2 and 1/3 respectively.
57u + 6v = 5
38u + 21v = 9
Here a1 = 57, b1 = 6, c1 = – 5
a2 = 38, b2 = 21, c2 = – 9
x + y = 19 ….. (i)
x - y = 3 … (ii)
2x = 22
x = 11
11 + y = 19
y = 8
The solution of the given system of equation is 11 and 8 respectively.
a2 = a, b2 = – b, c2 = b2 - a2
By cross multiplication method
x = a
y = b
Hence the solution of the given system of equation are a and b respectively.
x = a2
y = b2
The solution of the given system of equation are a2 and b2 respectively.
ax + by = a2 + b2
Here, a1 = a, b2 = b, Let c1 = – (a2 + b2)
The solution of the given system of equations are a and b respectively.
2ax + 3by = a + 2b
3ax + 2by = 2a + b
The given system of equation is
2ax + 3by = a + 2b …… (i)
3ax + 2by = 2a + b ….. (ii)
Here a1 = 2a, b1 = 3b, c1 = – (a + 2b)
a2 = 3a, b2 = 2b, c2 = – (2a + b)
The solutions of the system of equations are respectively.
5ax + 6by = 28
3ax + 4by = 18
The systems of equations are:
5ax + 6by = 28 …. (i)
3ax + 4by = 18 …. (ii)
Here a1 = 5a, b1 = 6b, c1 = – (28)
a2 = 3a, b2 = 4b, c2 = – (18)
The solution of the given system of equation is 2/a and 3/b.
(a + 2b)x + (2a - b)y = 2
(a – 2b)x + (2a + b)y = 3
(a + 2b)x + (2a - b)y = 2 ……. (i)
(a - 2b)x + (2a + b)y = 3 ….. (ii)
Here a1 = a + 2b, b1 = 2a - b, c1 = - (2)
a2 = a - 2b, b2 = 2a + b, c2 = - (3)
By cross multiplication method:
The solution of the system of equations are
The given systems of equations are:
From equation (i)
From equation (ii)
x + y - 2a2 = 0
a2 = 1, b2 = 1, c2 = – 2a2
The solutions of the given system of equations arerespectively.
The system of equation is given by:
bx + cy = a + b ……. (i)
bx + cy - (a + b) = 0
2abx - 2acy - 2a(a - b) = 0 …. (iv)
By cross multiplication
x = a/b
y = b/c
The solution of the system of equations are a/b and b/c.
(a - b)x +(a + b)y = 2a2 - 2b2(a + b)(x + y) = 4ab
(a - b)x + (a + b)y = 2a2 - 2b2 ….. (i)
(a + b)(x + y) = 4ab …… (ii)
(a - b)x + (a + b)y - 2a2 - 2b2 = 0
= (a - b)x + (a + b)y - 2(a2 - b2) = 0
(a - b)x + (a - b)y - 4ab = 0
Here, a1 = a - b, b1 = a + b, c1 = - 2(a2 + b2)
Here, a2 = a + b, b2 = a + b, c2 = - 4ab
The solution of the system of equations arerespectively.
a2x + b2y = c2
b2x + a2y = d2
a2x + b2y = c2 ….. (i)
b2x + a2y = d2 …… (ii)
Here, a1 = a2, b1 = b2, c1 = – c2
Here, a2 = b2, b2 = a2, c2 = – d2
The solution of the given system of equations arerespectively.
2(ax - by + a + 4b = 0
2(bx + ay) + b - 4a = 0
The given system of equation may be written as:
2(ax - by + a + 4b = 0 ….. (i)
2(bx + ay) + b - 4a = 0 …. (ii)
Here, a1 = 2a, b1 = -2b, c1 = a + 4b
Here, a2 = 2b, b2 = 2a, c2 = b - 4a
The solution of the given pair of equations are -1/2 and 2 respectively.
6(ax + by) = 3a + 2b
6(bx - ay) = 3b - 2a
The systems of equations are
6(ax + by) = 3a + 2b …… (i)
6(bx - ay) = 3b - 2a ……. (ii)
6ax + 6by - (3a + 2b) = 0 …… (iii)
6bx - 6ay - (3b - 2a) = 0 …… (iv)
Here, a1 = 6a, b1 = 6b, c1 = - (3a + 2b)
Here, a2 = 6b, b2 = -6a, c2 = - (3b - 2a)
The solution of the given pair of equations are 1/2 and 1/3 respectively.
The given systems of equations are
The pair of equations becomes:
a2u - b2v = 0
a2bu + b2av - (a + b) = 0
Here, a1 = a2, b1 = -b2, c1 = 0
Here, a2 = a2b, b2 = b2a, c2 = – (a + b)
The solution of the given pair of equations are 1/a2 and 1/b2 respectively.
mx - my = m2 + n2
x + y = 2m
mx - my = m2 + n2 …… (i)
x + y = 2m …….. (ii)
Here, a1 = m, b1 = -n, c1 = - (m2 + n2)
Here, a2 = 1, b2 = 1, c2 = - (2m)
x = m + n
y = m - n
The solutions of the given pair of equations are m+n and m-n respectively.
ax - by = 2ab
The given pair of equations are:
ax - by = 2ab ….. (ii)
Here, a2 = a, b2 = – b, c2 = - (2ab)
x = b
y = – a
The solution of the given pair of equations are b and – a respectively.
x + y - 2ab = 0 ……. (ii)
x = ab
y = ab
The solutions of the given pair of equations are ab and ab respectively.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Chapter 3: Pair of Linear Equations in Two...