## Advertisement

PRINTABLE FOR KIDS

XII (12) HSC

XI (11) FYJC
X (10) SSC

### Exercise 3.3 Pair Of Linear Equations In Two Variables Class 10th Mathematics KC Sinha Solution

Exercise 3.3
1. 3x - 5y - 4 = 0 9x = 2y + 7 Solve the following system of equations by…
2. 3x + 4y = 10 2x - 2y = 2 Solve the following system of equations by elimination…
3. x + y = 5 2x - 3y = 4 Solve the following system of equations by elimination…
4. 2x + 3y = 8 4x + 6y = 7 Solve the following system of equations by elimination…
5. 8x + 5y = 9 3x + 2y = 4 Solve the following system of equations by elimination…
6. 2x + 3y = 46 3x + 5y = 74 Solve the following system of equations by…
7. 0.4x - 1.5y = 6.5 0.3x + 0.2y = 0.9 Solve the following system of equations by…
8. √2x - √3y = 0 √5x + √2y = 0 Solve the following system of equations by…
9. 2x + 5y = 1 2x + 3y = 3 Solve the following system of equations by elimination…
10. 3x - 8/y = 5 x - y/3 = 3 Solve the following system of equations by elimination…
11. x/6 + y/15 = 4 x/3 - y/12 = 19/4 Solve the following system of equations by…
12. x + 6/y = 6 3x - 8/y = 5 Solve the following system of equations by elimination…
13. 37x + 43y = 123 43x + 37y = 117 Solve the following equations by elimination…
14. 217x + 131y = 913 131x + 217y = 827 Solve the following equations by…
15. 99x + 101y = 499 101x + 99y = 501 Solve the following equations by elimination…
16. 29x - 23y = 110 23x - 29y = 98 Solve the following equations by elimination…
17. 1/x - 1/y = 1 1/x + 1/y = 7 , x not equal 0 , y not equal 0 Solve the following…
18. 2/x + 3/y = 13 5/x - 4/y = - 2 , x not equal 0 , y not equal 0 Solve the…
19. 3a/x - 2b/y + 5 = 0 , a/x + 3b/y - 2 = 0 , (x not equal 0 , y not equal 0)…
20. 3a/x - 2b/y + 5 = 0 , a/x + 3b/y - 2 = 0 , (x not equal 0 , y not equal 0)…
21. 2x+5y/xy = 6 , 4x-5y/xy = - 3 , where x ≠ 0 and y ≠ 0 Solve the following…
22. x + y = 2xy x - y = 6xy Solve the following system of equations by elimination…
23. 5x + 3y = 19xy 7x - 2y = 8xy Solve the following system of equations by…
24. x + y = 7xy 2x - 3y = - xy Solve the following system of equations by…
25. 1/2 (2x+3y) + 12/7 (3x-2y) = 1/2 , 7/(2x+3y) + 4/(3x-2y) = 2 Where (2x + 3y) ≠…
26. 6/x-1 - 3/y-2 = 1 , x not equal 1 , y not equal 1 3/x-1 + 2/y+1 = 13/6 , x not…
27. 44/x+y + 30/x-y = 10 55/x+y + 40/x-y = 13 , x+y not equal 0 , x-y not equal 0…
28. 5/x-1 + 1/y-2 = 2 6/x-1 - 3/y-2 = 1 Solve for x and y the following system of…
29. Aftab tells his daughter, "seven years ago, I was seven times as old as you…
30. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be…
31. The difference between two numbers is 26 and one number is three times the…

###### Exercise 3.3
Question 1.

Solve the following system of equations by elimination method:

3x – 5y – 4 = 0

9x = 2y + 7

Answer:

Given pair of linear equations is

3x – 5y – 4 = 0 …(i)

And 9x = 2y + 7 …(ii)

On multiplying Eq. (i) by 3 to make the coefficients of x equal, we get the equation as

9x – 15y – 12 = 0 …(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

9x – 15y – 12 – 9x = 0 – 2y – 7

⇒ – 15y + 2y = – 7 + 12

⇒ – 13y = 5

On putting  in Eq. (ii), we get

Hence,  and  , which is the required solution.

