# RD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables Ex 3.3 And Ex 3.4

### RD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables Ex 3.3

Solve the following systems of equations:
Question 1.
11x + 15y + 23 = 0
7x – 2y – 20 = 0
Solution:
11x + 15y + 23 = 0 => 11x + 15y = -23 ……..(i)
7x – 2y – 20 = 0 => 7x – 2y = 20 ……….(ii)
Multiply (i) by 2 and (ii) 15, we get
22x + 30y = -46
105x – 30y = 300
127x = 254 => x = $\frac { 254 }{ 127 }$= 2
7 x 2 – 2y = 20 => 14 – 2y = 20
-2y = 20 – 14 = 6
y = -3

Question 2.
3x – 7y + 10 = 0
y – 2x – 3 = 0
Solution:

Question 3.
0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Solution:

Question 4.

Solution:

Question 5.
7(y + 3) – 2(x + 2) = 14
4(y – 2) + 3(x – 3) = 2
Solution:

Question 6.
$\frac { x }{ 7 } +\frac { y }{ 3 } =5$
$\frac { x }{ 2 } -\frac { y }{ 9 } =6$
Solution:

Question 7.
$\frac { x }{ 3 } +\frac { y }{ 4 } = 11$
$\frac { 5x }{ 6 } -\frac { y }{ 3 } = -7$
Solution:

Question 8.
$\frac { 4 }{ x }$+ 3y = 8
$\frac { 6 }{ x }$– 4y = -5
Solution:

Question 9.
x + $\frac { y }{ 2 }$= 4
$\frac { x }{ 3 }$+ 2y = 5
Solution:

Question 10.
x + 2y = $\frac { 3 }{ 2 }$
2x + y = $\frac { 3 }{ 2 }$
Solution:

Question 11.
√2x – √3y = 0
√3x – √8y = 0
Solution:

Question 12.
3x – $\frac { y + 7 }{ 11 }$+ 2 = 10
2y + $\frac { y + 11 }{ 7 }$= 10
Solution:

Question 13.
2x – $\frac { 3 }{ y }$= 9
3x + $\frac { 7 }{ y }$= 2, y ≠ 0
Solution:

Hence x = 3, y = -1

Question 14.
0.3x + 0.7y = 0.74
0.3x + 0.5y = 0.5
Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.

Solution:

Question 34.
x + y = 5xy
3x + 2y = 13xy
Solution:

Question 35.

Solution:

Question 36.
2 (3u – v) = 5uv
2 (u + 3v) = 5uv
Solution:

Question 37.

Solution:

Question 38.

Solution:

Question 39.

Solution:

Question 40.

Solution:

Question 41.

Solution:

Question 42.

Solution:

Question 43.
152x – 378y = -74
– 378x + 152y = -604
Solution:
152x – 378y = -74 ……..(i)
– 378x + 152y = -604 ………(ii)
Adding (i) and (ii), we get
– 226x – 226y = 678
Dividing by -226,
x + y = 3 ……..(iiii)
and subtracting (ii) from (i)
530x – 530y = 530
Dividing by 530,
x – y = 1 ……..(iv)
2x = 4
x = 2
From (iii), 2 + y = 3
y = 3 – 2 = 1
Hence x = 2, y = 1

Question 44.
99x + 101y = 499
101x + 99y = 501
Solution:
99x + 101 y = 499 ….(i)
101x + 99y = 501 ……(ii)
200x + 200y = 1000
x + y = 5 ……(iii)
(Dividing by 200)
Subtracting we get
-2x + 2y = -2
=> x – y = 1 ….(iv)
(Dividing by -2)
2x = 6 => x = 3
and subtracting (iv) from (iii)
2y = 4 => y = 2
Hence x = 3, y = 2

Question 45.
23x – 29y = 98
29x – 23y = 110
Solution:
23x – 29y = 98 ….(i)
29x – 23y = 110 ….(ii)
Adding (i) and (ii) we get
52x – 52y = 208
x – y = 4 ….(iii)
(Dividing by 52)
Subtracting (ii) from (i)
6x + 6y = 12
=> x + y = 2 ….(iv)
(Dividing by 6)
2x = 6 => x = 3
Subtracting (iv) from (iii)
2y = -2 => y = -1
Hence x = 3, y = -1

Question 46.
x – y + z = 4
x – 2y – 2z = 9
2x + y + 3z = 1
Solution:
x – y + z = 4 ……(i)
x – 2y – 2z = 9 ……(ii)
2x + y + 3z = 1 ……(iii)

