RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Other Exercises

• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.10
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

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
Adding we get
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)
Adding (iii) and (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)
Adding we get
200x + 200y = 1000
x + y = 5 ……(iii)
(Dividing by 200)
Subtracting we get
-2x + 2y = -2
=> x – y = 1 ….(iv)
(Dividing by -2)
Now adding (iii) and (iv)
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)
Adding (iii) and (iv)
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:

Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 are helpful to complete your math homework.

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