Question 2.

Solve the following system of equations by elimination method:

3x + 4y = 10

2x – 2y = 2

Answer:

Given pair of linear equations is

3x + 4y = 10 …(i)

And 2x – 2y = 2 …(ii)

On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of x equal, we get the equation as

6x + 8y = 20 …(iii)

6x – 6y = 6 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

6x – 6y – 6x – 8y = 6 – 20

⇒ – 14y = – 14

⇒ y = 1

On putting y = 1 in Eq. (ii), we get

2x – 2(1) = 2

⇒ 2x – 2 = 2

⇒ x = 2

Hence, x = 2 and y = 1 , which is the required solution.

Question 3.

Solve the following system of equations by elimination method:

x + y = 5

2x – 3y = 4

Answer:

Given pair of linear equations is

x + y = 5 …(i)

And 2x – 3y = 4 …(ii)

On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as

2x + 2y = 10 …(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

2x + 2y – 2x + 3y = 10 – 4

⇒ 5y = 6

On putting  in Eq. (i), we get

x + y = 5

Hence,  and  , which is the required solution.

Question 4.

Solve the following system of equations by elimination method:

2x + 3y = 8

4x + 6y = 7

Answer:

Given pair of linear equations is

2x + 3y = 8 …(i)

And 4x + 6y = 7 …(ii)

On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as

4x + 6y = 16 …(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

4x + 6y – 4x – 6y = 16 – 7

⇒ 0 = 9

Which is a false equation involving no variable.

So, the given pair of linear equations has no solution i.e. this pair of linear equations is inconsistent.

Question 5.

Solve the following system of equations by elimination method:

8x + 5y = 9

3x + 2y = 4

Answer:

Given pair of linear equations is

8x + 5y = 9 …(i)

And 3x + 2y = 4 …(ii)

On multiplying Eq. (i) by 2 and Eq. (ii) by 5 to make the coefficients of y equal, we get the equation as

16x + 10y = 18 …(iii)

15x + 10y = 20 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

15x + 10y – 16x – 10y = 20 – 18

⇒ – x = 2

⇒ x = – 2

On putting x = – 2 in Eq. (ii), we get

3x + 2y = 4

⇒ 3( – 2) + 2y = 4

⇒ – 6 + 2y = 4

⇒ 2y = 4 + 6

⇒ 2y = 10

Hence,  and  , which is the required solution.

Question 6.

Solve the following system of equations by elimination method:

2x + 3y = 46

3x + 5y = 74

Answer:

Given pair of linear equations is

2x + 3y = 46 …(i)

And 3x + 5y = 74 …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equation as

6x + 9y = 138 …(iii)

6x + 10y = 148 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

6x + 10y – 6x – 9y = 148 – 138

⇒ y = 10

On putting y = 10 in Eq. (ii), we get

3x + 5y = 74

⇒ 3x + 5(10) = 74

⇒ 3x + 50 = 74

⇒ 3x = 74 – 50

⇒ 3x = 24

⇒ x = 8

Hence, x = 8 and y = 10 , which is the required solution.

Question 7.

Solve the following system of equations by elimination method:

0.4x – 1.5y = 6.5

0.3x + 0.2y = 0.9

Answer:

Given pair of linear equations is

0.4x – 1.5y = 6.5 …(i)

And 0.3x + 0.2y = 0.9 …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as

1.2x – 4.5y = 19.5 …(iii)

1.2x + 0.8y = 3.6 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

1.2x + 0.8y – 1.2x + 4.5y = 3.6 – 19.5

⇒ 5.3y = – 15.9

⇒ y = – 3

On putting y = – 3 in Eq. (ii), we get

0.3x + 0.2y = 0.9

⇒ 0.3x + 0.2( – 3) = 0.9

⇒ 0.3x – 0.6 = 0.9

⇒ 0.3x = 1.5

⇒ x = 1.5/0.3

⇒ x = 5

Hence, x = 5 and y = – 3 , which is the required solution.

