**Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.4**

**Question: 1**

Solve the system of equations:

x + 2y + 1 = 0 and 2x – 3y – 12 = 0

**Solution:**

x + 2y + 1 = 0 …. (i)

2x - 3y - 12 = 0 …….. (ii)

Here a_{1 }= 1, b_{1 }= 2, c_{1 }= 1

a_{2 }= 2, b_{2 }= -3, c_{2 }= – 12

By cross multiplication method,

Now,

x = 3

And,

= y = - 2

The solution of the given system of equation is 3 and – 2 respectively.

**Question: 2**

Solve the system of equations:

3x + 2y + 25 = 0, 2x + y + 10 = 0

**Solution:**

3x + 2y + 25 = 0 …. (i)

2x + y + 10 = 0 ….. (ii)

Here a_{1 }= 3, b_{1 }= 2, c_{1 }= 25

a_{2 }= 2, b_{2 }= 1, c_{2 }= 10

By cross multiplication method,

Now,

x = 5

And,

y = – 20

The solution of the given system of equation is 5 and -20 respectively.

**Question: 3**

Solve the system of equations:

2x + y = 35, 3x + 4y = 65

**Solution:**

2x + y = 35 …. (i)

3x + 4y = 65 ….. (ii)

Here a_{1 }= 2, b_{1 }= 1, c_{1 }= 35

a_{2 }= 3, b_{2 }= 4, c_{2 }= 65

By cross multiplication method,

Now,

x = 15

And,

y = 5

The solution of the given system of equation is 15 and 5 respectively.

**Question: 4**

Solve the system of equations:

2x – y – 6 = 0, x – y – 2 = 0

**Solution:**

2x - y = 6 …. (i)

x - y = 2 ….. (ii)

Here a_{1 }= 2, b_{1 }= -1, c_{1 }= 6

a _{2}= 1, b_{2 }= -1, c_{2 }= 2

By cross multiplication method,

Now,

x = 4

And,

y = 2

The solution of the given system of equation is 4 and 2 respectively

**Question: 5**

Solve the system of equations:

**Solution:**

Taking 1/x = u

Taking 1/y = v

u + v = 2 …… (iii)

u - v = 6 …. (iv)

By cross multiplication method,

Now,

u/4 = 1/-2

u = - 2

And,

- 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.

**Question: 6**

Solve the system of equations:

ax + by = a – b, bx - ay = a + b

**Solution:**

ax + by = a - b …. (i)

bx - ay = a + b ….. (ii)

Here a_{1 }= a, b_{1 }= b, c_{1 }= a - b

a_{2 }= b, b_{2 }= -a, c_{2 }= a + b

By cross multiplication method,

Now,

x = 1

And,

y = – 1

The solution of the given system of equation is 1 and -1 respectively.

**Question: 7**

Solve the system of equations:

x + ay - b = 0, ax - by - c = 0

**Solution:**

x + ay - b = 0 ….. (i)

ax - by - c = 0 ……. (ii)

Here a_{1 }= 1, b_{1 }= a, c_{1 }= - b

a_{2 }= a, b_{2 }= - b , c_{2 }= - c

By cross multiplication method,

Now,

And,

The solution of the given system of equation isrespectively.

**Question: 8**

Solve the system of equations:

ax + by = a^{2}

bx + ay = b^{2}

**Solution:**

ax + by = a^{2 }…. (i)

bx + ay = b^{2 }….. (ii)

Here a_{1 }= a, b_{1 }= b, c_{1 }= a^{2}

a_{2 }= b, b_{2 }= a, c_{2 }= b^{2}

By cross multiplication method,

Now,

And,

The solution of the given system of equation is respectively.

**Question: 9**

Solve the system of equations:

**Solution:**

The given system of equations are:

5u - 2v = -1

15u + 7v = -10

Here a_{1 }= 5, b_{1 }= – 2, c_{1 }= 1

a_{2 }= 15, b_{2 }= 7, c_{2 }= 10

By cross multiplication method,

Now,

x + y = 5 …. (i)

And,

v = 1

x - y = 1 …… (ii)

Adding equation (i) and (ii)

2x = 6

x = 3

Putting the value of x in equation (i)

3 + y = 5

y = 2

The solution of the given system of equation is 3 and 2 respectively.

