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.


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