Question 8.

Solve the following system of equations by elimination method:

√2x – √3y = 0

√5x + √2y = 0

Answer:

Given pair of linear equations is

√2 x – √3 y = 0 …(i)

And √5 x + √2 y = 0 …(ii)

On multiplying Eq. (i) by √2 and Eq. (ii) by √3 to make the coefficients of y equal, we get the equation as

2x – √6 y = 0 …(iii)

√15 x + √6 y = 0 …(iv)

On adding Eq. (iii) and (iv), we get

2x – √6 y + √15 x + √6 y = 0

⇒ 2x + √15 x = 0

⇒ x(2 + √15) = 0

⇒ x = 0

On putting x = 0 in Eq. (i), we get

√2 x – √3 y = 0

⇒ √2(0) – √3 y = 0

⇒ – √3 y = 0

⇒ y = 0

Hence, x = 0 and y = 0 , which is the required solution.

Question 9.

Solve the following system of equations by elimination method:

2x + 5y = 1

2x + 3y = 3

Answer:

Given pair of linear equations is

2x + 5y = 1 …(i)

And 2x + 3y = 3 …(ii)

On subtracting Eq. (ii) from Eq. (i), we get

2x + 3y – 2x – 5y = 3 – 1

⇒ – 2y = 2

⇒ y = – 1

On putting y = – 1 in Eq. (ii), we get

2x + 3( – 1) = 3

⇒ 2x – 3 = 3

⇒ 2x = 6

⇒ x = 6/2

⇒ x = 3

Hence, x = 3 and y = – 1 , which is the required solution.

Question 10.

Solve the following system of equations by elimination method:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (ii) by 2 to make the coefficients of  equal, we get the equation as

…(iii)

On adding Eq. (i) and Eq. (ii), we get

⇒ x = 2

On putting in Eq. (ii), we get

⇒ y = – 3

Hence, x = 2 and y = – 3 , which is the required solution.

Question 11.

Solve the following system of equations by elimination method:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (ii) by  to make the coefficients of  equal, we get the equation as

…(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

⇒ y = 15

On putting y = 15 in Eq. (ii), we get

⇒ x = 18

Hence, x = 18 and y = 15 , which is the required solution.

Question 12.

Solve the following system of equations by elimination method:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (i) by 3 to make the coefficients of  equal, we get the equation as

…(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

⇒ y = 2

On putting y = 2 in Eq. (i), we get

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

Question 13.

Solve the following equations by elimination method:

37x + 43y = 123

43x + 37y = 117

Answer:

Given pair of linear equations is

37x + 43y = 123 …(i)

And 43x + 37y = 117 …(ii)

On multiplying Eq. (i) by 43 and Eq. (ii) by 37 to make the coefficients of x equal, we get the equation as

1591x + 1849y = 5289 …(iii)

1591x + 1369y = 4329 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 1591x + 1369y – 1591x – 1849y = 4329 – 5289

⇒ – 480y = – 960

⇒ y = 2

On putting y = 2 in Eq. (ii), we get

⇒ 43x + 37(2) = 117 ⇒ 43x + 74 = 117

⇒ 43x = 117 – 74

⇒ 43x = 43

⇒ x = 1

Hence, x = 1 and y = 2 , which is the required solution.

Question 14.