**Question: 10**

Solve the system of equations:

**Solution:**

Let 1/x = u

Let 1/y = v

The given system of equations becomes:

2u + 3v = 13 …… (i)

5u - 4v = – 2 …. (ii)

By cross multiplication method,

Now,

u = 2

And,

v = 3

The solutions of the given system of equations are 1/2 and 1/3 respectively.

**Question: 11**

Solve the system of equations:

**Solution:**

The given system of equations are:

57u + 6v = 5

38u + 21v = 9

Here a_{1 }= 57, b_{1 }= 6, c_{1 }= – 5

a_{2 }= 38, b_{2 }= 21, c_{2 }= – 9

By cross multiplication method,

Now,

x + y = 19 ….. (i)

And,

x - y = 3 … (ii)

Adding equation (i) and (ii)

2x = 22

x = 11

Putting the value of x in equation (i)

11 + y = 19

y = 8

The solution of the given system of equation is 11 and 8 respectively.

**Question: 12**

Solve the system of equations:

**Solution:**

a_{2 }= a, b_{2 }= – b, c_{2 }= b^{2 }- a^{2}

By cross multiplication method

x = a

y = b

Hence the solution of the given system of equation are a and b respectively.

**Question: 13**

**Solution:**

By cross multiplication method

x = a^{2}

y = b^{2}

The solution of the given system of equation are a^{2 }and b^{2 }respectively.

**Question: 14**

ax + by = a^{2 }+ b^{2}

**Solution:**

Here, a_{1 }= a, b_{2 }= b, Let c_{1 }= – (a^{2 }+ b^{2})

By cross multiplication method

x = a

y = b

The solution of the given system of equations are a and b respectively.

**Question: 15**

2ax + 3by = a + 2b

3ax + 2by = 2a + b

**Solution:**

The given system of equation is

2ax + 3by = a + 2b …… (i)

3ax + 2by = 2a + b ….. (ii)

Here a_{1 }= 2a, b_{1 }= 3b, c_{1 }= – (a + 2b)

a_{2 }= 3a, b_{2 }= 2b, c_{2 }= – (2a + b)

By cross multiplication method

Now,

The solutions of the system of equations are respectively.

**Question: 16**

5ax + 6by = 28

3ax + 4by = 18

**Solution:**

The systems of equations are:

5ax + 6by = 28 …. (i)

3ax + 4by = 18 …. (ii)

Here a_{1 }= 5a, b_{1 }= 6b, c_{1 }= – (28)

a_{2 }= 3a, b_{2 }= 4b, c_{2 }= – (18)

By cross multiplication method

Now,

The solution of the given system of equation is 2/a and 3/b.

**Question: 17**

(a + 2b)x + (2a - b)y = 2

(a – 2b)x + (2a + b)y = 3

**Solution:**

The given system of equations are:

(a + 2b)x + (2a - b)y = 2 ……. (i)

(a - 2b)x + (2a + b)y = 3 ….. (ii)

Here a_{1 }= a + 2b, b_{1 }= 2a - b, c_{1 }= - (2)

a_{2 }= a - 2b, b_{2 }= 2a + b, c_{2 }= - (3)

By cross multiplication method:

The solution of the system of equations are

**Question: 18**

Solve the system of equations:

**Solution:**

The given systems of equations are:

From equation (i)

From equation (ii)

x + y - 2a^{2 }= 0

a_{2 }= 1, b_{2 }= 1, c_{2 }= – 2a^{2}

By cross multiplication method:

The solutions of the given system of equations arerespectively.

**Question: 19**

Solve the system of equations:

**Solution:**

The system of equation is given by:

bx + cy = a + b ……. (i)

From equation (i)

bx + cy - (a + b) = 0

From equation (ii)

2abx - 2acy - 2a(a - b) = 0 …. (iv)

By cross multiplication

x = a/b

And,

y = b/c

The solution of the system of equations are a/b and b/c.