Question 47.
x – y + z = 4
x + y + z = 2
2x + y – 3z = 0
Solution:
x – y + z = 4 ….(i)
x + y + z = 2 ….(ii)
2x + y – 3z = 0 ….(iii)
From (i)
z = 4 – x + y
Substituting the values of z in (ii) and (iii)
x + y + 4 – x + y = 2
2y = 2 – 4 = -2
y = -1
and 2x + y – 3(4 – x + y) = O
2x + y – 12 + 3x – 3y = 0
5x – 2y = 12
5x – 2(-1) = 12
5x + 2 = 12
5x = 12 – 2 = 10
x = 2
From (i),
2 – (-1) + z = 4
2 + 1 + z = 4
3 + z = 4
z = 4 – 3 = 1
Hence x = 2, y = 1, z = 1

Question 48.
21x + 47y = 110
47x + 21y = 162
Solution:
We have,
21x + 47y = 110 …(i)
47x + 21y = 162 …(ii)
Multiplying equation (i) by 47 and Equation (ii) by 21, we get
987x + 2209y = 5170 …(iii)
987x + 441y = 3402 …(iv)
Subtracting equation (iv) from equation (iii),
we get
1768y = 1768
y = 1
Substituting value of y in equation (i), we get
21x + 47 = 110
or 21x = 63
or x = 3
So, x = 3, y = 1

Question 49.
If x + 1 is a factor of 2x3 + ax2 + 2bx + 1, then find the values of a and b given that 2a – 3b = 4
Solution:

Question 50.

Solution:

Question 51.
Find the values of x and y in the following rectangle

Solution:

Hence, the required values of x and y are 1 and 4, respectively.

Question 52.
Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x – y = 1. How many such lines can we find?
Solution:

Plotting the points A (2, 0) and B (0, 2), we get the straight line AB. Plotting the points C (0, -1) and D ($\frac { 1 }{ 2 }$, 0) we get the straight line CD. The lines AB and CD intersect at E (1, 1).
Hence, infinite lines can pass through the intersection point of linear equations x + y = 2 and 2x – y = 1
¡.e.,E(1, 1) like as y = x, 2x + y = 3, x + 2y = 3, so on.

Question 53.
Write a pair of linear equations which has the unique solution x = -1, y = 3. How many such pairs can you write?
Solution:

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## Pair of Linear Equations in Two Variables Class 10 Solutions Exercise 3.4

### RD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables Ex 3.4

Solve each of the following systems of equations by the method of cross multiplication.
Question 1.
x + 2y + 1 = 0
2x – 3y – 12 = 0 (C.B.S.E. 1992)
Solution:

Question 2.
3x + 2y + 25 = 0
2x + y + 10 = 0 (C.B.S.E. 1992)
Solution:

Question 3.
2x + y = 35
3x + 4y = 65 (C.B.S.E. 1993)
Solution:

Question 4.
2x – y = 6
x – y = 2 (C.B.S.E. 1994)
Solution:

Question 5.
$\frac { x+y }{ xy } =2,\quad \frac { x-y }{ xy } =6$
Solution:

Question 6.
ax + by = a – b
bx – ay = a + b
Solution:
ax + by = a – b
bx – ay = a + b

Question 7.
x + ay = b
ax – by = c
Solution:
x + ay = b

Question 8.
ax + by = a2
bx + ay = b2
Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.
$\frac { x }{ { a } } +\frac { y }{ b } =a+b$
$\frac { x }{ { { a }^{ 2 } } } +\frac { y }{ { b }^{ 2 } } =2$
Solution:

Question 14.
$\frac { x }{ { { a } } } =\frac { y }{ { b } }$
ax + by = a2 + b2
Solution:

Question 15.
2ax + 3by = a + 2b
3ax + 2by = 2a + b
Solution:

Question 16.
5ax + 6by = 28
3ax + 4by = 18
Solution:

Question 17.
(a + 2b) x + (2a – b) y = 2
(a – 2b) x + (2a + b) y = 3
Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.
(a – b) x + (a + b) y = 2a2 – 2b2
(a + b) (x + y) = 4ab
Solution:

Question 21.
a2x + b2y = c2
b2x + a2y = d2
Solution:

Question 22.
ax + by = $\frac { a + b }{ 2 }$
3x + 5y = 4
Solution:

Hence x = $\frac { 1 }{ 2 }$, y = $\frac { 1 }{ 2 }$

Question 23.
2 (ax – by) + a + 4b = 0
2 (bx + ay) + b – 4a = 0 (C.B.S.E. 2004)
Solution:

Question 24.
6 (ax + by) = 3a + 2b
6 (bx – ay) = 3b – 2a (C.B.S.E. 2004)
Solution:

Question 25.
$\frac { { a }^{ 2 } }{ x } -\frac { { b }^{ 2 } }{ y } =0$
$\frac { { a }^{ 2 }b }{ x } -\frac { { b }^{ 2 }a }{ y } =a+b\quad ,\quad x,y\quad \neq 0$
Solution:

Question 26.
mx – ny = m2 + n2
x + y = 2m (C.B.S.E. 2006C)
Solution:

Question 27.

Solution:

Question 28.

Solution:

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