Solve the following equations by elimination method:

217x + 131y = 913

131x + 217y = 827

Answer:

Given pair of linear equations is

217x + 131y = 913 …(i)

And 131x + 217y = 827 …(ii)

On multiplying Eq. (i) by 131 and Eq. (ii) by 217 to make the coefficients of x equal, we get the equation as

28427x + 17161y = 119603 …(iii)

28427x + 47089y = 179459 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 28427x + 47089y – 28427x – 17161y = 179459 – 119603

⇒ 47089y – 17161y = 179459 – 119603

⇒ 29928y = 59856

⇒ y = 2

On putting y = 2 in Eq. (ii), we get

⇒ 131x + 217(2) = 827 ⇒ 131x + 434 = 827

⇒ 131x = 393

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

Question 15.

Solve the following equations by elimination method:

99x + 101y = 499

101x + 99y = 501

Answer:

Given pair of linear equations is

99x + 101y = 499 …(i)

And 101x + 99y = 501 …(ii)

On multiplying Eq. (i) by 101 and Eq. (ii) by 99 to make the coefficients of x equal, we get the equation as

9999x + 10201y = 50399 …(iii)

9999x + 9801y = 49599 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 9999x + 9801y – 9999x – 10201y = 49599 – 50399

⇒ 9801y – 10201y = 49599 – 50399

⇒ – 400y = – 800

⇒ y = 2

On putting y = 2 in Eq. (i), we get

⇒ 99x + 101(2) = 499 ⇒ 99x + 202 = 499

⇒ 99x = 297

⇒ x = 297/99

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

Question 16.

Solve the following equations by elimination method:

29x – 23y = 110

23x – 29y = 98

Answer:

Given pair of linear equations is

29x – 23y = 110 …(i)

And 23x – 29y = 98 …(ii)

On multiplying Eq. (i) by 23 and Eq. (ii) by 29 to make the coefficients of x equal, we get the equation as

667x – 529y = 2530 …(iii)

667x – 841y = 2842 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 667x – 841y – 667x + 529y = 2842 – 2530

⇒ – 312y = 312

⇒ y = – 1

On putting y = 2 in Eq. (ii), we get

⇒ 29x – 23( – 1) = 110 ⇒ 29x + 23 = 110

⇒ 29x = 110 – 23

⇒ 29x = 87

⇒ x = 3

Hence, x = 3 and y = – 1 , which is the required solution.

Question 17.

Solve the following system of equations by elimination method:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

Adding Eq. (i) and Eq. (ii), we get

On putting  in Eq. (ii), we get

Hence,  and  , which is the required solution.

Question 18.

Solve the following system of equations by elimination method:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (i) by 5 and Eq. (ii) by 2 to make the coefficients of  equal, we get the equation as

…(iii)

…(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

On putting  in Eq. (ii), we get

Hence,  and  , which is the required solution.

Question 19.

Solve the following system of equations by elimination method:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of  equal, we get the equation as

…(iii)

…(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

On putting  in Eq. (ii), we get

Hence,  and  , which is the required solution.

Question 20.

Solve the following system of equations by elimination method:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (ii) by 3 to make the coefficients of  equal, we get the equation as

…(iii)

On subtracting Eq. (i) from Eq. (iii), we get

⇒ y = b

On putting y = b in Eq. (ii), we get

⇒ x = – a

Hence, x = – a and y = b , which is the required solution.

Question 21.

Solve the following system of equations by elimination method:

, where x ≠ 0 and y ≠ 0

Answer:

Given pair of linear equations is

Or 2x + 5y = 6xy …(i)

And

Or 4x – 5y = – 3xy …(ii)

On adding Eq. (i) and Eq. (ii), we get

2x + 5y + 4x – 5y = 6xy – 3xy

⇒ 6x = 3xy

⇒ y = 2 and x = 0

On putting y = 2 in Eq. (ii), we get

2x + 5(2) = 6xy

⇒ 2x + 10 = 6x(2)

⇒ 2x + 10 = 12x ⇒ 2x – 12x = – 10

⇒ – 10x = – 10

⇒ x = 1

On putting x = 0 , we get y = 0

Hence, x = 0,1 and y = 0,2 , which is the required solution.