**Question: 20**

Solve the system of equations:

(a - b)x +(a + b)y = 2a^{2 }- 2b^{2}(a + b)(x + y) = 4ab

**Solution:**

The given system of equations are:

(a - b)x + (a + b)y = 2a^{2 }- 2b^{2 }….. (i)

(a + b)(x + y) = 4ab …… (ii)

From equation (i)

(a - b)x + (a + b)y - 2a^{2 }- 2b^{2 }= 0

= (a - b)x + (a + b)y - 2(a^{2 }- b^{2}) = 0

From equation (ii)

(a - b)x + (a - b)y - 4ab = 0

Here, a_{1 }= a - b, b_{1 }= a + b, c_{1 }= - 2(a^{2 }+ b^{2})

Here, a_{2 }= a + b, b_{2 }= a + b, c_{2 }= - 4ab

By cross multiplication method

Now,

The solution of the system of equations arerespectively.

**Question: 21**

Solve the system of equations:

a^{2}x + b^{2}y = c^{2}

b^{2}x + a^{2}y = d^{2}

**Solution:**

The given system of equations are:

a^{2}x + b^{2}y = c^{2 }….. (i)

b^{2}x + a^{2}y = d^{2 }…… (ii)

Here, a_{1 }= a^{2}, b_{1 }= b^{2}, c_{1 }= – c^{2}

Here, a_{2 }= b^{2}, b_{2 }= a^{2}, c_{2 }= – d^{2}

By cross multiplication method

Now,

The solution of the given system of equations arerespectively.

**Question: 22**

Solve the system of equations:

2(ax - by + a + 4b = 0

2(bx + ay) + b - 4a = 0

**Solution:**

The given system of equation may be written as:

2(ax - by + a + 4b = 0 ….. (i)

2(bx + ay) + b - 4a = 0 …. (ii)

Here, a_{1 }= 2a, b_{1 }= -2b, c_{1 }= a + 4b

Here, a_{2 }= 2b, b_{2 }= 2a, c_{2 }= b - 4a

By cross multiplication method

y = 2

The solution of the given pair of equations are -1/2 and 2 respectively.

**Question: 23**

Solve the system of equations:

6(ax + by) = 3a + 2b

6(bx - ay) = 3b - 2a

**Solution:**

The systems of equations are

6(ax + by) = 3a + 2b …… (i)

6(bx - ay) = 3b - 2a ……. (ii)

From equation (i)

6ax + 6by - (3a + 2b) = 0 …… (iii)

From equation (ii)

6bx - 6ay - (3b - 2a) = 0 …… (iv)

Here, a_{1 }= 6a, b_{1 }= 6b, c_{1 }= - (3a + 2b)

Here, a_{2 }= 6b, b_{2 }= -6a, c_{2 }= - (3b - 2a)

By cross multiplication method

The solution of the given pair of equations are 1/2 and 1/3 respectively.

**Question: 24**

Solve the system of equations:

**Solution:**

The given systems of equations are

Taking 1/x = u

Taking 1/y = v

The pair of equations becomes:

a^{2}u - b^{2}v = 0

a^{2}bu + b^{2}av - (a + b) = 0

Here, a_{1 }= a^{2}, b_{1 }= -b^{2}, c_{1 }= 0

Here, a_{2 }= a^{2}b, b_{2 }= b^{2}a, c_{2 }= – (a + b)

By cross multiplication method

The solution of the given pair of equations are 1/a^{2} and 1/b^{2} respectively.

**Question: 25**

Solve the system of equations:

mx - my = m^{2 }+ n^{2}

x + y = 2m

**Solution:**

mx - my = m^{2 }+ n^{2 }…… (i)

x + y = 2m …….. (ii)

Here, a_{1 }= m, b_{1 }= -n, c_{1 }= - (m^{2 }+ n^{2})

Here, a_{2 }= 1, b_{2 }= 1, c_{2 }= - (2m)

By cross multiplication method

x = m + n

y = m - n

The solutions of the given pair of equations are m+n and m-n respectively.

**Question: 26**

Solve the system of equations:

ax - by = 2ab

**Solution:**

The given pair of equations are:

ax - by = 2ab ….. (ii)

Here, a_{2 }= a, b_{2 }= – b, c_{2 }= - (2ab)

By cross multiplication method

x = b

y = – a

The solution of the given pair of equations are b and – a respectively.

**Question: 27**

Solve the system of equations:

x + y - 2ab = 0 ……. (ii)

**Solution:**

x + y - 2ab = 0 ……. (ii)

By cross multiplication method

x = ab

y = ab

The solutions of the given pair of equations are ab and ab respectively.