Question 22.

Solve the following system of equations by elimination method:

x + y = 2xy

x – y = 6xy

Answer:

Given pair of linear equations is

x + y = 2xy …(i)

And x – y = 6xy …(ii)

On adding Eq. (i) and Eq. (ii), we get

x + y + x – y = 2xy + 6xy

⇒ 2x = 8xy

On putting  in Eq. (ii), we get

⇒ 4x – 1 = 6x ⇒ – 1 = 6x – 4x

⇒ – 1 = 2x

On putting x = 0 , we get y = 0

Hence,  and  , which is the required solution.

Question 23.

Solve the following system of equations by elimination method:

5x + 3y = 19xy

7x – 2y = 8xy

Answer:

Given pair of linear equations is

5x + 3y = 19xy …(i)

And 7x – 2y = 8xy …(ii)

On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of y equal, we get the equation as

10x + 6y = 38xy …(iii)

And 21x – 6y = 24xy …(iv)

On adding Eq. (i) and Eq. (ii), we get

10x + 6y + 21x – 6y = 38xy + 24xy

⇒ 31x = 62xy

On putting  in Eq. (ii), we get

7x – 2y = 8xy

⇒ 7x – 1 = 4x ⇒ – 1 = 4x – 7x

⇒ – 1 = – 3x

On putting x = 0 , we get y = 0

Hence,  and  , which is the required solution.

Question 24.

Solve the following system of equations by elimination method:

x + y = 7xy

2x – 3y = – xy

Answer:

Given pair of linear equations is

x + y = 7xy …(i)

And 2x – 3y = – xy …(ii)

On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as

2x + 2y = 14xy …(iii)

On subtracting Eq. (ii) and Eq. (iii), we get

2x + 2y – 2x + 3y = 14xy + xy

⇒ 2y + 3y = 15xy

⇒ 5y = 15xy

On putting  in Eq. (ii), we get

2x – 3y = – xy

On putting x = 0 , we get x = 0

Hence,  and  , which is the required solution.

Question 25.

Solve for x and y the following system of equations:

Where (2x + 3y) ≠ 0 and (3x – 2y) ≠ 0

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (i) by 7 and Eq. (ii) by  to make the coefficients equal of first term, we get the equation as

…(iii)

…(iv)

On substracting Eq. (iii) from Eq. (iv), we get

…(a)

On multiplying Eq. (ii) by  to make the coefficients equal of second term, we get the equation as

…(v)

On substracting Eq. (i) from Eq. (iv), we get

…(b)

From Eq. (a) and (b), we get

⇒ 2(4 + 2y) = 3(7 – 3y)

⇒ 8 + 4y = 21 – 9y

⇒ 4y + 9y = 21 – 8

⇒ 13y = 13

⇒ y = 1

On putting the value of y = 1 in Eq. (b), we get

Hence, x = 2 and y = 1 , which is the required solution.

Question 26.

Solve for x and y the following system of equations:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients equal of first term, we get the equation as

…(iii)

…(iv)

On substracting Eq. (iii) from Eq. (iv), we get

⇒ y + 1 = 3

⇒ y = 3 – 1

⇒ y = 2

On putting the value of y = 2 in Eq. (ii), we get

⇒ x – 1 = 2

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

Question 27.

Solve for x and y the following system of equations:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (i) by 4 and Eq. (ii) by 3 to make the coefficients equal of second term, we get the equation as

…(iii)

…(iv)

On substracting Eq. (iii) from Eq. (iv), we get

⇒ x + y = 11 …(a)

On putting the value of x + y = 11 in Eq. (1), we get

⇒ 6(x – y) = 30

⇒ x – y = 5 …(b)

Adding Eq. (a) and (b), we get

⇒ 2x = 16

⇒ x = 8

On putting value of x = 8 in eq. (a), we get

8 + y = 11

⇒ y = 3

Hence, x = 8 and y = 3 , which is the required solution.

Question 28.

Solve for x and y the following system of equations:

Answer:

Given pair of linear equations is

…(i)

And  …(ii)

On multiplying Eq. (i) by 3 to make the coefficients equal of second term, we get the equation as

…(iii)

On adding Eq. (ii) and Eq. (iii), we get

⇒ x – 1 = 3

⇒ x = 3 + 1

⇒ x = 4

On putting the value of x = 4 in Eq. (ii), we get

⇒ (y – 2) = 3

⇒ y = 3 + 2

⇒ y = 5

Hence, x = 4 and y = 5 , which is the required solution.

Question 29.

Form the pair of linear equations for the following problems and find their solution by elimination method:

Aftab tells his daughter, "seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." Find their present ages.

Answer:

Let the present age of father i.e. Aftab = x yr

And the present age of his daughter = y yr

Seven years ago,

Aftab’s age = (x – 7)yr

Daughter’s age = (y – 7)yr

According to the question,

(x – 7) = 7(y – 7)

⇒ x – 7 = 7y – 49

⇒ x – 7y = – 42 …(i)

After three years,

Aftab’s age = (x + 3)yr

Daughter’s age = (y + 3)yr

According to the question,

(x + 3) = 3(y + 3)

⇒ x + 3 = 3y + 9

⇒ x – 3y = 6 …(ii)

Now, we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

x – 3y – x + 7y = 6 – ( – 42)

⇒ – 3y + 7y = 6 + 42

⇒ 4y = 48

⇒ y = 12

On putting y = 12 in Eq. (ii) we get

x – 3(12) = 6

⇒ x – 36 = 6

⇒ x = 42

Hence, the age of Aftab is 42years and age of his daughter is 12years.

Question 30.

Form the pair of linear equations for the following problems and find their solution by elimination method:

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer:

Let the present age of Nuri = x yr

And the present age of Sonu = y yr

Five years ago,

Nuri’s age = (x – 5)yr

Sonu’s age = (y – 5)yr

According to the question,

(x – 5) = 3(y – 5)

⇒ x – 5 = 3y – 15

⇒ x – 3y = – 10 …(i)

After ten years,

Aftab’s age = (x + 10)yr

Daughter’s age = (y + 10)yr

According to the question,

(x + 10) = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x – 2y = 10 …(ii)

Now, we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

x – 2y – x + 3y = 10 – ( – 10)

⇒ – 2y + 3y = 10 + 10

⇒ y = 20

On putting y = 20 in Eq. (i) we get

x – 3(20) = – 10

⇒ x – 60 = – 10

⇒ x = 50

Hence, the age of Nuri is 50 years and age of Sonu is 20 years.

Question 31.

Form the pair of linear equations for the following problems and find their solution by elimination method:

The difference between two numbers is 26 and one number is three times the other. Find them.

Answer:

Let the one number = x

And the other number = y

According to the question,

x – y = 26 …(i)

and x = 3y

or x – 3y = 0 …(ii)

Now, we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

x – 3y – x + y = 0 – 26

⇒ – 3y + y = – 26

⇒ – 2y = – 26

⇒ y = 13

On putting y = 13 in Eq. (ii) we get

x – 3(13) = 0

⇒ x – 39 = 0

⇒ x = 39

Hence, the two numbers are 39 and 13.

## PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

SUBJECTS

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION
SSC MATHS I PAPER SOLUTION
SSC MATHS II PAPER SOLUTION
SSC SCIENCE I PAPER SOLUTION
SSC SCIENCE II PAPER SOLUTION
SSC ENGLISH PAPER SOLUTION
SSC & HSC ENGLISH WRITING SKILL
HSC ACCOUNTS NOTES
HSC OCM NOTES
HSC ECONOMICS NOTES
HSC SECRETARIAL PRACTICE NOTES

2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Important-formula

THANKS